CALCULUS I
Solutions to Practice Problems
Derivatives
Paul Dawkins
Calculus I
Table of Contents
Preface ............................................................................................................................................ 1
Derivatives ...................................................................................................................................... 1
The Definition of the Derivative .............................................................................................................. 2
Interpretations of the Derivative ...........................................................................................................10
Differentiation Formulas ........................................................................................................................27
Product and Quotient Rule .....................................................................................................................40
Derivatives of Trig Functions .................................................................................................................46
Derivatives of Exponential and Logarithm Functions ..........................................................................52
Derivatives of Inverse Trig Functions ....................................................................................................56
Derivatives of Hyperbolic Functions ......................................................................................................58
Chain Rule ................................................................................................................................................59
Implicit Differentiation ...........................................................................................................................78
Related Rates ...........................................................................................................................................88
Higher Order Derivatives ......................................................................................................................101
Logarithmic Differentiation ..................................................................................................................108
Preface
Here are the solutions to the practice problems for my Calculus I notes. Some solutions will have
more or less detail than other solutions. The level of detail in each solution will depend up on
several issues. If the section is a review section, this mostly applies to problems in the first
chapter, there will probably not be as much detail to the solutions given that the problems really
should be review. As the difficulty level of the problems increases less detail will go into the
basics of the solution under the assumption that if you’ve reached the level of working the harder
problems then you will probably already understand the basics fairly well and won’t need all the
explanation.
This document was written with presentation on the web in mind. On the web most solutions are
broken down into steps and many of the steps have hints. Each hint on the web is given as a
popup however in this document they are listed prior to each step. Also, on the web each step can
be viewed individually by clicking on links while in this document they are all showing. Also,
there are liable to be some formatting parts in this document intended for help in generating the
web pages that haven’t been removed here. These issues may make the solutions a little difficult
to follow at times, but they should still be readable.
Derivatives
© 2007 Paul Dawkins
i
http://tutorial.math.lamar.edu/terms.aspx
Calculus I
The Definition of the Derivative
1. Use the definition of the derivative to find the derivative of,
( )
6
f x
=
Solution
There really isn’t much to do for this problem other than to plug the function into the definition of
the derivative and do a little algebra.
( )
(
)
( )
0
0
0
0
6 6
0
lim
lim
lim
lim 0
0
h
h
h
h
f x
h
f x
f
x
h
h
h
→
→
→
→
+
−
−
′
=
=
=
=
=
So, the derivative for this function is,
( )
0
f
x
′
=
2. Use the definition of the derivative to find the derivative of,
( )
3 14
V t
t
= −
Step 1
First we need to plug the function into the definition of the derivative.
( )
(
)
( )
(
) (
)
0
0
3 14
3 14
lim
lim
h
h
V t
h
V t
t
h
t
V t
h
h
→
→
+
−
−
+
− −
′
=
=
Make sure that you properly evaluate the first function evaluation. This is one of the more
common errors that students make with these problems.
Also watch for the parenthesis on the second function evaluation. You are subtracting off the
whole function and so you need to make sure that you deal with the minus sign properly. Either
put in the parenthesis as we’ve done here or make sure the minus sign get distributed through
properly. This is another very common error and one that if you make will often make the
problem impossible to complete.
Step 2
Now all that we need to do is some quick algebra and we’ll be done.
( )
(
)
0
0
0
3 14
14
3 14
14
lim
lim
lim
14
14
h
h
h
t
h
t
h
V t
h
h
→
→
→
−
−
− +
−
′
=
=
=
−
= −
The derivative for this function is then,
( )
14
V t
′
= −
© 2007 Paul Dawkins
2
http://tutorial.math.lamar.edu/terms.aspx
Calculus I
3. Use the definition of the derivative to find the derivative of,
( )
2
g x
x
=
Step 1
First we need to plug the function into the definition of the derivative.
( )
(
) ( )
(
)
2
2
0
0
lim
lim
h
h
g x
h
g x
x
h
x
g x
h
h
→
→
+
−
+
−
′
=
=
Make sure that you properly evaluate the first function evaluation. This is one of the more
common errors that students make with these problems.
Step 2
Now all that we need to do is some quick algebra and we’ll be done.
( )
(
)
(
)
2
2
2
0
0
0
2
2
lim
lim
lim 2
2
h
h
h
h
x
h
x
xh
h
x
g x
x
h
x
h
h
→
→
→
+
+
+
−
′
=
=
=
+
=
The derivative for this function is then,
( )
2
g x
x
′
=
4. Use the definition of the derivative to find the derivative of,
( )
2
10 5
Q t
t t
=
+ −
Step 1
First we need to plug the function into the definition of the derivative.
( )
(
)
( )
(
) (
)
(
)
2
2
0
0
10 5
10 5
lim
lim
h
h
t
h
t
h
t
t
Q t
h
Q t
Q t
h
h
→
→
+
+
− +
−
+ −
+
−
′
=
=
Make sure that you properly evaluate the first function evaluation. This is one of the more
common errors that students make with these problems.
Also watch for the parenthesis on the second function evaluation. You are subtracting off the
whole function and so you need to make sure that you deal with the minus sign properly. Either
put in the parenthesis as we’ve done here or make sure the minus sign get distributed through
properly. This is another very common error and one that if you make will often make the
problem impossible to complete.
© 2007 Paul Dawkins
3
http://tutorial.math.lamar.edu/terms.aspx
Calculus I
Step 2
Now all that we need to do is some algebra (and it might get a little messy here, but that is
somewhat common with these types of problems) and we’ll be done.
( )
(
)
(
)
2
2
2
0
0
0
10 5
5
2
10 5
lim
5 2
lim
lim 5 2
5 2
h
h
h
t
h t
th h
t
t
Q t
h
h
t
h
t
h
t
h
→
→
→
+ +
− −
−
− − +
′
=
− −
=
=
− −
= −
The derivative for this function is then,
( )
5 2
Q t
t
′
= −
5. Use the definition of the derivative to find the derivative of,
( )
2
4
9
W z
z
z
=
−
Step 1
First we need to plug the function into the definition of the derivative.
( )
(
)
( )
(
)
(
)
(
)
2
2
0
0
4
9
4
9
lim
lim
h
h
z
h
z
h
z
z
W z
h
W z
W
z
h
h
→
→
+
−
+
−
−
+
−
′
=
=
Make sure that you properly evaluate the first function evaluation. This is one of the more
common errors that students make with these problems.
Also watch for the parenthesis on the second function evaluation. You are subtracting off the
whole function and so you need to make sure that you deal with the minus sign properly. Either
put in the parenthesis as we’ve done here or make sure the minus sign get distributed through
properly. This is another very common error and one that if you make will often make the
problem impossible to complete.
Step 2
Now all that we need to do is some algebra (and it might get a little messy here, but that is
somewhat common with these types of problems) and we’ll be done.
© 2007 Paul Dawkins
4
http://tutorial.math.lamar.edu/terms.aspx
Calculus I
( )
(
)
(
)
(
)
2
2
2
0
2
2
2
0
0
0
4
2
9
9
4
9
lim
4
8
4
9
9
4
9
lim
8
4
9
lim
lim 8
4
9
8
9
h
h
h
h
z
zh
h
z
h
z
z
W
z
h
z
zh
h
z
h
z
z
h
h
z
h
z
h
z
h
→
→
→
→
+
+
−
−
−
+
′
=
+
+
−
−
−
+
=
+
−
=
=
+
−
=
−
The derivative for this function is then,
( )
8
9
W
z
z
′
=
−
6. Use the definition of the derivative to find the derivative of,
( )
3
2
1
f x
x
=
−
Step 1
First we need to plug the function into the definition of the derivative.
( )
(
)
( )
(
)
(
)
3
3
0
0
2
1
2
1
lim
lim
h
h
x
h
x
f x
h
f x
f
x
h
h
→
→
+
− −
−
+
−
′
=
=
Make sure that you properly evaluate the first function evaluation. This is one of the more
common errors that students make with these problems.
Also watch for the parenthesis on the second function evaluation. You are subtracting off the
whole function and so you need to make sure that you deal with the minus sign properly. Either
put in the parenthesis as we’ve done here or make sure the minus sign get distributed through
properly. This is another very common error and one that if you make will often make the
problem impossible to complete.
Step 2
Now all that we need to do is some algebra (and it might get a little messy here, but that is
somewhat common with these types of problems) and we’ll be done.
( )
(
)
(
)
(
)
3
2
2
3
3
0
3
2
2
3
3
0
2
2
2
2
2
0
0
2
3
3
1 2
1
lim
2
6
6
2
1 2
1
lim
6
6
2
lim
lim 6
6
2
6
h
h
h
h
x
x h
xh
h
x
f
x
h
x
x h
xh
h
x
h
h
x
xh
h
x
xh
h
x
h
→
→
→
→
+
+
+
− −
+
′
=
+
+
+
− −
+
=
+
+
=
=
+
+
=
© 2007 Paul Dawkins
5
http://tutorial.math.lamar.edu/terms.aspx
Calculus I
The derivative for this function is then,
( )
2
6
f
x
x
′
=
7. Use the definition of the derivative to find the derivative of,
( )
3
2
2
1
g x
x
x
x
=
−
+ −
Step 1
First we need to plug the function into the definition of the derivative.
( )
(
)
( )
(
)
(
)
(
)
3
2
3
2
0
0
2
1
2
1
lim
lim
h
h
x
h
x
h
x
h
x
x
x
g x
h
g x
g x
h
h
→
→
+
−
+
+ + − −
−
+ −
+
−
′
=
=
Make sure that you properly evaluate the first function evaluation. This is one of the more
common errors that students make with these problems.
Also watch for the parenthesis on the second function evaluation. You are subtracting off the
whole function and so you need to make sure that you deal with the minus sign properly. Either
put in the parenthesis as we’ve done here or make sure the minus sign get distributed through
properly. This is another very common error and one that if you make will often make the
problem impossible to complete.
Step 2
Now all that we need to do is some algebra (and it will get a little messy here, but that is
somewhat common with these types of problems) and we’ll be done.
( )
(
)
(
)
(
)
(
)
3
2
2
3
2
2
3
2
0
3
2
2
3
2
2
3
2
0
2
2
0
2
2
2
0
3
3
2
2
1
2
1
lim
3
3
2
4
2
1
2
1
lim
3
3
4
2
1
lim
lim 3
3
4
2
1
3
4
1
h
h
h
h
x
x h
xh
h
x
xh
h
x
h
x
x
x
g x
h
x
x h
xh
h
x
xh
h
x
h
x
x
x
h
h
x
xh
h
x
h
h
x
xh
h
x
h
x
x
→
→
→
→
+
+
+
−
+
+
+ + − −
−
+ −
′
=
+
+
+
−
−
−
+ + − −
+
− +
=
+
+
−
−
+
=
=
+
+
−
−
+ =
−
+
The derivative for this function is then,
( )
2
3
4
1
g x
x
x
′
=
−
+
© 2007 Paul Dawkins
6
http://tutorial.math.lamar.edu/terms.aspx
Calculus I
8. Use the definition of the derivative to find the derivative of,
( )
5
R z
z
=
Step 1
First we need to plug the function into the definition of the derivative.
( )
(
)
( )
0
0
1
5
5
lim
lim
h
h
R z
h
R z
R z
h
h z
h
z
→
→
+
−
′
=
=
−
+
Make sure that you properly evaluate the first function evaluation. This is one of the more
common errors that students make with these problems.
Also note that in order to make the problem a little easier to read rewrote the rational expression
in the definition a little bit. This doesn’t need to be done, but will make things a little nicer to
look at.
Step 2
Next we need to combine the two rational expressions into a single rational expression.
( )
(
)
(
)
0
5
5
1
lim
h
z
z
h
R z
h
z z
h
→
−
+
′
=
+
Step 3
Now all that we need to do is some algebra and we’ll be done.
( )
(
)
(
)
(
)
2
0
0
0
1 5
5
5
1
5
5
5
lim
lim
lim
h
h
h
z
z
h
h
R z
h
z z
h
h z z
h
z z
h
z
→
→
→
−
−
−
−
′
=
=
=
= −
+
+
+
The derivative for this function is then,
( )
2
5
R z
z
′
= −
9. Use the definition of the derivative to find the derivative of,
( )
1
4
t
V t
t
+
=
+
Step 1
First we need to plug the function into the definition of the derivative.
( )
(
)
( )
0
0
1
1
1
lim
lim
4
4
h
h
V t
h
V t
t
h
t
V t
h
h t
h
t
→
→
+
−
+ +
+
′
=
=
−
+ +
+
© 2007 Paul Dawkins
7
http://tutorial.math.lamar.edu/terms.aspx
Calculus I
Make sure that you properly evaluate the first function evaluation. This is one of the more
common errors that students make with these problems.
Also note that in order to make the problem a little easier to read rewrote the rational expression
in the definition a little bit. This doesn’t need to be done, but will make things a little nicer to
look at.
Step 2
Next we need to combine the two rational expressions into a single rational expression.
( )
(
)(
) (
)(
)
(
)(
)
0
1
4
1
4
1
lim
4
4
h
t
h
t
t
t
h
V t
h
t
h
t
→
+ +
+
− +
+ +
′
=
+ +
+
Step 3
Now all that we need to do is some algebra (and it will get a little messy here, but that is
somewhat common with these types of problems) and we’ll be done.
( )
(
)
(
)(
)
(
)(
)
(
)(
)
(
)(
) (
)
2
2
0
2
2
0
2
0
0
5
4
4
5
4
1
lim
4
4
1
5
4
4
5
4
lim
4
4
1
3
3
3
lim
lim
4
4
4
4
4
h
h
h
h
t
th
t
h
t
th
t
h
V t
h
t
h
t
t
th
t
h
t
th
t
h
h
t
h
t
h
h
t
h
t
t
h
t
t
→
→
→
→
+ + +
+ −
+ + + +
′
=
+ +
+
+ + +
+ − − − − −
=
+ +
+
=
=
=
+ +
+
+ +
+
+
The derivative for this function is then,
( )
(
)
2
3
4
V t
t
′
=
+
10. Use the definition of the derivative to find the derivative of,
( )
3
4
Z t
t
=
−
Step 1
First we need to plug the function into the definition of the derivative.
( )
(
)
( )
(
)
0
0
3
4
3
4
lim
lim
h
h
t
h
t
Z t
h
Z t
Z t
h
h
→
→
+
− −
−
+
−
′
=
=
© 2007 Paul Dawkins
8
http://tutorial.math.lamar.edu/terms.aspx
Calculus I
Make sure that you properly evaluate the first function evaluation. This is one of the more
common errors that students make with these problems.
Step 2
Next we need to rationalize the numerator.
( )
(
)
(
)
(
)
(
)
(
)
(
)
0
3
4
3
4
3
4
3
4
lim
3
4
3
4
h
t
h
t
t
h
t
Z t
h
t
h
t
→
+
− −
−
+
− +
−
′
=
+
− +
−
Step 3
Now all that we need to do is some algebra (and it will get a little messy here, but that is
somewhat common with these types of problems) and we’ll be done.
( )
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
0
0
0
0
3
4
3
4
3
3
4 3
4
lim
lim
3
4
3
4
3
4
3
4
3
3
3
lim
lim
2 3
4
3
4
3
4
3
4
3
4
h
h
h
h
t
h
t
t
h
t
Z t
h
t
h
t
h
t
h
t
h
t
t
h
t
h
t
h
t
→
→
→
→
+
− −
−
+
− − +
′
=
=
+
− +
−
+
− +
−
=
=
=
−
+
− +
−
+
− +
−
Be careful when multiplying out the numerator here. It is easy to lose track of the minus sign (or
parenthesis for that matter) on the second term. This is a very common mistake that students
make.
The derivative for this function is then,
( )
3
2 3
4
Z t
t
′
=
−
11. Use the definition of the derivative to find the derivative of,
( )
1 9
f x
x
=
−
Step 1
First we need to plug the function into the definition of the derivative.
( )
(
)
( )
(
)
0
0
1 9
1 9
lim
lim
h
h
x
h
x
f x
h
f x
f
x
h
h
→
→
−
+
−
−
+
−
′
=
=
Make sure that you properly evaluate the first function evaluation. This is one of the more
common errors that students make with these problems.
Step 2
© 2007 Paul Dawkins
9
http://tutorial.math.lamar.edu/terms.aspx
Calculus I
Next we need to rationalize the numerator.
( )
(
)
(
)
(
)
(
)
(
)
(
)
0
1 9
1 9
1 9
1 9
lim
1 9
1 9
h
x
h
x
x
h
x
f
x
h
x
h
x
→
−
+
−
−
−
+
+
−
′
=
−
+
+
−
Step 3
Now all that we need to do is some algebra (and it will get a little messy here, but that is
somewhat common with these types of problems) and we’ll be done.
( )
(
) (
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
0
0
0
0
1 9
1 9
1 9
9
1 9
lim
lim
1 9
1 9
1 9
1 9
9
9
9
lim
lim
2 1 9
1 9
1 9
1 9
1 9
h
h
h
h
x
h
x
x
h
x
f
x
h
x
h
x
h
x
h
x
h
x
x
h
x
h
x
h
x
→
→
→
→
−
+
− −
−
−
− +
′
=
=
−
+
+
−
−
+
+
−
−
−
−
=
=
=
−
−
+
+
−
−
+
+
−
Be careful when multiplying out the numerator here. It is easy to lose track of the minus sign (or
parenthesis for that matter) on the second term. This is a very common mistake that students
make.
The derivative for this function is then,
( )
9
2 1 9
f
x
x
−
′
=
−
Interpretations of the Derivative
1. Use the graph of the function,
( )
f x
, estimate the value of
( )
f
a
′
for
(a)
2
a
= −
(b)
3
a
=
© 2007 Paul Dawkins
10
http://tutorial.math.lamar.edu/terms.aspx
Calculus I
Solution
Hint : Remember that one of the interpretations of the derivative is the slope of the tangent line to
the function.
(a)
2
a
= −
Step 1
Given that one of the interpretations of the derivative is that it is the slope of the tangent line to
the function at a particular point let’s first sketch in a tangent line at the point on the graph.
Step 2
The function is clearly decreasing here and so we know that the derivative at this point will be
negative. Now, from this sketch of the tangent line it looks like if we run over 1 we go down 4
and so we can estimate that,
( )
2
4
f ′
− = −
(b)
3
a
=
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Calculus I
Step 1
Given that one of the interpretations of the derivative is that it is the slope of the tangent line to
the function at a particular point. Let’s first sketch in a tangent line at the point.
Step 2
The function is clearly increasing here and so we know that the derivative at this point will be
positive. Now, from this sketch of the tangent line it looks like if we run over 1 we go up 2 and
so we can estimate that,
( )
3
2
f ′
=
2. Use the graph of the function,
( )
f x
, estimate the value of
( )
f
a
′
for
(a)
1
a
=
(b)
4
a
=
Solution
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Calculus I
Hint : Remember that one of the interpretations of the derivative is the slope of the tangent line to
the function.
(a)
1
a
=
Step 1
Given that one of the interpretations of the derivative is that it is the slope of the tangent line to
the function at a particular point let’s first sketch in a tangent line at the point on the graph.
Step 2
The function is clearly decreasing here and so we know that the derivative at this point will be
positive. Now, from this sketch of the tangent line it looks like if we run over 1 we go up 1 and
so we can estimate that,
( )
1
1
f ′
=
(b)
4
a
=
Step 1
Given that one of the interpretations of the derivative is that it is the slope of the tangent line to
the function at a particular point. Let’s first sketch in a tangent line at the point.
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Calculus I
Step 2
The function is clearly decreasing here and so we know that the derivative at this point will be
positive. Now, from this sketch of the tangent line it looks like if we run over 1 we go up 5 and
so we can estimate that,
( )
4
5
f ′
=
3. Sketch the graph of a function that satisfies
( )
1
3
f
=
,
( )
1
1
f ′
=
,
( )
4
5
f
=
,
( )
4
2
f ′
= −
.
Solution
Hint : Remember that one of the interpretations of the derivative is the slope of the tangent line to
the function.
Step 1
First, recall that one of the interpretations of the derivative is that it is the slope of the tangent line
to the function at a particular point. So, let’s start off with a graph that has the given points on it
and a sketch of a tangent line at the points whose slope is the value of the derivative at the points.
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Calculus I
Step 2
Now, all that we need to do is sketch in a graph that goes through the indicated points and at the
same time it must be parallel to the tangents that we sketched. There are many possible sketches
that we can make here and so don’t worry if your sketch is not the same as the one here. This is
just one possible sketch that meets the given conditions.
While, it’s not really needed here is a sketch of the function without all the extra bits that we put
in to help with the sketch.
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Calculus I
4. Sketch the graph of a function that satisfies
( )
3
5
f
− =
,
( )
3
2
f ′
− = −
,
( )
1
2
f
=
,
( )
1
0
f ′
=
,
( )
4
2
f
= −
,
( )
4
3
f ′
= −
.
Solution
Hint : Remember that one of the interpretations of the derivative is the slope of the tangent line to
the function.
Step 1
First, recall that one of the interpretations of the derivative is that it is the slope of the tangent line
to the function at a particular point. So, let’s start off with a graph that has the given points on it
and a sketch of a tangent line at the points whose slope is the value of the derivative at the points.
Step 2
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Calculus I
Now, all that we need to do is sketch in a graph that goes through the indicated points and at the
same time it must be parallel to the tangents that we sketched. There are many possible sketches
that we can make here and so don’t worry if your sketch is not the same as the one here. This is
just one possible sketch that meets the given conditions.
While, it’s not really needed here is a sketch of the function without all the extra bits that we put
in to help with the sketch.
5. Below is the graph of some function,
( )
f x
. Use this to sketch the graph of the derivative,
( )
f
x
′
.
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Calculus I
Solution
Hint 1 : Where will the derivative be zero? This gives us a couple of starting points for our
sketch.
Step 1
From the graph of the function it is pretty clear that we will have horizontal tangent lines at
2
x
= −
,
1
x
=
and
5
x
=
. Because we will have horizontal tangents here we also know that the
derivative at these points must be zero. Therefore, we know the following derivative evaluations.
( )
( )
( )
2
0
1
0
5
0
f
f
f
′
′
′
− =
=
=
Hint 2 : Recall that the derivative can also be used to tell us where the function is increasing and
decreasing. Knowing this we can use the graph to determine where the derivative will be positive
and where it will be negative.
Step 2
The points we found above break the x-axis up into regions where the function is increasing and
decreasing. Recall that if the derivative is positive then the function is increasing and likewise if
the derivative is negative then the function is decreasing. Using these ideas we can easily identify
the sign of the derivative on each of the regions. Doing this gives,
( )
( )
( )
( )
2
0
2
1
0
1
5
0
5
0
x
f
x
x
f
x
x
f
x
x
f
x
′
< −
>
′
− < <
<
′
< <
>
′
>
<
Hint 3 : At this point all we have to do is try and put all this together and come up with a sketch
of the derivative.
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Calculus I
Step 3
This is the tricky part of this problem.
In the range
2
x
< −
we know the derivative must be positive and that it must be zero at
2
x
= −
so it makes sense that just to the left of
2
x
= −
the derivative must be decreasing.
In the range
2
1
x
− < <
we know that the derivative will be negative and that it will be zero at the
endpoints of the range. So, to the right of
2
x
= −
the derivative will have to be decreasing (goes
from zero to a negative number). Likewise, to the left of
1
x
=
the derivative will have to be
increasing (goes from a negative number to zero).
Note that we don’t really know just how the derivative will behave everywhere in this range, but
we can use the general behavior near the endpoints and go with the simplest way to connect the
two up to get an idea of what the derivative should look like.
Following similar reasoning we can see that the derivative should be increasing just to the right of
1
x
=
(goes from zero to a positive number), decreasing just to the left of
5
x
=
(goes from a
positive number to zero) and decreasing just to the right of
5
x
=
(goes from zero to a negative
number).
Step 4
So, putting all of this together here is a sketch of the derivative. Note that we included the a scale
on the vertical axis if you would like to try and estimate some specific values of the derivative as
we did in
Example 4
of this section.
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Calculus I
6. Below is the graph of some function,
( )
f x
. Use this to sketch the graph of the derivative,
( )
f
x
′
.
Solution
Hint 1 : Because the derivative of a function is also the slope of the tangent line. We can
therefore determine actual values of the derivative at almost every spot.
Step 1
Because the three portions of the function are actually lines and the tangent line to a line would
just be the line itself we can easily compute the derivative on each portion of the curve.
On each of the portions we can use the grid included on the graph to compute the slope of each
part. Knowing the slope of the graph on each portion will in turn tell us the slope of the tangent
line for each portion. This in turn tells us that the derivative on each of the three portions is then,
( )
( )
( )
1
1
2
1
2
3
2
2
x
f
x
x
f
x
x
f
x
′
< −
= −
′
− < <
=
′
>
= −
Hint 2 : What is the derivative at the “sharp points”?
Step 2
Recall
Example 4
from the previous section. In that example we showed that the derivative of the
absolute value function does not exist at
0
x
=
. The limit on the left side of
0
x
=
(which gives
the slope of the line on the left) and the limit on the right side of
0
x
=
(which gives the slope of
the line on the right) were different and so the overall limit did not exist. This in turn tells us that
the derivative doesn’t exist at that point.
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Calculus I
Here we have the same problem. We’ll leave it to you to verify that the right and left handed
limits at
1
x
= −
are not the same and so the derivative does not exist at
1
x
= −
. Likewise, the
derivative does not exist at
2
x
=
.
There will therefore be open dots on the graph at these two points.
Step 3
Here is the sketch of the derivative of this function.
7. Answer the following questions about the function
( )
2
4
9
W z
z
z
=
−
.
(a) Is the function increasing or decreasing at
1
z
= −
?
(b) Is the function increasing or decreasing at
2
z
=
?
(c) Does the function ever stop changing? If yes, at what value(s) of z does the
function stop changing?
(a) Is the function increasing or decreasing at
1
z
= −
?
We know that the derivative of a function gives us the rate of change of the function and so we’ll
first need the derivative of this function. We computed this derivative in
Problem 5
from the
previous section and so we won’t show the work here. If you need the practice you should go
back and redo that problem before proceeding.
So, from our previous work we know that the derivative is,
( )
8
9
W
z
z
′
=
−
Now all that we need to do is to compute :
( )
1
17
W ′
− = −
. This is negative and so we know that
the function must be decreasing at
1
z
= −
.
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Calculus I
(b) Is the function increasing or decreasing at
2
z
=
?
Again, all we need to do is compute a derivative and since we’ve got the derivative written down
in the first part there’s no reason to redo that here.
The evaluation is :
( )
2
7
W ′
=
. This is positive and so we know that the function must be
increasing at
2
z
=
.
(c) Does the function ever stop changing? If yes, at what value(s) of z does the function stop
changing?
Here all that we’re really asking is if the derivative is ever zero. So we need to solve,
( )
9
0
8
9
0
8
W
z
z
z
′
=
→
− =
⇒
=
So, the function will stop changing at
9
8
z
=
.
8. What is the equation of the tangent line to
( )
3 14
f x
x
= −
at
8
x
=
.
Solution
We know that the derivative of a function gives us the slope of the tangent line and so we’ll first
need the derivative of this function. We computed this derivative in
Problem 2
from the previous
section and so we won’t show the work here. If you need the practice you should go back and
redo that problem before proceeding.
Note that we did use a different set of letters in the previous problem, but the work is identical.
So, from our previous work (with a corresponding change of variables) we know that the
derivative is,
( )
14
f
x
′
= −
This tells us that the slope of the tangent line at
8
x
=
is then :
( )
8
14
m
f ′
=
= −
. We also know
that a point on the tangent line is :
( )
(
)
(
)
8,
8
8, 109
f
=
−
.
The tangent line is then,
(
)
109 14
8
3 14
y
x
x
= −
−
− = −
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Calculus I
Note that, in this case the tangent is the same as the function. This should not be surprising
however as the function is a line and so any tangent line (i.e. parallel line) will in fact be the same
as the line itself.
9. The position of an object at any time t is given by
( )
1
4
t
s t
t
+
=
+
.
(a) Determine the velocity of the object at any time t.
(b) Does the object ever stop moving? If yes, at what time(s) does the object stop
moving?
(a) Determine the velocity of the object at any time t.
We know that the derivative of a function gives is the velocity of the object and so we’ll first
need the derivative of this function. We computed this derivative in
Problem 9
from the previous
section and so we won’t show the work here. If you need the practice you should go back and
redo that problem before proceeding.
Note that we did use a different letter for the function in the previous problem, but the work is
identical. So, from our previous work we know that the derivative is,
( )
(
)
2
3
4
s t
t
′
=
+
(b) Does the object ever stop moving? If yes, at what time(s) does the object stop moving?
We know that the object will stop moving if the velocity (i.e. the derivative) is zero. In this case
the derivative is a rational expression and clearly the numerator will never be zero. Therefore, the
derivative will not be zero and therefore the object never stops moving.
10. What is the equation of the tangent line to
( )
5
f x
x
=
at
1
2
x
=
?
Solution
We know that the derivative of a function gives us the slope of the tangent line and so we’ll first
need the derivative of this function. We computed this derivative in
Problem 8
from the previous
section and so we won’t show the work here. If you need the practice you should go back and
redo that problem before proceeding.
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Calculus I
Note that we did use a different set of letters in the previous problem, but the work is identical.
So, from our previous work (with a corresponding change of variables) we know that the
derivative is,
( )
2
5
f
x
x
′
= −
This tells us that the slope of the tangent line at
1
2
x
=
is then :
1
20
2
m
f
′
=
= −
. We also
know that a point on the tangent line is :
1
1
1
,
,10
2
2
2
f
=
.
The tangent line is then,
1
10 20
20 20
2
y
x
x
=
−
−
=
−
11. Determine where, if anywhere, the function
( )
3
2
2
1
g x
x
x
x
=
−
+ −
stops changing.
Solution
We know that the derivative of a function gives us the rate of change of the function and so we’ll
first need the derivative of this function. We computed this derivative in
Problem 7
from the
previous section and so we won’t show the work here. If you need the practice you should go
back and redo that problem before proceeding.
From our previous work (with a corresponding change of variables) we know that the derivative
is,
( )
2
3
4
1
g x
x
x
′
=
−
+
If the function stops changing at a point then the derivative will be zero at that point. So,
to determine if we function stops changing we will need to solve,
( )
(
)(
)
2
0
3
4
1
0
1
3
1
1
0
,
1
3
g x
x
x
x
x
x
x
′
=
−
+ =
−
− =
⇒
=
=
So, the function will stop changing at
1
3
x
=
and
1
x
=
.
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Calculus I
12. Determine if the function
( )
3
4
Z t
t
=
−
increasing or decreasing at the given points.
(a)
5
t
=
(b)
10
t
=
(c)
300
t
=
(a)
5
t
=
We know that the derivative of a function gives us the rate of change of the function and so we’ll
first need the derivative of this function. We computed this derivative in
Problem 10
from the
previous section and so we won’t show the work here. If you need the practice you should go
back and redo that problem before proceeding.
So, from our previous work we know that the derivative is,
( )
3
2 3
4
Z t
t
′
=
−
Now all that we need to do is to compute :
( )
3
5
2 11
Z ′
=
. This is positive and so we know that
the function must be increasing at
5
t
=
.
(b)
10
t
=
Again, all we need to do is compute a derivative and since we’ve got the derivative written down
in the first part there’s no reason to redo that here.
The evaluation is :
( )
3
10
2 26
Z ′
=
. This is positive and so we know that the function must be
increasing at
10
t
=
.
(c)
300
t
=
Again, all we need to do is compute a derivative and since we’ve got the derivative written down
in the first part there’s no reason to redo that here.
The evaluation is :
( )
3
300
2 896
Z ′
=
. This is positive and so we know that the function must
be increasing at
300
t
=
.
Final Note
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Calculus I
As a final note to all the parts of this problem let’s notice that we did not really need to do any
evaluations. Because we know that square roots will always be positive it is clear that the
derivative will always be positive regardless of the value of t we plug in.
13. Suppose that the volume of water in a tank for
0
6
t
≤ ≤
is given by
( )
2
10 5
Q t
t t
=
+ −
.
(a) Is the volume of water increasing or decreasing at
0
t
=
?
(b) Is the volume of water increasing or decreasing at
6
t
=
?
(c) Does the volume of water ever stop changing? If yes, at what times(s) does the
volume stop changing?
(a) Is the volume of water increasing or decreasing at
0
t
=
?
We know that the derivative of a function gives us the rate of change of the function and so we’ll
first need the derivative of this function. We computed this derivative in
Problem 4
from the
previous section and so we won’t show the work here. If you need the practice you should go
back and redo that problem before proceeding.
So, from our previous work we know that the derivative is,
( )
5 2
Q t
t
′
= −
Now all that we need to do is to compute :
( )
0
5
Q′
=
. This is positive and so we know that the
volume of water in the tank must be increasing at
0
t
=
.
(b) Is the volume of water increasing or decreasing at
6
t
=
?
Again, all we need to do is compute a derivative and since we’ve got the derivative written down
in the first part there’s no reason to redo that here.
The evaluation is :
( )
6
7
Q′
= −
. This is negative and so we know that the volume of water in the
tank must be decreasing at
6
t
=
.
(c) Does the volume of water ever stop changing? If yes, at what times(s) does the volume stop
changing?
Here all that we’re really asking is if the derivative is ever zero. So we need to solve,
( )
5
0
5 2
0
2
Q t
t
t
′
=
→
−
=
⇒
=
© 2007 Paul Dawkins
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Calculus I
So, the volume of water will stop changing at
5
2
.
Differentiation Formulas
1. Find the derivative of
( )
3
6
9
4
f x
x
x
=
−
+
.
Solution
There isn’t much to do here other than take the derivative using the rules we discussed in this
section.
( )
2
18
9
f
x
x
′
=
−
2. Find the derivative of
4
2
2
10
13
y
t
t
t
=
−
+
.
Solution
There isn’t much to do here other than take the derivative using the rules we discussed in this
section.
3
8
20
13
dy
t
t
dt
=
−
+
3. Find the derivative of
( )
7
7
4
3
9
g z
z
z
z
−
=
−
+
.
Solution
There isn’t much to do here other than take the derivative using the rules we discussed in this
section.
( )
6
8
28
21
9
g z
z
z
−
′
=
+
+
© 2007 Paul Dawkins
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Calculus I
4. Find the derivative of
( )
4
3
2
9
8
12
h y
y
y
y
−
−
−
=
−
+
+
.
Solution
There isn’t much to do here other than take the derivative using the rules we discussed in this
section.
( )
5
4
3
4
27
16
h y
y
y
y
−
−
−
′
= −
+
−
5. Find the derivative of
3
4
8
2
y
x
x
x
=
+
−
.
Solution
There isn’t much to do here other than take the derivative using the rules we discussed in this
section.
Remember that you’ll need to convert the roots to fractional exponents before you start taking the
derivative. Here is the rewritten function.
1
1
1
3
2
4
8
2
y
x
x
x
=
+
−
The derivative is,
2
1
3
3
2
4
1
8
1
2
3
2
dy
x
x
x
dx
−
−
−
=
+
−
6. Find the derivative of
( )
5
3
3
7
8
10
6
3
f x
x
x
x
=
−
+
−
.
Solution
There isn’t much to do here other than take the derivative using the rules we discussed in this
section.
Remember that you’ll need to convert the roots to fractional exponents before you start taking the
derivative. Here is the rewritten function.
( )
( ) ( )
( )
3
8
7
1
1
1
3
7
8
5
3
2
5
2
3
10
6
3 10
6
3
f x
x
x
x
x
x
x
=
−
+
− =
−
+
−
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Calculus I
The derivative is,
( )
2
5
2
5
5
5
5
3
5
3
2
2
3
7
8
7
10
6
6
16
5
2
3
2
f
x
x
x
x
x
x
x
−
−
′
=
−
+
=
−
+
7. Find the derivative of
( )
3
5
4
1
8
6
f t
t
t
t
= −
+
.
Solution
There isn’t much to do here other than take the derivative using the rules we discussed in this
section.
Remember that you’ll need to rewrite the terms so that each of the t’s are in the numerator with
negative exponents before taking the derivative. Here is the rewritten function.
( )
1
3
5
1
4
8
6
f t
t
t
t
−
−
−
=
−
+
The derivative is,
( )
2
4
6
1
4
40
2
f
t
t
t
t
−
−
−
′
= −
+
−
8. Find the derivative of
( )
4
10
3
6
1
1
8
3
R z
z
z
z
=
+
−
.
Solution
There isn’t much to do here other than take the derivative using the rules we discussed in this
section.
Remember that you’ll need to rewrite the terms so that each of the z’s are in the numerator with
negative exponents and rewrite the root as a fractional exponent before taking the derivative.
Here is the rewritten function.
( )
3
4
10
2
1
1
6
8
3
R z
z
z
z
−
−
−
=
+
−
The derivative is,
© 2007 Paul Dawkins
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Calculus I
( )
( )
(
)
5
5
5
11
5
11
2
2
3
1
1
1
10
6
4
10
9
2
8
3
2
3
R z
z
z
z
z
z
z
−
−
−
−
−
−
′
=
−
+
−
−
−
= −
−
+
9. Find the derivative of
(
)
2
3
9
z
x
x
=
−
.
Solution
There isn’t much to do here other than take the derivative using the rules we discussed in this
section.
Remember that in order to do this derivative we’ll first need to multiply the function out before
we take the derivative. Here is the rewritten function.
3
3
9
z
x
x
=
−
The derivative is,
2
9
9
dz
x
dx
=
−
10. Find the derivative of
( ) (
)
(
)
2
4 2
g y
y
y
y
=
−
+
.
Solution
There isn’t much to do here other than take the derivative using the rules we discussed in this
section.
Remember that in order to do this derivative we’ll first need to multiply the function out before
we take the derivative. Here is the rewritten function.
( )
3
2
2
8
g y
y
y
y
=
−
−
The derivative is,
( )
2
3
4
8
g
y
y
y
′
=
−
−
© 2007 Paul Dawkins
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Calculus I
11. Find the derivative of
( )
3
4
7
8
x
x
h x
x
−
+
=
.
Solution
There isn’t much to do here other than take the derivative using the rules we discussed in this
section.
Remember that in order to do this derivative we’ll first need to divide the function out and
simplify before we take the derivative. Here is the rewritten function.
( )
3
2
1
4
7
8
4
7 8
x
x
h x
x
x
x
x
x
−
=
−
+ =
− +
The derivative is,
( )
2
8
8
h x
x
x
−
′
=
−
12. Find the derivative of
( )
5
3
3
5
2
y
y
y
f y
y
−
+
=
.
Solution
There isn’t much to do here other than take the derivative using the rules we discussed in this
section.
Remember that in order to do this derivative we’ll first need to divide the function out and
simplify before we take the derivative. Here is the rewritten function.
( )
5
3
2
2
3
3
3
5
2
5 2
y
y
y
f y
y
y
y
y
y
−
=
−
+
=
− +
The derivative is,
( )
3
2
4
f
y
y
y
−
′
=
−
13. Determine where, if anywhere, the function
( )
3
2
9
48
2
f x
x
x
x
=
+
−
+
is not changing.
Hint : Recall the various interpretations of the derivative. One of them is exactly what we need to
do this problem.
© 2007 Paul Dawkins
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Calculus I
Solution
Step 1
Recall that one of the interpretations of the derivative is that it gives the rate of change of the
function. So, the function won’t be changing if its rate of change is zero and so all we need to do
is find the derivative and set it equal to zero to determine where the rate of change is zero and
hence the function will not be changing.
First the derivative, and we’ll do a little factoring while we are at it.
( )
(
)
(
)(
)
2
2
3
18
48
3
6
16
3
8
2
f
x
x
x
x
x
x
x
′
=
+
−
=
+
−
=
+
−
Step 2
Now all that we need to do is set this equation to zero and solve.
( )
(
)(
)
0
3
8
2
0
f
x
x
x
′
=
⇒
+
−
=
We can easily see from this that the derivative will be zero at
8
x
= −
and
2
x
=
. The function
therefore not be changing at,
8
and
2
x
x
= −
=
14. Determine where, if anywhere, the function
4
3
2
2
3
y
z
z
z
=
− −
is not changing.
Hint : Recall the various interpretations of the derivative. One of them is exactly what we need to
do this problem.
Solution
Step 1
Recall that one of the interpretations of the derivative is that it gives the rate of change of the
function. So, the function won’t be changing if its rate of change is zero and so all we need to do
is find the derivative and set it equal to zero to determine where the rate of change is zero and
hence the function will not be changing.
First the derivative, and we’ll do a little factoring while we are at it.
(
)
3
2
2
8
3
6
8
3
6
dy
z
z
z
z
z
z
dz
=
−
−
=
−
−
Step 2
© 2007 Paul Dawkins
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Calculus I
Now all that we need to do is set this equation to zero and solve.
(
)
2
2
0
8
3
6
0
0,
8
3
6
0
dy
dz
z
z
z
z
z
z
=
−
−
=
→
=
−
− =
We can easily see from this that the derivative will be zero at
0
z
=
, however, because the
quadratic doesn’t factor we’ll need to use the quadratic formula to determine where, if anywhere,
that will be zero.
( )
( )( )
( )
2
3
3
4 8
6
3
201
2 8
16
z
±
−
−
−
±
=
=
The function therefore not be changing at,
3
201
3
201
0
1.07359
0.69859
16
16
z
z
z
+
−
=
=
=
=
= −
15. Find the tangent line to
( )
16
4
g x
x
x
=
−
at
4
x
=
.
Hint : Recall the various interpretations of the derivative. One of them will help us do this
problem.
Solution
Step 1
Recall that one of the interpretations of the derivative is that it gives slope of the tangent line to
the graph of the function.
So, we’ll need the derivative of the function. However before doing that we’ll need to do a little
rewrite. Here is that work as well as the derivative.
( )
( )
1
1
1
2
2
2
2
16
2
16
4
16
2
g x
x
x
g x
x
x
x
x
−
−
−
′
=
−
⇒
= −
−
= −
−
Note that we rewrote the derivative back into rational expressions with roots to help with the
evaluation.
Step 2
Next we need to evaluate the function and derivative at
4
x
=
.
© 2007 Paul Dawkins
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Calculus I
( )
( )
2
16
16
2
4
4 4
4
4
2
4
4
4
g
g′
=
−
= −
= −
−
−
Step 3
Now all that we need to do is write down the equation of the tangent line.
( )
( )(
)
(
)
4
4
4
4 2
4
2
4
y
g
g
x
x
y
x
′
=
+
−
= − −
−
→
= − +
16. Find the tangent line to
( )
4
6
7
8
2
f x
x
x
x
−
=
+
+
at
1
x
= −
.
Hint : Recall the various interpretations of the derivative. One of them will help us do this
problem.
Solution
Step 1
Recall that one of the interpretations of the derivative is that it gives slope of the tangent line to
the graph of the function.
So, we’ll need the derivative of the function.
( )
3
7
3
7
48
28
48
2
28
2
f
x
x
x
x
x
−
′
=
−
+ =
−
+
Note that we rewrote the derivative back into rational expressions help a little with the evaluation.
Step 2
Next we need to evaluate the function and derivative at
1
x
= −
.
( )
( )
1
7 8 2 13
1
28 48 2
22
f
f ′
− = + − =
− = − +
+ =
Step 3
Now all that we need to do is write down the equation of the tangent line.
( )
( )(
)
(
)
1
1
1
13 22
1
22
35
y
f
f
x
x
y
x
′
=
− +
−
+ =
+
+
→
=
+
17. The position of an object at any time t is given by
( )
4
3
2
3
40
126
9
s t
t
t
t
=
−
+
−
.
(a) Determine the velocity of the object at any time t.
(b) Does the object ever stop changing?
(c) When is the object moving to the right and when is the object moving to the left?
© 2007 Paul Dawkins
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Calculus I
Solution
Hint : Recall the various interpretations of the derivative. One of them is exactly what we need
for this part.
(a) Determine the velocity of the object at any time t.
Recall that one of the interpretations of the derivative is that it gives the velocity of an object if
we know the position function of the object.
We’ve been given the position function of the object and so all we need to do is find its derivative
and we’ll have the velocity of the object at any time t.
The velocity of the object is then,
( )
(
)(
)
3
2
12
120
252
12
3
7
s t
t
t
t
t t
t
′
=
−
+
=
−
−
Note that the derivative was factored for later parts and doesn’t really need to be done in general.
Hint : If the object isn’t moving what is the velocity?
(b) Does the object ever stop changing?
The object will not be moving if the velocity is ever zero and so all we need to do is set the
derivative equal to zero and solve.
( )
(
)(
)
0
12
3
7
0
s t
t t
t
′
=
⇒
−
−
=
From this it is pretty easy to see that the derivative will be zero, and hence the object will not be
moving, at,
0
3
7
t
t
t
=
=
=
Hint : How does the direction (right vs. left) of movement relate to the sign (positive or negative)
of the derivative?
(c) When is the object moving to the right and when is the object moving to the left?
To answer this part all we need to know is where the derivative is positive (and hence the object
is moving to the right) or negative (and hence the object is moving to the left). Because the
derivative is continuous we know that the only place it can change sign is where the derivative is
zero. So, as we did in this section a quick number line will give us the sign of the derivative for
the various intervals.
© 2007 Paul Dawkins
35
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Calculus I
Here is the number line for this problem.
From this we get the following right/left movement information.
Moving Right : 0
3, 7
Moving Left :
0, 3
7
t
t
t
t
< <
< < ∞
− ∞ < <
< <
Note that depending upon your interpretation of t as time you may or may not have included the
interval
0
t
−∞ < <
in the “Moving Left” portion.
18. Determine where the function
( )
3
4
5
6 40
5
4
h z
z
z
z
= +
−
−
is increasing and decreasing.
Solution
Hint : Recall the various interpretations of the derivative. One of them is exactly what we need to
get the problem started.
Step 1
Recall that one of the interpretations of the derivative is that it gives the rate of change of the
function. Since we are talking about where the function is increasing and decreasing we are
clearly talking about the rate of change of the function.
So, we’ll need the derivative.
( )
(
)(
)
2
3
4
2
120
20
20
20
3
2
h z
z
z
z
z
z
z
′
=
−
−
= −
+
−
Note that the derivative was factored for later steps and doesn’t really need to be done in general.
Hint : Where is the function not changing?
Step 2
© 2007 Paul Dawkins
36
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Calculus I
Next, we need to know where the function is not changing and so all we need to do is set the
derivative equal to zero and solve.
( )
(
)(
)
2
0
20
3
2
0
h z
z
z
z
′
=
⇒
−
+
−
=
From this it is pretty easy to see that the derivative will be zero, and hence the function will not
be moving, at,
0
3
2
z
z
z
=
= −
=
Hint : How does the increasing/decreasing behavior of the function relate to the sign (positive or
negative) of the derivative?
Step 3
To get the answer to this problem all we need to know is where the derivative is positive (and
hence the function is increasing) or negative (and hence the function is decreasing). Because the
derivative is continuous we know that the only place it can change sign is where the derivative is
zero. So, as we did in this section a quick number line will give us the sign of the derivative for
the various intervals.
Here is the number line for this problem.
From this we get the following increasing/decreasing information.
Increasing :
3
0, 0
2
Decreasing :
3, 2
z
z
z
z
− < <
< <
− ∞ < < −
< < ∞
19. Determine where the function
( ) (
)(
)
2
1
2
R x
x
x
=
+
−
is increasing and decreasing.
Solution
Hint : Recall the various interpretations of the derivative. One of them is exactly what we need to
get the problem started.
© 2007 Paul Dawkins
37
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Calculus I
Step 1
Recall that one of the interpretations of the derivative is that it gives the rate of change of the
function. Since we are talking about where the function is increasing and decreasing we are
clearly talking about the rate of change of the function.
So, we’ll need the derivative. First however we’ll need to multiply out the function so we can
actually take the derivative. Here is the rewritten function and the derivative.
( )
( )
(
)
3
2
2
3
4
3
6
3
2
R x
x
x
R x
x
x
x x
′
=
−
+
=
−
=
−
Note that the derivative was factored for later steps and doesn’t really need to be done in general.
Hint : Where is the function not changing?
Step 2
Next, we need to know where the function is not changing and so all we need to do is set the
derivative equal to zero and solve.
( )
(
)
0
3
2
0
R x
x x
′
=
⇒
−
=
From this it is pretty easy to see that the derivative will be zero, and hence the function will not
be moving, at,
0
2
x
x
=
=
Hint : How does the increasing/decreasing behavior of the function relate to the sign (positive or
negative) of the derivative?
Step 3
To get the answer to this problem all we need to know is where the derivative is positive (and
hence the function is increasing) or negative (and hence the function is decreasing). Because the
derivative is continuous we know that the only place it can change sign is where the derivative is
zero. So, as we did in this section a quick number line will give us the sign of the derivative for
the various intervals.
Here is the number line for this problem.
© 2007 Paul Dawkins
38
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Calculus I
From this we get the following increasing/decreasing information.
Increasing :
0, 2
Decreasing : 0
2
x
z
z
− ∞ < <
< < ∞
< <
20. Determine where, if anywhere, the tangent line to
( )
3
2
5
f x
x
x
x
=
−
+
is parallel to the line
4
23
y
x
=
+
.
Solution
Step 1
The first thing that we’ll need of course is the slope of the tangent line. So, all we need to do is
take the derivative of the function.
( )
2
3
10
1
f
x
x
x
′
=
−
+
Hint : What is the relationship between the slope of two parallel lines?
Step 2
Two lines that are parallel will have the same slope and so all we need to do is determine where
the slope of the tangent line will be 4, the slope of the given line. In other words, we’ll need to
solve,
( )
2
2
4
3
10
1
4
3
10
3
0
f
x
x
x
x
x
′
=
→
−
+ =
→
−
− =
This quadratic doesn’t factor and so a quick use of the quadratic formula will solve this for us.
10
136
10 2 34
5
34
6
6
3
x
±
±
±
=
=
=
So, the tangent line will be parallel to
4
23
y
x
=
+
at,
5
34
5
34
0.276984
3.61032
3
3
x
x
−
+
=
= −
=
=
© 2007 Paul Dawkins
39
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Calculus I
Product and Quotient Rule
1. Use the Product Rule to find the derivative of
( )
(
)(
)
2
3
2
4
8
12
f t
t
t t
t
=
−
−
+
.
Solution
There isn’t much to do here other than take the derivative using the product rule.
( ) (
)
(
) (
)(
)
3
2
2
2
4
3
2
8
1
8
12
4
3
16
20
132
24
96
12
f
t
t
t
t
t
t
t
t
t
t
t
t
′
=
−
−
+
+
−
−
=
−
+
+
−
Note that we multiplied everything out to get a “simpler” answer.
2. Use the Product Rule to find the derivative of
(
)
(
)
3
3
3
1
2
y
x
x
x
−
= +
−
.
Solution
There isn’t much to do here other than take the derivative using the product rule. We’ll also need
to convert the roots to fractional exponents.
1
3
3
3
2
1
2
y
x
x
x
−
= +
−
The derivative is then,
1
2
2
5
1
3
5
3
4
4
3
3
3
6
2
2
2
3
2
3
2
11
2
1
3
3
2
3
2
3
3
dy
x
x
x
x
x
x
x
x
x
x
dx
−
−
−
−
−
−
=
−
+ +
−
−
= −
−
−
−
Note that we multiplied everything out to get a “simpler” answer.
3. Use the Product Rule to find the derivative of
( )
(
)(
)
2
2
3
1 2
3
5
8
h z
z
z
z
z
z
= +
+
+
−
.
Solution
There isn’t much to do here other than take the derivative using the product rule.
© 2007 Paul Dawkins
40
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Calculus I
( ) (
)
(
) (
)(
)
2
3
2
2
2
3
4
2 6
5
8
1 2
3
5 16
3
5 36
90
88
15
h z
z
z
z
z
z
z
z
z
z
z
z
z
′
=
+
+
−
+ +
+
+
−
= +
+
+
−
Note that we multiplied everything out to get a “simpler” answer.
4. Use the Quotient Rule to find the derivative of
( )
2
6
2
x
g x
x
=
−
.
Solution
There isn’t much to do here other than take the derivative using the quotient rule.
( )
(
)
( )
(
)
(
)
2
2
2
2
12
2
6
1
24
6
2
2
x
x
x
x
x
g x
x
x
−
−
−
−
′
=
=
−
−
5. Use the Quotient Rule to find the derivative of
( )
4
2
3
2
1
w w
R w
w
+
=
+
.
Solution
There isn’t much to do here other than take the derivative using the quotient rule.
( )
(
)(
) (
)
( )
(
)
(
)
3
2
4
5
3
2
2
2
2
2
3 4
2
1
3
4
4
4
6
3
2
1
2
1
w
w
w
w
w
w
w
w
R w
w
w
+
+ −
+
+
−
+
′
=
=
+
+
6. Use the Quotient Rule to find the derivative of
( )
2
2
7
4
x
x
f x
x
x
+
=
−
.
Solution
There isn’t much to do here other than take the derivative using the quotient rule.
© 2007 Paul Dawkins
41
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Calculus I
( )
(
)
(
)
(
)
(
)
(
)
1
1
2
1
2
2
2
2
2
7
4
7 8
2
2
7
4
x
x
x
x
x
x
f
x
x
x
−
−
−
−
+
+
′
=
−
7. If
( )
2
8
f
= −
,
( )
2
3
f ′
=
,
( )
2
17
g
=
and
( )
2
4
g′
= −
determine the value of
(
) ( )
2
f g ′
.
Solution
We know that the product rule is,
(
) ( )
( ) ( )
( ) ( )
f g
x
f
x g x
f x g x
′
′
′
=
+
Now, we want to know the value of this at
2
x
=
and so all we need to do is plug this into the
derivative. Doing this gives,
(
) ( )
( ) ( )
( ) ( )
2
2
2
2
2
f g
f
g
f
g
′
′
′
=
+
Now, we were given values for all these quantities and so all we need to do is plug these
into our “formula” above.
(
) ( ) ( )( ) ( )( )
2
3 17
8
4
83
f g ′
=
+ −
− =
8. If
( )
( )
3
f x
x g x
=
,
( )
7
2
g
− =
,
( )
7
9
g′
− = −
determine the value of
( )
7
f ′
−
.
Hint : Even though we don’t know what
( )
g x
is we can still use the product rule to take the
derivative and then we can use the given information to get the value of
( )
7
f ′
−
.
Solution
Even though we don’t know what
( )
g x
is we do have a product of two functions here and so we
can use the product rule to determine the derivative of
( )
f x
.
( )
( )
( )
2
3
3
f
x
x g x
x g x
′
′
=
+
Now all we need to do is plug
7
x
= − into this and use the given information to
determine the value of
( )
7
f ′
− .
© 2007 Paul Dawkins
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Calculus I
( ) ( ) ( ) ( ) ( ) ( )( ) (
)( )
2
3
7
3
7
7
7
7
3 49
2
343
9
3381
f
g
g
′
′
− = −
− + −
− =
+ −
− =
9. Find the equation of the tangent line to
( )
(
)
(
)
2
1 12
4
f x
x
x
= +
−
at
9
x
=
.
Solution
Step 1
We know that the derivative of the function will give us the slope of the tangent line so we’ll
need the derivative of the function. We’ll use the product rule to get the derivative.
( )
(
)
(
)
(
)
(
)
(
)
1
2
2
2
6
6
4
1 12
2
4
2
1 12
f
x
x
x
x
x
x
x
x
x
−
′
=
−
+ +
−
=
−
−
+
Step 2
Note that we didn’t bother to “simplify” the derivative (other than converting the fractional
exponent back to a root) because all we really need this for is a quick evaluation.
Speaking of which here are the evaluations that we’ll need for this problem.
( ) ( )(
)
( ) ( )(
)
( )
9
37
77
2849
9
2
77
18 37
820
f
f ′
=
−
= −
=
−
−
= −
Step 3
Now all that we need to do is write down the equation of the tangent line.
( )
( )(
)
(
)
9
9
9
2849 820
9
820
4531
y
f
f
x
x
y
x
′
=
+
−
= −
−
−
→
= −
+
10. Determine where
( )
2
2
1 8
x
x
f x
x
−
=
+
is increasing and decreasing.
Solution
Step 1
We’ll first need the derivative, which will require the quotient rule, because we know that the
derivative will give us the rate of change of the function. Here is the derivative.
( )
(
)
(
) (
)
( )
(
)
(
)
2
2
2
2
2
2
2
1 2
1 8
16
1 2
8
1 8
1 8
x
x
x
x
x
x
x
f
x
x
x
−
+
− −
−
−
′
=
=
+
+
© 2007 Paul Dawkins
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Calculus I
Step 2
Next, we need to know where the function is not changing and so all we need to do is set the
derivative equal to zero and solve. In this case it is clear that the denominator will never be zero
for any real number and so the derivative will only be zero where the numerator is zero.
Therefore, setting the numerator equal to zero and solving gives,
(
)
(
)(
)
2
2
1 2
8
8
2
1
4
1 2
1
0
x
x
x
x
x
x
−
−
= −
+
− = −
−
+ =
From this it is pretty easy to see that the derivative will be zero, and hence the function will not
be changing, at,
1
1
2
4
x
x
= −
=
Step 3
To get the answer to this problem all we need to know is where the derivative is positive (and
hence the function is increasing) or negative (and hence the function is decreasing). Because the
derivative is continuous we know that the only place it can change sign is where the derivative is
zero. So, as we did in this section a quick number line will give us the sign of the derivative for
the various intervals.
Here is the number line for this problem.
From this we get the following increasing/decreasing information.
1
1
Increasing :
2
4
1
1
Decreasing :
,
2
4
x
x
x
− < <
− ∞ < < −
< < ∞
11. Determine where
( )
(
)(
)
2
2
4
1 5
V t
t
t
=
−
+
is increasing and decreasing.
© 2007 Paul Dawkins
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Calculus I
Solution
Step 1
We’ll first need the derivative, for which we will use the product rule, because we know that the
derivative will give us the rate of change of the function. Here is the derivative.
( ) ( )
(
) (
)
( )
(
)
2
2
3
2
2
1 5
4
10
38
20
2 19 10
V t
t
t
t
t
t
t
t
t
′
= −
+
+ −
=
−
=
−
Step 2
Next, we need to know where the function is not changing and so all we need to do is set the
derivative equal to zero and solve. From the factored form of the derivative it is easy to see that
the derivative will be zero at,
19
0
1.3784
10
t
t
=
= ±
= ±
Step 3
To get the answer to this problem all we need to know is where the derivative is positive (and
hence the function is increasing) or negative (and hence the function is decreasing). Because the
derivative is continuous we know that the only place it can change sign is where the derivative is
zero. So, as we did in this section a quick number line will give us the sign of the derivative for
the various intervals.
Here is the number line for this problem.
From this we get the following increasing/decreasing information.
19
19
Increasing :
,
0
10
10
19
19
Decreasing :
0,
10
10
t
t
t
x
− ∞ < < −
< <
−
< <
< < ∞
© 2007 Paul Dawkins
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Calculus I
Derivatives of Trig Functions
1. Evaluate
( )
0
sin 10
lim
z
z
z
→
.
Solution
All we need to do is set this up to allow us to use the fact from the notes in this section.
( )
( )
( )
( )
0
0
0
sin 10
10 sin 10
sin 10
lim
lim
10 lim
10 1
10
10
10
z
z
z
z
z
z
z
z
z
→
→
→
=
=
=
=
2. Evaluate
(
)
( )
0
sin 12
lim
sin 5
α
α
α
→
.
Solution
All we need to do is set this up to allow us to use the fact from the notes in this section.
(
)
( )
(
)
( )
(
)
( )
(
)
( )
(
)
( )
( )( )
0
0
0
0
0
0
sin 12
12 sin 12
sin 12
5
12
5
lim
lim
lim
sin 5
12
5 sin 5
5
12
sin 5
sin 12
sin 12
12
5
12
5
lim
lim
lim
5
12
sin 5
5
12
sin 5
12
12
1 1
5
5
α
α
α
α
α
α
α
α
α
α
α
α
α
α
α
α
α
α
α
α
α
α
α
α
α
α
α
α
→
→
→
→
→
→
=
=
=
=
=
=
3. Evaluate
( )
0
cos 4
1
lim
x
x
x
→
−
.
Solution
All we need to do is set this up to allow us to use the fact from the notes in this section.
© 2007 Paul Dawkins
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Calculus I
( )
( )
(
)
( )
( )
0
0
0
4 cos 4
1
cos 4
1
cos 4
1
lim
lim
4 lim
4 0
0
4
4
x
x
x
x
x
x
x
x
x
→
→
→
−
−
−
=
=
=
=
4. Differentiate
( )
( )
( )
2 cos
6 sec
3
f x
x
x
=
−
+
.
Solution
Not much to do here other than take the derivative.
( )
( )
( ) ( )
2 sin
6 sec
tan
f
x
x
x
x
′
= −
−
5. Differentiate
( )
( )
( )
10 tan
2 cot
g z
z
z
=
−
.
Solution
Not much to do here other than take the derivative.
( )
( )
( )
2
2
10 sec
2 csc
g z
z
z
′
=
+
6. Differentiate
( )
( ) ( )
tan
sec
f w
w
w
=
.
Solution
Not much to do here other than take the derivative, which will require the product rule.
( )
( )
( )
( )
( ) ( )
( )
( )
( )
2
3
2
sec
sec
tan
sec
tan
sec
sec
tan
f
w
w
w
w
w
w
w
w
w
′
=
+
=
+
7. Differentiate
( )
( )
3
2
sin
h t
t
t
t
= −
.
Solution
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Calculus I
Not much to do here other than take the derivative, which will require the product rule for the
second term.
You’ll need to be careful with the minus sign on the second term. You can either use a set of
parenthesis around the derivative of the second term or you can think of the minus sign as part of
the “first” function. We’ll think of the minus sign as part of the first function for this problem.
( )
( )
( )
2
2
3
2 sin
cos
h t
t
t
t
t
t
′
=
−
−
8. Differentiate
( )
6 4
csc
y
x
x
= +
.
Solution
Not much to do here other than take the derivative, which will require the product rule for the
second term.
( )
( )
( ) ( )
(
)
( )
( ) ( )
1
1
1
2
2
2
4
csc
4
csc
cot
2
csc
4
csc
cot
y
x
x
x
x
x
x
x
x
x
x
−
−
′ =
+
−
=
−
9. Differentiate
( )
( )
( )
1
2 sin
4 cos
R t
t
t
=
−
.
Solution
Not much to do here other than take the derivative, which will require the quotient rule.
( ) ( )
( )
( )
(
)
( )
( )
( )
(
)
( )
( )
(
)
( )
( )
( )
( )
(
)
2
2
0 2 sin
4 cos
1 2 cos
4 sin
2 cos
4 sin
2 sin
4 cos
2 sin
4 cos
t
t
t
t
t
t
R t
t
t
t
t
−
−
+
−
−
′
=
=
−
−
10. Differentiate
( )
( )
( )
tan
1 csc
v
v
Z v
v
+
=
+
.
Solution
Not much to do here other than take the derivative, which will require the quotient rule.
© 2007 Paul Dawkins
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Calculus I
( )
( )
(
)
( )
(
)
( )
(
)
( ) ( )
(
)
( )
(
)
( )
(
)
( )
(
)
( ) ( )
( )
(
)
( )
(
)
2
2
2
2
1 sec
1 csc
tan
csc
cot
1 csc
1 sec
1 csc
csc
cot
tan
1 csc
v
v
v
v
v
v
Z v
v
v
v
v
v v
v
v
+
+
− +
−
′
=
+
+
+
+
+
=
+
11. Find the tangent line to
( )
( )
( )
tan
9 cos
f x
x
x
=
+
at
x
π
=
.
Solution
Step 1
We know that the derivative of the function will give us the slope of the tangent line so we’ll
need the derivative of the function.
( )
( )
( )
2
sec
9 sin
f
x
x
x
′
=
−
Step 2
Now all we need to do is evaluate the function and the derivative at the point in question.
( )
( )
( )
( )
( )
( )
2
tan
9 cos
9
sec
9 sin
1
f
f
π
π
π
π
π
π
′
=
+
= −
=
−
=
Step 3
Now all that we need to do is write down the equation of the tangent line.
( )
( )(
)
( )(
)
9
1
9
y
f
f
x
x
y
x
π
π
π
π
π
′
=
+
−
= − +
−
→
= − −
Don’t get excited about the presence of the
π
in the answer. It is just a number like the 9 is and
so is nothing to worry about.
12. The position of an object is given by
( )
( )
2 7 cos
s t
t
= +
determine all the points where the
object is not moving.
Solution
We know that the object will not be moving if its velocity, which is simply the derivative of the
position function, is not zero. So all we need to do is take the derivative, set it equal to zero and
solve.
© 2007 Paul Dawkins
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Calculus I
( )
( )
( )
7 sin
7 sin
0
s t
t
t
′
= −
⇒
−
=
So, for this problem the object will not be moving anywhere that sine is zero. From our
recollection of the unit circle we know that will be at,
0
2
2
and
2
0, 1, 2, 3,
t
n
n
t
n
n
π
π
π
π
= +
=
= +
= ± ± ±
13. Where in the range
[
]
2, 7
−
is the function
( )
( )
4 cos
f x
x
x
=
−
is increasing and
decreasing.
Solution
Step 1
We’ll first need the derivative because we know that the derivative will give us the rate of change
of the function. Here is the derivative.
( )
( )
4 sin
1
f
x
x
′
= −
−
Step 2
Next, we need to know where the function is not changing and so all we need to do is set the
derivative equal to zero and solve.
( )
( )
1
4 sin
1 0
sin
4
x
x
−
− =
⇒
= −
A quick calculator computation tells us that,
1
1
sin
0.2527
4
x
−
=
−
= −
Recalling our work in the Review chapter on solving trig equations we know that a positive angle
corresponding to this solution is :
2
0.2527
6.0305
x
π
=
−
=
. Either can be used, but we will
use the positive angle.
Also, from a quick check on a unit circle we can see that
0.2527
3.3943
x
π
= +
=
will be a
second solution.
Putting all of this together and we can see that the derivative will be zero at,
6.0305 2
and
3.3943 2
0, 1, 2, 3,
x
n
x
n
n
π
π
=
+
=
+
= ± ± ±
© 2007 Paul Dawkins
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Calculus I
Finally, all we need to so is plug in some n’s to determine which solutions fall in the interval we
are working on,
[
]
2, 7
−
.
1:
0.2527
2.8889
n
x
x
= −
= −
= −
0 :
6.0305
3.3943
1:
12.3137
n
x
x
n
x
=
=
=
=
=
9.6775
x
=
So, in the interval
[
]
2, 7
−
the function will stop changing at the following three points.
0.2527, 3.3943, 6.0305
x
= −
Step 3
To get the answer to this problem all we need to know is where the derivative is positive (and
hence the function is increasing) or negative (and hence the function is decreasing). Because the
derivative is continuous we know that the only place it can change sign is where the derivative is
zero. So, as we did in this section a quick number line will give us the sign of the derivative for
the various intervals.
Here is the number line for this problem.
From this we get the following increasing/decreasing information.
Increasing :
2
0.2527, 3.3943
6.0305
Decreasing :
0.2527
3.3943, 6.0305
7
x
x
x
x
− ≤ < −
< <
−
< <
< ≤
Note that because we’ve only looked at what is happening in the interval
[
]
2, 7
−
we can’t say
anything about the increasing/decreasing nature of the function outside of this interval.
© 2007 Paul Dawkins
51
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Calculus I
Derivatives of Exponential and Logarithm Functions
1. Differentiate
( )
2
8
x
x
f x
=
−
e
.
Solution
Not much to do here other than take the derivative using the formulas from class.
( )
( )
2
8 ln 8
x
x
f
x
′
=
−
e
2. Differentiate
( )
( )
( )
3
4 log
ln
g t
t
t
=
−
.
Solution
Not much to do here other than take the derivative using the formulas from class.
( )
( )
4
1
ln 3
g t
t
t
′
=
−
3. Differentiate
( )
( )
3 log
w
R w
w
=
.
Solution
Not much to do here other than take the derivative using the formulas from class.
( )
( ) ( )
( )
3
3 ln 3 log
ln 10
w
w
R w
w
w
′
=
+
Recall that
( )
log x
is the common logarithm and so is really
( )
10
log
x
.
4. Differentiate
( )
5
ln
z
y
z
z
=
− e
.
© 2007 Paul Dawkins
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Calculus I
Solution
Not much to do here other than take the derivative using the formulas from class.
( )
4
5
ln
z
z
y
z
z
z
′ =
−
−
e
e
5. Differentiate
( )
1
y
y
h y
=
− e
.
Solution
Not much to do here other than take the derivative using the formulas from class.
( )
( )
(
) ( )
(
)
(
)
2
2
1 1
1
1
1
y
y
y
y
y
y
y
y
h y
−
−
−
−
+
′
=
=
−
−
e
e
e
e
e
e
6. Differentiate
( )
( )
1 5
ln
t
f t
t
+
=
.
Solution
Not much to do here other than take the derivative using the formulas from class.
( )
( )
( ) (
)
( )
( )
( )
2
2
1
1
5 ln
1 5
5 ln
5
1 5
ln
ln
ln
t
t
t
t
t
t
f t
t
t
t
− +
− −
+
=
=
=
7. Find the tangent line to
( )
7
4
x
x
f x
=
+ e
at
0
x
=
.
Solution
Step 1
© 2007 Paul Dawkins
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Calculus I
We know that the derivative of the function will give us the slope of the tangent line so we’ll
need the derivative of the function.
( )
( )
7 ln 7
4
x
x
f
x
′
=
+ e
Step 2
Now all we need to do is evaluate the function and the derivative at the point in question.
( )
( )
( )
0
5
0
ln 7
4
5.9459
f
f ′
=
=
+ =
Step 3
Now all that we need to do is write down the equation of the tangent line.
( )
( )(
)
( )
(
)
0
0
0
5
ln 7
4
5 5.9459
y
f
f
x
x
x
′
=
+
−
= +
+
= +
8. Find the tangent line to
( )
( )
( )
2
ln
log
f x
x
x
=
at
2
x
=
.
Solution
Step 1
We know that the derivative of the function will give us the slope of the tangent line so we’ll
need the derivative of the function.
( )
( )
( )
( )
2
log
ln
ln 2
x
x
f
x
x
x
′
=
+
Step 2
Now all we need to do is evaluate the function and the derivative at the point in question.
( )
( )
( )
( )
( )
( )
( )
( )
2
2
log
2
ln 2
2
ln 2 log
2
ln 2
2
1
2
2 ln 2
f
f ′
=
=
=
+
=
Step 3
Now all that we need to do is write down the equation of the tangent line.
( )
( )(
)
( ) ( )(
)
( )
2
2
2
ln 2
1
2
2
ln 2
y
f
f
x
x
x
′
=
+
−
=
+
−
= − +
9. Determine if
( )
t
t
V t
=
e
is increasing or decreasing at the following points.
(a)
4
t
= −
(b)
0
t
=
(c)
10
t
=
Solution
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Calculus I
(a)
4
t
= −
We know that the derivative of the function will give us the rate of change for the function and so
we’ll need that.
( )
( )
( )
( )
( )
2
2
1
1
t
t
t
t
t
t
t
t
t
t
V t
−
−
−
′
=
=
=
e
e
e
e
e
e
e
Now, all we need to do is evaluate the derivative at the point in question. So,
( )
4
5
4
272.991
0
V
−
′ − =
=
>
e
( )
4
0
V ′
− >
and so the function must be increasing at
4
t
= −
.
(b)
0
t
=
We found the derivative of the function in the first part so here all we need to do is the evaluation.
( )
0
1
0
1 0
V ′
=
= >
e
( )
0
0
V ′
>
and so the function must be increasing at
0
t
=
.
(c)
10
t
=
We found the derivative of the function in the first part so here all we need to do is the evaluation.
( )
10
9
10
0.0004086
0
V
−
′
=
= −
<
e
( )
10
0
V ′
<
and so the function must be decreasing at
10
t
=
.
10. Determine if
( ) (
) ( )
6 ln
G z
z
z
=
−
is increasing or decreasing at the following points.
(a)
1
z
=
(b)
5
z
=
(c)
20
z
=
Solution
(a)
1
z
=
We know that the derivative of the function will give us the rate of change for the function and so
we’ll need that.
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Calculus I
( )
( )
6
ln
z
G z
z
z
−
′
=
+
Now, all we need to do is evaluate the derivative at the point in question. So,
( )
( )
1
ln 1
5
5
0
G′
=
− = − <
( )
1
0
G′
<
and so the function must be decreasing at
1
z
=
.
(b)
5
z
=
We found the derivative of the function in the first part so here all we need to do is the evaluation.
( )
( )
1
5
ln 5
1.40944
0
5
G′
=
− =
>
( )
5
0
G′
>
and so the function must be increasing at
5
z
=
.
(c)
20
z
=
We found the derivative of the function in the first part so here all we need to do is the evaluation.
( )
( )
7
20
ln 20
3.69573
10
G′
=
+
=
( )
20
0
G′
>
and so the function must be increasing at
20
z
=
.
Derivatives of Inverse Trig Functions
1. Differentiate
( )
( )
( )
1
2 cos
6 cos
T z
z
z
−
=
+
.
Solution
Not much to do here other than take the derivative using the formulas from class.
( )
( )
2
6
2 sin
1
T
z
z
z
′
= −
−
−
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Calculus I
2. Differentiate
( )
( )
( )
1
1
csc
4 cot
g t
t
t
−
−
=
−
.
Solution
Not much to do here other than take the derivative using the formulas from class.
( )
2
2
1
4
1
1
g t
t
t t
′
= −
+
+
−
3. Differentiate
( )
6
1
5
sec
y
x
x
−
=
−
.
Solution
Not much to do here other than take the derivative using the formulas from class.
5
2
1
30
1
dy
x
dx
x x
=
−
−
4. Differentiate
( )
( )
( )
2
1
sin
tan
f w
w
w
w
−
=
+
.
Solution
Not much to do here other than take the derivative using the formulas from class.
( )
( )
( )
2
1
2
cos
2 tan
1
w
f
w
w
w
w
w
−
′
=
+
+
+
5. Differentiate
( )
( )
1
sin
1
x
h x
x
−
=
+
.
Solution
Not much to do here other than take the derivative using the formulas from class.
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Calculus I
( )
( )
(
)
( )
(
)
1
2
1
2
2
2
2
1
sin
1
1
sin
1
1
1
1
x
x
x
x
x
x
h x
x
x
x
−
−
+
−
+ −
−
−
′
=
=
+
−
+
Derivatives of Hyperbolic Functions
1. Differentiate
( )
( )
( )
( )
sinh
2 cosh
sech
f x
x
x
x
=
+
−
.
Solution
Not much to do here other than take the derivative using the formulas from class.
( )
( )
( )
( )
( )
cosh
2 sinh
sech
tanh
f
x
x
x
x
x
′
=
+
+
2. Differentiate
( )
( )
( )
2
tan
csch
R t
t
t
t
=
+
.
Solution
Not much to do here other than take the derivative using the formulas from class.
( )
( )
( )
( )
( )
2
2
sec
2 csch
csch
coth
R t
t
t
t
t
t
t
′
=
+
−
3. Differentiate
( )
( )
1
tanh
z
g z
z
+
=
.
Solution
Not much to do here other than take the derivative using the formulas from class.
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Calculus I
( )
( ) (
)
( )
( )
2
2
tanh
1 sech
tanh
z
z
z
g z
z
− +
′
=
Chain Rule
1. Differentiate
( )
(
)
4
2
6
7
f x
x
x
=
+
.
Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside”
functions and then apply the chain rule.
Solution
For this problem the outside function is (hopefully) clearly the exponent of 4 on the parenthesis
while the inside function is the polynomial that is being raised to the power. The derivative is
then,
( )
(
)
(
)
(
)
(
)
3
3
2
2
4 6
7
12
7
4 12
7 6
7
f
x
x
x
x
x
x
x
′
=
+
+
=
+
+
2. Differentiate
( )
(
)
2
2
4
3
2
g t
t
t
−
=
− +
.
Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside”
functions and then apply the chain rule.
Solution
For this problem the outside function is (hopefully) clearly the exponent of -2 on the parenthesis
while the inside function is the polynomial that is being raised to the power. The derivative is
then,
( )
(
)
(
)
(
)
(
)
3
3
2
2
2 4
3
2
8
3
2 8
3 4
3
2
g t
t
t
t
t
t
t
−
−
′
= −
− +
− = −
−
− +
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Calculus I
3. Differentiate
3
1 8
y
z
=
−
.
Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside”
functions and then apply the chain rule.
Solution
For this problem, after converting the root to a fractional exponent, the outside function is
(hopefully) clearly the exponent of
1
3
while the inside function is the polynomial that is being
raised to the power (or the polynomial inside the root – depending upon how you want to think
about it). The derivative is then,
(
)
(
) ( )
(
)
1
2
2
3
3
3
1
8
1 8
1 8
8
1 8
3
3
dy
y
z
z
z
dz
−
−
= −
⇒
=
−
− = −
−
4. Differentiate
( )
( )
csc 7
R w
w
=
.
Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside”
functions and then apply the chain rule.
Solution
For this problem the outside function is (hopefully) clearly the trig function and the inside
function is the stuff inside of the trig function. The derivative is then,
( )
( ) ( )
7 csc 7
cot 7
R w
w
w
′
= −
In dealing with functions like cosecant (or secant for that matter) be careful to make sure that the
inside function gets substituted into both terms of the derivative of the outside function. One of
the more common mistakes with this kind of problem is to only substitute the
7w
into only the
cosecant or only the cotangent instead of both as it should be.
5. Differentiate
( )
( )
(
)
2 sin 3
tan
G x
x
x
=
+
.
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Calculus I
Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside”
functions and then apply the chain rule.
Solution
For this problem the outside function is (hopefully) clearly the sine function and the inside
function is the stuff inside of the trig function. The derivative is then,
( )
( )
(
)
( )
(
)
2
2 3 sec
cos 3
tan
G x
x
x
x
′
=
+
+
6. Differentiate
( )
(
)
tan 4 10
h u
u
=
+
.
Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside”
functions and then apply the chain rule.
Solution
For this problem the outside function is (hopefully) clearly the trig function and the inside
function is the stuff inside of the trig function. The derivative is then,
( )
(
)
2
10 sec
4 10
h u
u
′
=
+
7. Differentiate
( )
7
4
5
t t
f t
+
= + e
.
Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside”
functions and then apply the chain rule.
Solution
Note that we only need to use the Chain Rule on the second term as we can differentiate the first
term without the Chain Rule.
Now, recall that for exponential functions outside function is the exponential function itself and
the inside function is the exponent. The derivative is then,
( )
(
)
7
6
4
4 7
t t
f
t
t
+
′
=
+
e
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Calculus I
8. Differentiate
( )
( )
1
cos x
g x
−
= e
.
Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside”
functions and then apply the chain rule.
Solution
For exponential functions remember that the outside function is the exponential function itself
and the inside function is the exponent. The derivative is then,
( )
( )
( )
1
cos
sin
x
g x
x
−
′
=
e
9. Differentiate
( )
1 6
2
z
H z
−
=
.
Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside”
functions and then apply the chain rule.
Solution
For exponential functions remember that the outside function is the exponential function itself
and the inside function is the exponent. The derivative is then,
( )
(
)
( )
1 6
6 2
ln 2
z
H
z
−
′
= −
10. Differentiate
( )
(
)
1
tan
3
1
u t
t
−
=
−
.
Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside”
functions and then apply the chain rule.
Solution
For this problem the outside function is (hopefully) clearly the inverse tangent and the inside
function is the stuff inside of the inverse tangent. The derivative is then,
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Calculus I
( )
(
)
2
3
3
1
1
u t
t
′
=
−
+
11. Differentiate
( )
(
)
2
3
ln 1 5
F y
y
y
=
−
+
.
Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside”
functions and then apply the chain rule.
Solution
For this problem the outside function is (hopefully) clearly the logarithm and the inside function
is the stuff inside of the logarithm. The derivative is then,
( )
(
)
2
2
2
3
2
3
1
10
3
10
3
1 5
1 5
y
y
F y
y
y
y
y
y
y
−
+
=
−
+
=
−
+
−
+
With logarithm problems remember that after differentiating the logarithm (i.e. the outside
function) you need to substitute the inside function into the derivative. So, instead of getting just,
1
y
we get the following (i.e. we plugged the inside function into the derivative),
2
3
1
1 5 y
y
−
+
Then, we can’t forget of course to multiply by the derivative of the inside function.
12. Differentiate
( )
( )
( )
(
)
ln sin
cot
V x
x
x
=
−
.
Hint : Recall that with Chain Rule problems you need to identify the “inside” and “outside”
functions and then apply the chain rule.
Solution
For this problem the outside function is (hopefully) clearly the logarithm and the inside function
is the stuff inside of the logarithm. The derivative is then,
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Calculus I
( )
( )
( )
( )
( )
(
)
( )
( )
( )
( )
2
2
cos
csc
1
cos
csc
sin
cot
sin
cot
x
x
V x
x
x
x
x
x
x
+
=
+
=
−
−
With logarithm problems remember that after differentiating the logarithm (i.e. the outside
function) you need to substitute the inside function into the derivative. So, instead of getting just,
1
x
we get the following (i.e. we plugged the inside function into the derivative),
( )
( )
1
sin
cot
x
x
−
Then, we can’t forget of course to multiply by the derivative of the inside function.
13. Differentiate
( )
( )
( )
6
6
sin
sin
h z
z
z
=
+
.
Hint : Don’t get too locked into problems only requiring a single use of the Chain Rule.
Sometimes separate terms will require different applications of the Chain Rule, or maybe only
one of the terms will require the Chain Rule.
Solution
For this problem each term will require a separate application of the Chain Rule and don’t forget
that,
( )
( )
6
6
sin
sin
z
z
=
So, in the first term the outside function is the sine function, while the sine function is the inside
function in the second term. The derivative is then,
( )
( )
( ) ( )
5
6
5
6
cos
6 sin
cos
h z
z
z
z
z
′
=
+
14. Differentiate
( )
7
w
S w
w
−
=
+ e
.
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Calculus I
Hint : Don’t get too locked into problems only requiring a single use of the Chain Rule.
Sometimes separate terms will require different applications of the Chain Rule, or maybe only
one of the terms will require the Chain Rule.
Solution
For this problem each term will require a separate application of the Chain Rule and make sure
you are careful with parenthesis in dealing with the root in the first term.
The derivative is then,
( ) ( )
( )
( )( )
( )
1
1
1
2
2
2
1
7
7
7
7
7
2
2
w
w
w
S w
w
S w
w
w
−
−
−
−
−
′
=
+
⇒
=
−
=
−
e
e
e
15. Differentiate
( )
(
)
7
2
3
sin
6
g z
z
z
=
−
+
.
Hint : Don’t get too locked into problems only requiring a single use of the Chain Rule.
Sometimes separate terms will require different applications of the Chain Rule, or maybe only
one of the terms will require the Chain Rule.
Solution
For this problem the first term requires no Chain Rule and the second term will require the Chain
Rule. The derivative is then,
( )
(
)
6
2
21
2 cos
6
g z
z
z
z
′
=
−
+
16. Differentiate
( )
( )
(
)
(
)
10
4
ln sin
3
f x
x
x
x
=
−
−
.
Hint : Don’t get too locked into problems only requiring a single use of the Chain Rule.
Sometimes separate terms will require different applications of the Chain Rule, or maybe only
one of the terms will require the Chain Rule.
Solution
For this problem each term will require a separate application of the Chain Rule. The derivative
is then,
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Calculus I
( )
( )
( )
(
)(
)
( )
(
)(
)
9
9
3
4
3
4
cos
10 4
3
3
cot
10 4
3
3
sin
x
f
x
x
x
x
x
x
x
x
x
′
=
−
−
−
=
−
−
−
17. Differentiate
( )
6
2
5
h t
t
t
t
=
−
.
Hint : Don’t forget the Product and Quotient Rule. Sometimes, in the process of doing the
Product or Quotient Rule you’ll need to use the Chain Rule when differentiating one or both of
the terms in the product or quotient.
Solution
For this problem we’ll need to do the Product Rule to start off the derivative. In the process we’ll
need to use the Chain Rule when we differentiate the second term.
The derivative is then,
( )
(
)
( )
(
)
(
)
(
)
(
)
(
)
(
)
1
6
2
2
1
1
1
1
5
2
6
2
5
2
6
2
2
2
2
2
5
1
1
6
5
5
10
1
6
5
10
1 5
2
2
h t
t
t
t
h t
t
t
t
t
t
t
t
t
t
t
t
t
t
t
−
−
=
−
′
=
−
+
−
− =
−
+
−
−
18. Differentiate
( )
( )
2
5
ln
q t
t
t
=
.
Hint : Don’t forget the Product and Quotient Rule. Sometimes, in the process of doing the
Product or Quotient Rule you’ll need to use the Chain Rule when differentiating one or both of
the terms in the product or quotient.
Solution
For this problem we’ll need to do the Product Rule to start off the derivative. In the process we’ll
need to use the Chain Rule when we differentiate the second term.
The derivative is then,
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Calculus I
( )
( )
( )
4
5
2
5
5
5
2 ln
2 ln
5
t
q t
t
t
t
t
t
t
t
′
=
+
=
+
19. Differentiate
( )
( ) (
)
cos 3
sec 1
g w
w
w
=
−
.
Hint : Don’t forget the Product and Quotient Rule. Sometimes, in the process of doing the
Product or Quotient Rule you’ll need to use the Chain Rule when differentiating one or both of
the terms in the product or quotient.
Solution
For this problem we’ll need to do the Product Rule to start off the derivative. In the process we’ll
need to use the Chain Rule when we differentiate each term.
The derivative is then,
( )
( )( ) (
)
( ) (
) (
)( )
( ) (
)
( ) (
) (
)
sin 3
3 sec 1
cos 3
sec 1
tan 1
1
3sin 3
sec 1
cos 3
sec 1
tan 1
g w
w
w
w
w
w
w
w
w
w
w
′
= −
−
+
−
−
−
= −
−
−
−
−
20. Differentiate
( )
2
sin 3
1
t
y
t
=
+
.
Hint : Don’t forget the Product and Quotient Rule. Sometimes, in the process of doing the
Product or Quotient Rule you’ll need to use the Chain Rule when differentiating one or both of
the terms in the product or quotient.
Solution
For this problem we’ll need to do the Quotient Rule to start off the derivative. In the process
we’ll need to use the Chain Rule when we differentiate the numerator.
The derivative is then,
( )
(
)
( )( )
(
)
( )
(
)
( )
(
)
2
2
2
2
2
2
3cos 3
1
sin 3
2
3cos 3
1
2 sin 3
1
1
t
t
t
t
t
t
t
t
dy
dt
t
t
+
−
+
−
=
=
+
+
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Calculus I
21. Differentiate
( )
( )
2
1
tan 12
x
K x
x
x
−
+
=
+
e
.
Hint : Don’t forget the Product and Quotient Rule. Sometimes, in the process of doing the
Product or Quotient Rule you’ll need to use the Chain Rule when differentiating one or both of
the terms in the product or quotient.
Solution
For this problem we’ll need to do the Quotient Rule to start off the derivative. In the process
we’ll need to use the Chain Rule when we differentiate both the numerator and the denominator.
The derivative is then,
( )
( )
(
)
(
)
( )
(
)
( )
(
)
2
2
2
2
2
tan 12
1
1 12 sec 12
tan 12
x
x
x
x
x
K x
x
x
−
−
−
+
− +
+
′
=
+
e
e
22. Differentiate
( )
( )
2
cos
x
f x
x
=
e
.
Hint : Don’t forget the Product and Quotient Rule. Sometimes, in the process of using the Chain
Rule, you’ll also need the Product and/or Quotient Rule.
Solution
For this problem we’ll start off using the Chain Rule, however when we differentiate the inside
function we’ll need to do the Product Rule.
The derivative is then,
( )
(
) (
)
2
2
2
sin
x
x
x
f
x
x
x
x
′
= −
+
e
e
e
23. Differentiate
( )
5
tan 4
z
x
x
=
+
.
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Calculus I
Hint : Sometimes the Chain Rule will need to be done multiple times before we finish taking the
derivative.
Step 1
This problem will require multiple uses of the Chain Rule and so we’ll step though the derivative
process to make each use clear.
Here is the first step of the derivative and we’ll need to use the Chain Rule in this step.
( )
(
)
( )
(
)
( )
(
)
1
2
1
2
5
tan 4
1
5
tan 4
5
tan 4
2
z
x
x
dz
d
x
x
x
x
dx
dx
−
=
+
=
+
+
Step 2
In this step we can see that we’ll need to use the Chain Rule on the second term.
The derivative is then,
( )
(
)
( )
(
)
1
2
2
1
5
tan 4
5 4 sec
4
2
dz
x
x
x
dx
−
=
+
+
In this step we were using the Chain Rule on the second term and so when multiplying by the
derivative of the inside function we only multiply the second term by the derivative of the inside
function and not both terms.
24. Differentiate
( )
(
)
(
)
3
6
sin 2
t
f t
t
−
=
+
−
e
.
Hint : Sometimes the Chain Rule will need to be done multiple times before we finish taking the
derivative.
Step 1
This problem will require multiple uses of the Chain Rule and so we’ll step though the derivative
process to make each use clear.
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Calculus I
Here is the first step of the derivative and we’ll need to use the Chain Rule in this step.
( )
(
)
(
)
(
)
(
)
2
6
6
3
sin 2
sin 2
t
t
d
f
t
t
t
dt
−
−
′
=
+
−
+
−
e
e
Step 2
In this step we can see that we’ll need to use the Chain Rule on each of the terms.
The derivative is then,
( )
(
)
(
)
(
)
(
)
2
6
6
3
sin 2
6
cos 2
t
t
f
t
t
t
−
−
′
=
+
−
−
−
−
e
e
25. Differentiate
( )
(
)
( )
(
)
10
2
1
ln
1
tan
6
g x
x
x
−
=
+ −
.
Hint : Sometimes the Chain Rule will need to be done multiple times before we finish taking the
derivative.
Step 1
This problem will require multiple uses of the Chain Rule and so we’ll step though the derivative
process to make each use clear.
Here is the first step of the derivative and we’ll need to use the Chain Rule in this step.
( )
(
)
( )
(
)
(
)
( )
(
)
9
2
1
2
1
10 ln
1
tan
6
ln
1
tan
6
d
g x
x
x
x
x
dx
−
−
′
=
+ −
+ −
Step 2
In this step we can see that we’ll need to use the Chain Rule on each of the terms.
The derivative is then,
( )
(
)
( )
(
)
9
2
1
2
2
2
6
10 ln
1
tan
6
1
36
1
x
g x
x
x
x
x
−
′
=
+ −
−
+
+
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Calculus I
26. Differentiate
( )
(
)
4
2
tan
1
h z
z
=
+
.
Hint : Sometimes the Chain Rule will need to be done multiple times before we finish taking the
derivative.
Step 1
This problem will require multiple uses of the Chain Rule and so we’ll step though the derivative
process to make each use clear. Also, recall that,
( )
( )
4
4
tan
tan
x
x
=
Here is the first step of the derivative and we’ll need to use the Chain Rule in this step.
( )
(
)
(
)
3
2
2
4 tan
1
tan
1
d
h z
z
z
dz
′
=
+
+
Step 2
As we can see the derivative from the previous step will also require the Chain Rule.
The derivative is then,
( )
(
) (
)
( )
(
) (
)
3
2
2
2
3
2
2
2
4 tan
1 sec
1 2
8 tan
1 sec
1
h z
z
z
z
z
z
z
′
=
+
+
=
+
+
27. Differentiate
( )
( )
(
)
1
2
3
12
sin
3
f x
x
x
−
=
+
.
Hint : Sometimes the Chain Rule will need to be done multiple times before we finish taking the
derivative.
Step 1
This problem will require multiple uses of the Chain Rule and so we’ll step though the derivative
process to make each use clear.
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Calculus I
Here is the first step of the derivative and we’ll need to use the Chain Rule in this step.
( )
( )
(
)
( )
( )
1
2
2
2
3
3
12
sin
3
12
sin
3
d
f
x
x
x
x
x
dx
−
′
= −
+
+
Step 2
As we can see the derivative from the previous step will also require the Chain Rule on each of
the terms.
The derivative from this step is,
( )
( )
(
)
( ) ( )
( )
( )
(
)
2
2
2
3
3
1
12
sin
3
12
12
2 sin 3
sin 3
3
d
f
x
x
x
x
x
x
dx
−
−
′
= −
+
+
Step 3
The second term will again use the Chain Rule as we can see.
The derivative is then,
( )
( )
(
)
( )
( ) ( )
2
2
2
3
3
12
sin
3
4 12
6 sin 3
cos 3
f
x
x
x
x
x
x
−
−
′
= −
+
+
28. Find the tangent line to
( )
2
4 2
6
x
f x
x
−
=
− e
at
2
x
=
.
Solution
Step 1
We know that the derivative of the function will give us the slope of the tangent line so we’ll
need the derivative of the function. Differentiating each term will require the Chain Rule as well.
( ) ( )
( )
( ) ( )
( ) ( )
1
2
2
1
1
2
2
2
2
2
4 2
6
1
4
4
2
2
6
1
4 2
6
6
2
2
x
x
x
x
f x
x
f
x
x
x
x
−
−
−
−
−
−
=
−
′
=
−
− =
+
=
+
e
e
e
e
Step 2
Now all we need to do is evaluate the function and the derivative at the point in question.
( ) ( )
( )
0
0
4
2
4 2
6
2
2
6
8
2
f
f ′
=
−
=
= +
=
e
e
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Calculus I
Step 3
Now all that we need to do is write down the equation of the tangent line.
( )
( )(
)
(
)
2
2
2
2 8
2
8
14
y
f
f
x
x
y
x
′
=
+
−
= +
−
→
=
−
29. Determine where
( )
(
)
3
4
2
8
V z
z
z
=
−
is increasing and decreasing.
Solution
Step 1
We’ll first need the derivative because we know that the derivative will give us the rate of change
of the function. Here is the derivative.
( )
(
)
( )(
) ( )
(
) (
)
(
) (
)
3
2
3
4
2
2
3
3
4
2
8
3 2
8
2
2
2
8
2 2
8
3
2
2
8
7
16
V z
z
z
z
z
z
z
z
z
z
z
z
′
=
−
+
−
=
−
− +
=
−
−
Note that we factored the derivative to help with the next step. In general we don’t need to do
this.
Step 2
Next, we need to know where the function is not changing and so all we need to do is set the
derivative equal to zero and solve. In this case it’s pretty easy to spot where the derivative will be
zero.
(
) (
)
2
3
16
2
2
8
7
16
0
0,
4,
2.2857
7
z
z
z
z
z
z
−
−
=
⇒
=
=
=
=
Step 3
To get the answer to this problem all we need to know is where the derivative is positive (and
hence the function is increasing) or negative (and hence the function is decreasing). Because the
derivative is continuous we know that the only place it can change sign is where the derivative is
zero. So, as we did in this section a quick number line will give us the sign of the derivative for
the various intervals.
Here is the number line for this problem.
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Calculus I
From this we get the following increasing/decreasing information.
16
Increasing :
0,
4, 4
7
16
Decreasing : 0
7
x
x
x
x
− ∞ < <
< <
< < ∞
< <
30. The position of an object is given by
( )
( )
sin 3
2
4
s t
t
t
=
− +
. Determine where in the
interval
[ ]
0, 3
the object is moving to the right and moving to the left.
Solution
Step 1
We’ll first need the derivative because we know that the derivative will give us the velocity of the
object. Here is the derivative.
( )
( )
3cos 3
2
s t
t
′
=
−
Step 2
Next, we need to know where the object stops moving and so all we need to do is set the
derivative equal to zero and solve.
( )
( )
2
3cos 3
2
0
cos 3
3
t
t
− =
⇒
=
A quick calculator computation tells us that,
1
2
3
cos
0.8411
3
t
−
=
=
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Calculus I
Recalling our work in the Review chapter and a quick check on a unit circle we can see that either
3
0.8411
t
= −
or
3
2
0.8411 5.4421
t
π
=
−
=
can be used for the second angle. Note that either
will work, but we’ll use the second simply because it is the positive angle.
Putting all of this together and dividing by 3 we can see that the derivative will be zero at,
3
0.8411 2
and
3
5.4421 2
0, 1, 2, 3,
2
2
0.2804
and
1.8140
0, 1, 2, 3,
3
3
t
n
t
n
n
n
n
t
t
n
π
π
π
π
=
+
=
+
= ± ± ±
=
+
=
+
= ± ± ±
Finally, all we need to so is plug in some n’s to determine which solutions fall in the interval we
are working on,
[ ]
0, 6
.
0 :
0.2804
1.8140
1:
2.3748
3.9084
n
t
t
n
t
t
=
=
=
=
=
=
So, in the interval
[ ]
0, 6
, the object stops moving at the following three points.
0.2804, 1.8140, 2.3748
t
=
Step 3
To get the answer to this problem all we need to know is where the derivative is positive (and
hence the object is moving to the right) or negative (and hence the object is moving to the left).
Because the derivative is continuous we know that the only place it can change sign is where the
derivative is zero. So, as we did in this section a quick number line will give us the sign of the
derivative for the various intervals.
Here is the number line for this problem.
From this we get the following moving right/moving left information.
Moving Right : 0
0.2804, 1.8140
2.3748
Moving Left : 0.2804
1.8140, 2.3748
3
t
x
x
x
≤ <
< <
< <
< ≤
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Calculus I
Note that because we’ve only looked at what is happening in the interval
[ ]
0, 3
we can’t say
anything about the moving right/moving left behavior of the object outside of this interval.
31. Determine where
( )
2 5
t
A t
t
−
= e
is increasing and decreasing.
Solution
Step 1
We’ll first need the derivative because we know that the derivative will give us the rate of change
of the function. Here is the derivative.
( )
(
)
5
2 5
5
2
2
t
t
t
A t
t
t
t
t
−
−
−
′
=
−
=
−
e
e
e
Note that we factored the derivative to help with the next step. In general we don’t need to do
this.
Step 2
Next, we need to know where the function is not changing and so all we need to do is set the
derivative equal to zero and solve. In this case it’s pretty easy to spot where the derivative will be
zero.
(
)
5
2
0
0,
2
t
t
t
t
t
−
− = ⇒
=
=
e
Step 3
To get the answer to this problem all we need to know is where the derivative is positive (and
hence the function is increasing) or negative (and hence the function is decreasing). Because the
derivative is continuous we know that the only place it can change sign is where the derivative is
zero. So, as we did in this section a quick number line will give us the sign of the derivative for
the various intervals.
Here is the number line for this problem.
From this we get the following increasing/decreasing information.
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Calculus I
Increasing :0
2
Decreasing :
0, 2
t
t
t
< <
− ∞ < <
< < ∞
32. Determine where in the interval
[
]
1, 20
−
the function
( )
(
)
4
3
ln
20
100
f x
x
x
=
+
+
is
increasing and decreasing.
Solution
Step 1
We’ll first need the derivative because we know that the derivative will give us the rate of change
of the function. Here is the derivative.
( )
(
)
2
3
2
4
3
4
3
4
15
4
60
20
100
20
100
x
x
x
x
f x
x
x
x
x
+
+
=
=
+
+
+
+
Note that we factored the numerator to help with the next step. In general we don’t need to do
this.
Step 2
Next, we need to know where the function is not changing and so all we need to do is set the
derivative equal to zero and solve.
(
)
(
)
2
2
4
3
4
15
0
4
15
0
0,
15
20
100
x
x
x
x
x
x
x
x
+
=
→
+
=
⇒
=
= −
+
+
Note, that in general, we also need to be concerned with where the derivative is not defined as
well. Functions can (and often do) change sign where they are not defined. In this case however
we’ve restricted the interval down to a range where the function exists and is continuous on the
given interval and so this is something we need to worry about for this problem.
In the next Chapter we will start also looking at what happens if the derivative is also not defined
at particular points.
Note as well that we really should at this point step back and recall that we are working on the
interval
[
]
1, 20
−
. We are only interested in what is happening on this interval and so we should
make sure that the points found above are inside the interval.
In this case only
0
x
=
is in the interval and so we’ll need to exclude
15
x
= −
from our work for
the rest of this problem.
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Calculus I
Step 3
To get the answer to this problem all we need to know is where the derivative is positive (and
hence the function is increasing) or negative (and hence the function is decreasing). Because the
derivative is continuous we know that the only place it can change sign is where the derivative is
zero. So, as we did in this section a quick number line will give us the sign of the derivative for
the various intervals.
Here is the number line for this problem.
So, we can see that, in this case function is increasing everywhere in the interval
[
]
1, 20
−
except
0
x
=
. Recall that at this point the derivative was zero and hence the function is not changing
(and therefore can’t be increasing).
So, the formal answer for this problem is,
Increasing : 1
0, 0
20
x
x
− ≤ <
< ≤
Note that because we’ve only looked at what is happening in the interval
[
]
1, 20
−
we can’t say
anything about the increasing/decreasing nature of the function outside of this interval.
Implicit Differentiation
1. For
3
1
x
y
=
do each of the following.
(a) Find
y′
by solving the equation for y and differentiating directly.
(b) Find
y′
by implicit differentiation.
(c) Check that the derivatives in (a) and (b) are the same.
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Calculus I
(a) Find
y′
by solving the equation for y and differentiating directly.
Step 1
First we just need to solve the equation for y.
1
3
3
y
x
y
x
=
⇒
=
Step 2
Now differentiate with respect to x.
2
1
3
3
y
x
−
′ =
(b) Find
y′
by implicit differentiation.
Hint : Don’t forget that y is really
( )
y x
and so we’ll need to use the Chain Rule when taking the
derivative of terms involving y! Also, don’t forget that because y is really
( )
y x
we may well
have a Product and/or a Quotient Rule buried in the problem.
Step 1
First, we just need to take the derivative of everything with respect to x and we’ll need to recall
that y is really
( )
y x
and so we’ll need to use the Chain Rule when taking the derivative of terms
involving y.
Also, prior to taking the derivative a little rewrite might make this a little easier.
3
1
x y
−
=
Now take the derivative and don’t forget that we actually have a product of functions of x here
and so we’ll need to use the Product Rule when differentiating the left side.
3
4
3
0
y
x y y
−
−
′
−
=
Step 2
Finally all we need to do is solve this for
y′
.
3
4
3
3
y
y
y
xy
x
−
−
′ =
=
(c) Check that the derivatives in (a) and (b) are the same.
Hint : To show they are the same all we need is to plug the formula for y (which we already
have….) into the derivative we found in (b) and, potentially with a little work, show that we get
the same derivative as we got in (a).
From (a) we have a formula for y written explicitly as a function of x so plug that into the
derivative we found in (b) and, with a little simplification/work, show that we get the same
derivative as we got in (a).
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Calculus I
1
2
3
1
3
3
3
3
y
x
y
x
x
x
−
′ =
=
=
So, we got the same derivative as we should.
2. For
2
3
4
x
y
+
=
do each of the following.
(d) Find
y′
by solving the equation for y and differentiating directly.
(e) Find
y′
by implicit differentiation.
(f) Check that the derivatives in (a) and (b) are the same.
(a) Find
y′
by solving the equation for y and differentiating directly.
Step 1
First we just need to solve the equation for y.
(
)
1
3
2
2
3
4
4
y
x
y
x
= −
⇒
=
−
Step 2
Now differentiate with respect to x.
(
)
2
2
2
3
3
4
y
x
x
−
′ = −
−
(b) Find
y′
by implicit differentiation.
Hint : Don’t forget that y is really
( )
y x
and so we’ll need to use the Chain Rule when taking the
derivative of terms involving y!
Step 1
First, we just need to take the derivative of everything with respect to x and we’ll need to recall
that y is really
( )
y x
and so we’ll need to use the Chain Rule when taking the derivative of terms
involving y.
Taking the derivative gives,
2
2
3
0
x
y y′
+
=
Step 2
Finally all we need to do is solve this for
y′
.
2
2
3
x
y
y
′ = −
(c) Check that the derivatives in (a) and (b) are the same.
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Calculus I
Hint : To show they are the same all we need is to plug the formula for y (which we already
have….) into the derivative we found in (b) and, potentially with a little work, show that we get
the same derivative as we got in (a).
From (a) we have a formula for y written explicitly as a function of x so plug that into the
derivative we found in (b) and, with a little simplification/work, show that we get the same
derivative as we got in (a).
(
)
(
)
2
2
2
3
3
2
2
2
3
2
2
4
3
3 4
x
x
y
x
x
y
x
−
′ = −
= −
= −
−
−
So, we got the same derivative as we should.
3. For
2
2
2
x
y
+
=
do each of the following.
(g) Find
y′
by solving the equation for y and differentiating directly.
(h) Find
y′
by implicit differentiation.
(i) Check that the derivatives in (a) and (b) are the same.
(a) Find
y′
by solving the equation for y and differentiating directly.
Step 1
First we just need to solve the equation for y.
(
)
1
2
2
2
2
2
2
y
x
y
x
= −
⇒
= ± −
Note that because we have no restriction on y (i.e. we don’t know if y is positive or negative) we
really do need to have the “
±
” there and that does lead to issues when taking the derivative.
Hint : Two formulas for y and so two derivatives.
Step 2
Now, because there are two formulas for y we will also have two formulas for the derivative, one
for each formula for y.
The derivatives are then,
(
)
(
)
(
)
(
)
(
)
(
)
1
1
2
2
2
2
1
1
2
2
2
2
2
2
0
2
2
0
y
x
y
x
x
y
y
x
y
x
x
y
−
−
′
=
−
⇒
= −
−
>
′
= − −
⇒
=
−
<
As noted above the first derivative will hold for
0
y
>
while the second will hold for
0
y
<
and
we can use either for
0
y
=
as the plus/minus won’t affect that case.
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Calculus I
(b) Find
y′
by implicit differentiation.
Hint : Don’t forget that y is really
( )
y x
and so we’ll need to use the Chain Rule when taking the
derivative of terms involving y!
Step 1
First, we just need to take the derivative of everything with respect to x and we’ll need to recall
that y is really
( )
y x
and so we’ll need to use the Chain Rule when taking the derivative of terms
involving y.
Taking the derivative gives,
2
2
0
x
y y′
+
=
Step 2
Finally all we need to do is solve this for
y′
.
x
y
y
′ = −
(c) Check that the derivatives in (a) and (b) are the same.
Hint : To show they are the same all we need is to plug the formula for y (which we already
have….) into the derivative we found in (b) and, potentially with a little work, show that we get
the same derivative as we got in (a). Again, two formulas for y so two derivatives…
From (a) we have a formula for y written explicitly as a function of x so plug that into the
derivative we found in (b) and, with a little simplification/work, show that we get the same
derivative as we got in (a).
Also, because we have two formulas for y we will have two formulas for the derivative.
First, if
0
y
>
we will have,
(
)
(
)
(
)
1
1
2
2
2
2
1
2
2
2
2
2
x
x
y
x
y
x
x
y
x
−
′
=
−
⇒
= − = −
= −
−
−
Next, if
0
y
<
we will have,
(
)
(
)
(
)
1
1
2
2
2
2
1
2
2
2
2
2
x
x
y
x
y
x
x
y
x
−
′
= − −
⇒
= − = −
=
−
− −
So, in both cases, we got the same derivative as we should.
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Calculus I
4. Find
y′
by implicit differentiation for
3
2
6
2
4
y
x
y
x
+
− =
.
Hint : Don’t forget that y is really
( )
y x
and so we’ll need to use the Chain Rule when taking the
derivative of terms involving y!
Step 1
First, we just need to take the derivative of everything with respect to x and we’ll need to recall
that y is really
( )
y x
and so we’ll need to use the Chain Rule when taking the derivative of terms
involving y.
Differentiating with respect to x gives,
2
5
6
8
6
y y
x
y
x
′
′
+
− =
Step 2
Finally all we need to do is solve this for
y′
.
(
)
5
2
5
2
6
8
6
1
6
8
6
1
x
x
y
y
x
x
y
y
−
′
′
−
=
−
⇒
=
−
5. Find
y′
by implicit differentiation for
( )
2
4
7
sin 3
12
y
x
y
+
=
−
.
Hint : Don’t forget that y is really
( )
y x
and so we’ll need to use the Chain Rule when taking the
derivative of terms involving y!
Step 1
First, we just need to take the derivative of everything with respect to x and we’ll need to recall
that y is really
( )
y x
and so we’ll need to use the Chain Rule when taking the derivative of terms
involving y.
Differentiating with respect to x gives,
( )
3
14
3cos 3
4
y y
x
y y
′
′
+
= −
Step 2
Finally all we need to do is solve this for
y′
.
(
)
( )
( )
3
3
3cos 3
14
4
3cos 3
14
4
x
y
y
y
x
y
y
y
−
′
′
+
= −
⇒
=
+
6. Find
y′
by implicit differentiation for
( )
sin
x
y
x
−
=
e
.
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Calculus I
Hint : Don’t forget that y is really
( )
y x
and so we’ll need to use the Chain Rule when taking the
derivative of terms involving y!
Step 1
First, we just need to take the derivative of everything with respect to x and we’ll need to recall
that y is really
( )
y x
and so we’ll need to use the Chain Rule when taking the derivative of terms
involving y.
Differentiating with respect to x gives,
( )
cos
1
x
y y′
−
=
e
Don’t forget the
y′
on the cosine after differentiating. Again, y is really
( )
y x
and so when
differentiating
( )
sin y
we really differentiating
( )
sin y x
and so we are differentiating using
the Chain Rule!
Step 2
Finally all we need to do is solve this for
y′
.
( )
(
)
( )
1
1 sec
cos
x
x
y
y
y
−
′ =
=
−
−
e
e
7. Find
y′
by implicit differentiation for
2
7
5
3
4
2
4
x y
x
x
y
−
=
+
.
Hint : Don’t forget that y is really
( )
y x
and so we’ll need to use the Chain Rule when taking the
derivative of terms involving y! Also, don’t forget that because y is really
( )
y x
we may well
have a Product and/or a Quotient Rule buried in the problem.
Step 1
First, we just need to take the derivative of everything with respect to x and we’ll need to recall
that y is really
( )
y x
and so we’ll need to use the Chain Rule when taking the derivative of terms
involving y. This also means that the first term on the left side is really a product of functions of
x and hence we will need to use the Product Rule when differentiating that term.
Differentiating with respect to x gives,
7
2
6
4
2
8
28
2
5
12
xy
x y y
x
y y
′
′
+
− =
+
Step 2
Finally all we need to do is solve this for
y′
.
(
)
7
4
7
4
2
2
6
2
2
6
8
5
2
8
5
2
12
28
12
28
xy
x
xy
x
y
x y
y
y
y
x y
−
−
′
′
−
− =
−
⇒
=
−
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Calculus I
8. Find
y′
by implicit differentiation for
(
)
2
2
cos
2
1
y
x
y
x
+
+
=
e
.
Hint : Don’t forget that y is really
( )
y x
and so we’ll need to use the Chain Rule when taking the
derivative of terms involving y! Also, don’t forget that because y is really
( )
y x
we may well
have a Product and/or a Quotient Rule buried in the problem.
Step 1
First, we just need to take the derivative of everything with respect to x and we’ll need to recall
that y is really
( )
y x
and so we’ll need to use the Chain Rule when taking the derivative of terms
involving y. This also means that the second term on the left side is really a product of functions
of x and hence we will need to use the Product Rule when differentiating that term.
Differentiating with respect to x gives,
(
)
(
)
2
2
2
2
2
sin
2
2
0
y
y
x
y
x
y
y y x
′
′
−
+
+
+
+
=
e
e
Step 2
Finally all we need to do is solve this for
y′
(with some potentially messy algebra).
(
)
(
)
(
)
(
)
(
)
(
)
(
)
2
2
2
2
2
2
2
2
2
2
2
2
2 sin
2
2
sin
2
2
0
2
2 sin
2
0 2 sin
2
2 sin
2
2
2 sin
2
y
y
y
y
y
y
x
x
y
y
x
y
y y x
yx
x
y
y
x
x
y
x
x
y
y
yx
x
y
′
′
−
+
−
+
+
+
=
′
−
+
= +
+
−
+
−
′ =
−
+
e
e
e
e
e
e
9. Find
y′
by implicit differentiation for
(
)
2
4
2
tan
3
x y
x
y
=
+
.
Hint : Don’t forget that y is really
( )
y x
and so we’ll need to use the Chain Rule when taking the
derivative of terms involving y! Also, don’t forget that because y is really
( )
y x
we may well
have a Product and/or a Quotient Rule buried in the problem.
Step 1
First, we just need to take the derivative of everything with respect to x and we’ll need to recall
that y is really
( )
y x
and so we’ll need to use the Chain Rule when taking the derivative of terms
involving y. This also means that the when doing Chain Rule on the first tangent on the left side
we will need to do Product Rule when differentiating the “inside term”.
Differentiating with respect to x gives,
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Calculus I
(
) (
)
4
2
3
2
2
4
2
4
sec
3 2
x y
x y y
x y
y y
′
′
+
= +
Step 2
Finally all we need to do is solve this for
y′
(with some potentially messy algebra).
(
)
(
)
(
)
(
)
(
)
(
)
(
)
4
2
2
4
2
3
2
2
4
2
3
2
2
4
4
2
2
4
4
2
2
4
2
3
2
2
4
2
sec
4
sec
3 2
4
sec
2
3 2
sec
3 2
sec
4
sec
2
x y
x y
x y y
x y
y y
x y
x y
y y
x y
x y
x y
x y
y
x y
x y
y
′
′
+
= +
′
−
= −
−
′ =
−
10. Find the equation of then tangent line to
4
2
3
x
y
+
=
at
(
)
1,
2
−
.
Hint : We know how to compute the slope of tangent lines and with implicit differentiation that
shouldn’t be too hard at this point.
Step 1
The first thing to do is use implicit differentiation to find
y′
for this function.
3
3
2
4
2
0
x
x
y y
y
y
′
′
+
=
⇒
= −
Step 2
Evaluating the derivative at the point in question to get the slope of the tangent line gives,
1,
2
2
2
2
x
y
m
y
=
=−
′
=
= −
=
−
Step 3
Now, we just need to write down the equation of the tangent line.
( )
(
)
(
)
(
)
2
2
1
2
1
2
2
2
y
x
y
x
x
− −
=
−
⇒
=
− −
=
−
11. Find the equation of then tangent line to
2 2
2
3
x
y
y
x
=
+
e
at
( )
0, 3
.
Hint : We know how to compute the slope of tangent lines and with implicit differentiation that
shouldn’t be too hard at this point.
Step 1
The first thing to do is use implicit differentiation to find
y′
for this function.
2 2
2
2 2
2
2
2
2
2
3
2
2
3
x
x
x
x
x
y
yy
y
y
x
y
y
−
′
′
′
+
=
+
⇒
=
−
e
e
e
e
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Calculus I
Step 2
Evaluating the derivative at the point in question to get the slope of the tangent line gives,
0,
3
18
6
3
x
y
m
y
=
=
−
′
=
=
= −
Step 3
Now, we just need to write down the equation of the tangent line.
(
)
3
6
0
6
3
y
x
y
x
− = −
−
⇒
= − +
12. Assume that
( )
x
x t
=
,
( )
y
y t
=
and
( )
z
z t
=
and differentiate
2
3
4
1
x
y
z
−
+
=
with
respect to t.
Hint : This is just implicit differentiation like we’ve been doing to this point. The only difference
is that now all the functions are functions of some fourth variable, t. Outside of that there is
nothing different between this and the previous problems.
Solution
Differentiating with respect to t gives,
2
3
2
3
4
0
x x
y y
z z
′
′
′
−
+
=
Note that because we were not asked to give the formula for a specific derivative we don’t need
to go any farther. We could however, if asked, solved this for any of the three derivatives that are
present.
13. Assume that
( )
x
x t
=
,
( )
y
y t
=
and
( )
z
z t
=
and differentiate
( )
(
)
2
3
cos
sin
4
x
y
y
z
=
+
with respect to t.
Hint : This is just implicit differentiation like we’ve been doing to this point. The only difference
is that now all the functions are functions of some fourth variable, t. Outside of that there is
nothing different between this and the previous problems.
Solution
Differentiating with respect to t gives,
( )
( )
(
) (
)
2
2
3
2
cos
sin
3
4
cos
4
x x
y
x
y y
y y
z
y
z
′
′
′
′
−
=
+
+
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Calculus I
Note that because we were not asked to give the formula for a specific derivative we don’t need
to go any farther. We could however, if asked, solved this for any of the three derivatives that are
present.
Related Rates
1. In the following assume that x and y are both functions of t. Given
2
x
= −
,
1
y
=
and
4
x′
= −
determine
y′
for the following equation.
2
2
3 4 4
6
2
y
y
x
x
−
+
= − e
Hint : This is just like the problems worked in the section notes. The only difference is that
you’ve been given the equation and all the needed information and so you won’t have to worry
about finding that.
Step 1
The first thing that we need to do here is use implicit differentiation to differentiate the equation
with respect to t.
2
4 4
3 4 4
12
2
3
4
y
y
y y
x x
x x
x
y
−
−
′
′
′
′
+
= −
+
e
e
Step 2
All we need to do now is plug in the given information and solve for
y′
.
8
11
12
16
48 32
y
y
y
′
′
′
+
=
−
⇒
=
2. In the following assume that x, y and z are all functions of t. Given
4
x
=
,
2
y
= −
,
1
z
=
,
9
x′
=
and
3
y′
= −
determine
z′
for the following equation.
(
)
3
2
2
2
1
5
3
x
y
z
y z
x
−
+
=
+
−
Hint : This is just like the problems worked in the section notes. The only difference is that
you’ve been given the equation and all the needed information and so you won’t have to worry
about finding that.
Step 1
The first thing that we need to do here is use implicit differentiation to differentiate the equation
with respect to t.
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Calculus I
(
)
2
2
2
1
15
2
2
2
x
y
x y
z z
y y z
y z z
x x
′
′
′
′
′
′
−
−
+
=
+
+
Step 2
All we need to do now is plug in the given information and solve for
z′
.
45
7
27 12 15
12 8
72
z
z
z
′
′
′
+
+
=
+
+
⇒
=
3. For a certain rectangle the length of one side is always three times the length of the other side.
(a) If the shorter side is decreasing at a rate of 2 inches/minute at what rate is the
longer side decreasing?
(b) At what rate is the enclosed area decreasing when the shorter side is 6 inches long
and is decreasing at a rate of 2 inches/minute?
Hint : The equation needed here is a really simple equation. In fact, so simple it might be easy to
miss…
(a) If the shorter side is decreasing at a rate of 2 inches/minute at what rate is the longer side
decreasing?
Step 1
Let’s call the shorter side x and the longer side y. We know that
2
x′
= −
and want to find
y′
.
Now all we need is an equation that relates these two quantities and from the problem statement
we know the longer side is three times shorter side and so the equation is,
3
y
x
=
Step 2
Next step is to simply differentiate the equation with respect to t.
3
y
x
′
′
=
Step 3
Finally, plug in the known quantity and solve for what we want :
6
y′
= −
Hint : Once we have the equation for the area we can either simplify the equation as we did in this
section or we can use the result from the previous step and the equation directly.
(b) At what rate is the enclosed area decreasing when the shorter side is 6 inches long and is
decreasing at a rate of 2 inches/minute?
Step 1
Again we’ll call the shorter side x and the longer side y as with the last part. We know that
6
x
=
,
2
x′
= −
and want to find
A′
.
The equation we’ll need is just the area formula for a rectangle :
A
xy
=
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Calculus I
At this point we can either leave the equation as is and differentiate it or we can plug in
3
y
x
=
to simplify the equation down to a single variable then differentiate. Doing this gives,
( )
2
3
A x
x
=
Step 2
Now we need to differentiate with respect to t.
If we use the equation in terms of only x, which is probably the easiest to use we get,
6
A
x x
′
′
=
If we use the equation in terms of both x and y we get,
A
x y
x y
′
′
′
=
+
Step 3
Now all we need to do is plug in the known quantities and solve for
A′
.
Using the equation in terms of only x is the “easiest” because we already have all the known
quantities from the problem statement itself. Doing this gives,
( )( )
6 6
2
72
A′
=
− = −
Now let’s use the equation in terms of x and y. We know that
6
x
=
and
2
x′
= −
from the
problem statement. From part (a) we have
6
y′
= −
and we also know that
( )
3 6
18
y
=
=
.
Using these gives,
( )( ) ( )( )
6
6
2 18
72
A′
=
− + −
= −
So, as we can see both gives the same result but the second method is slightly more work,
although not much more.
4. A thin sheet of ice is in the form of a circle. If the ice is melting in such a way that the area of
the sheet is decreasing at a rate of 0.5 m
2
/sec at what rate is the radius decreasing when the area
of the sheet is 12 m
2
?
Step 1
We’ll call the area of the sheet A and the radius r and we know that the area of a circle is given
by,
2
A
r
π
=
We know that
0.5
A′
= −
and want to determine
r′
when
12
A
=
.
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Calculus I
Step 2
Next step is to simply differentiate the equation with respect to t.
2
A
r r
π
′
′
=
Step 3
Now, to finish this problem off we’ll first need to go back to the equation of the area and use the
fact that we know the area at the point we are interested in and determine the radius at that time.
2
12
12
1.9544
r
r
π
π
=
⇒
=
=
The rate of change of the radius is then,
(
)
0.5
2
1.9544
0.040717
r
r
π
′
′
−
=
⇒
= −
5. A person is standing 350 feet away from a model rocket that is fired straight up into the air at a
rate of 15 ft/sec. At what rate is the distance between the person and the rocket increasing (a) 20
seconds after liftoff? (b) 1 minute after liftoff?
Step 1
Here is a sketch for this situation that will work for both parts so we’ll put it here.
Step 2
In both parts we know that
15
y′
=
and want to determine
z′
for each given time. Using the
Pythagorean Theorem we get the following equation to relate y and z.
2
2
2
2
350
122500
z
y
y
=
+
=
+
Step 3
Finally, let’s differentiate this with respect to t and we can even solve it for
z′
so the actual
solution will be quick and simple to find.
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Calculus I
2
2
y y
z z
y y
z
z
′
′
′
′
=
⇒
=
We have now reached a point where the process will differ for each part.
(a) At what rate is the distance between the person and the rocket increasing 20 seconds after
liftoff?
To finish off this problem all we need to do is determine y (from the speed of the rocket and given
time) and z (reusing the Pythagorean Theorem).
( )( )
2
2
15 20
300
300
350
212500
50 85
460.9772
y
z
=
=
=
+
=
=
=
The rate of change of the distance between the two is then,
( )( )
300 15
9.76187
460.9772
z′
=
=
(b) At what rate is the distance between the person and the rocket increasing 1 minute after
liftoff?
This part is nearly identical to the first part with the exception that the time is now 60 seconds
(and note that we MUST be in seconds because the speeds are in time of seconds).
Here is the work for this problem.
( )( )
( )( )
2
2
15 60
900
900
350
932500
50 373
965.6604
900 15
13.98007
965.6604
y
z
z
=
=
=
+
=
=
=
′ =
=
6. A plane is 750 meters in the air flying parallel to the ground at a speed of 100 m/s and is
initially 2.5 kilometers away from a radar station. At what rate is the distance between the plane
and the radar station changing (a) initially and (b) 30 seconds after it passes over the radar
station? See the (probably bad) sketch below to help visualize the problem.
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Calculus I
(a) At what rate is the distance between the plane and the radar station changing initially?
Step 1
For this part we know that
100
x′
= −
when
2500
x
=
. In this case note that
x′
must be
negative because x will be decreasing in this part. Also note that we converted x to meters since
all the other quantities are in meters.
Here is a sketch for this part.
Step 2
We want to determine
z′
in this part so using the Pythagorean Theorem we get the following
equation to relate x and z.
2
2
2
2
750
562500
z
x
x
=
+
=
+
Step 3
Finally, let’s differentiate this with respect to t and we can even solve it for
z′
so the actual
solution will be quick and simple to find.
2
2
x x
z z
x x
z
z
′
′
′
′
=
⇒
=
Step 4
To finish off this problem all we need to do is determine z (reusing the Pythagorean Theorem)
and then plug into the equation from Step 3 above.
2
2
2500
750
6812500
250 109
2610.0766
z
=
+
=
=
=
The rate of change of the distance between the two for this part is,
(
)(
)
2500
100
95.7826
2610.0766
z
−
′ =
= −
(b) At what rate is the distance between the plane and the radar station changing 30 seconds after
it passes over the radar station?
Step 1
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Calculus I
For this part we know that
100
x′
=
and it will be positive in this case because x will now be
increasing as we can see in the sketch below.
Step 2
As with the previous part we want to determine
z′
and equation we’ll need is identical to the
previous part so we’ll just rewrite both it and its derivative here.
2
2
2
2
750
562500
2
2
z
x
x
x x
z z
x x
z
z
=
+
=
+
′
′
′
′
=
⇒
=
Step 3
To finish off this problem all we need to do is determine both x and z. For x we know the speed
of the plane and the fact that it has flown for 30 seconds after passing over the radar station. So x
is,
( )( )
100 30
3000
x
=
=
For z we just need to reuse the Pythagorean Theorem.
2
2
3000
750
9562500
750 17
3092.3292
z
=
+
=
=
=
The rate of change of the distance between the two for this part is then,
(
)( )
3000 100
97.0143
3092.3292
z′
=
=
7. Two people are at an elevator. At the same time one person starts to walk away from the
elevator at a rate of 2 ft/sec and the other person starts going up in the elevator at a rate of 7
ft/sec. What rate is the distance between the two people changing 15 seconds later?
Step 1
Here is a sketch for this part.
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Calculus I
We want to determine
z′
after 15 seconds given that
2
x′
=
,
7
y′
=
and assuming that they start
at the same point.
Step 2
Hopefully it’s clear that we’ll need the Pythagorean Theorem to solve this problem so here is that.
2
2
2
z
x
y
=
+
Step 3
Finally, let’s differentiate this with respect to t and we can even solve it for
z′
so the actual
solution will be quick and simple to find.
2
2
2
x x
y y
z z
x x
y y
z
z
′
′
+
′
′
′
′
=
+
⇒
=
Step 4
To finish off this problem all we need to do is determine all three lengths of the triangle in the
sketch above. We can find x and y using their speeds and time while we can find z by reusing the
Pythagorean Theorem.
( )( )
( )( )
2
2
2 15
30
7 15
105
30
105
11925
15 53
109.2016
x
y
z
=
=
=
=
=
+
=
=
=
The rate of change of the distance between the two people is then,
( )( ) ( )( )
30 2
105 7
7.2801
109.2016
z
+
′ =
=
8. Two people on bikes are at the same place. One of the bikers starts riding directly north at a
rate of 8 m/sec. Five seconds after the first biker started riding north the second starts to ride
directly east at a rate of 5 m/sec. At what rate is the distance between the two riders increasing 20
seconds after the second person started riding?
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Calculus I
Step 1
Here is a sketch of this situation.
We want to determine
z′
after 20 seconds after the second biker starts riding east given that
5
x′
=
,
8
y′
=
and assuming that they start at the same point.
Step 2
Hopefully it’s clear that we’ll need the Pythagorean Theorem to solve this problem so here is that.
2
2
2
z
x
y
=
+
Step 3
Finally, let’s differentiate this with respect to t and we can even solve it for
z′
so the actual
solution will be quick and simple to find.
2
2
2
x x
y y
z z
x x
y y
z
z
′
′
+
′
′
′
′
=
+
⇒
=
Step 4
To finish off this problem all we need to do is determine all three lengths of the triangle in the
sketch above. We can find x and y using their speeds and time while we can find z by reusing the
Pythagorean Theorem. Note that the biker riding east will be riding for 20 seconds and the biker
riding north will be riding for 25 seconds (this biker started 5 seconds earlier…).
( )( )
( )( )
2
2
5 20
100
8 25
200
100
200
50000
100 5
223.6068
x
y
z
=
=
=
=
=
+
=
=
=
The rate of change of the distance between the two people is then,
( )( ) (
)( )
100 5
200 8
9.3915
223.6068
z
+
′ =
=
9. A light is mounted on a wall 5 meters above the ground. A 2 meter tall person is initially 10
meters from the wall and is moving towards the wall at a rate of 0.5 m/sec. After 4 seconds of
moving is the tip of the shadow moving (a) towards or away from the person and (b) towards or
away from the wall?
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Calculus I
Step 1
Here is a sketch for this situation that will work for both parts so we’ll put it here. Also note that
we know that
0.5
p
x′
= −
for both parts.
(a) After 4 seconds of moving is the tip of the shadow towards or away from the person?
Step 2
In this case we want to determine
s
x′
when
( )
10 4 0.5
8
p
x
=
−
=
(although it will turn out that
we simply don’t need this piece of information for this problem….).
We can use the idea of similar triangles to get the following equation.
2
5
s
s
p
s
x
x
x
x
x
=
=
+
If we solve this for
s
x
we arrive at,
(
)
2
5
2
2
2
5
5
3
p
p
s
s
p
s
s
s
x
x
x
x
x
x
x
x
+
=
+
=
⇒
=
This equation will work perfectly for us.
Step 3
Differentiation with respect to t will give us,
2
3
p
s
x
x
′
′
=
Step 4
Finishing off this problem is very simple as all we need to do is plug in the known speed.
(
)
2
1
3
3
0.5
s
x′
= −
= −
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Calculus I
Because this rate is negative we can see that the tip of the shadow is moving towards the person
at a rate of
1
3
m/s.
(b) After 4 seconds of moving is the tip of the shadow towards or away from the wall?
Step 2
In this case we want to determine
x′
and the equation is really simple. All we need is,
p
s
x
x
x
=
+
Step 3
Differentiation with respect to t will give us,
p
s
x
x
x
′
′
′
=
+
Step 4
Finishing off this problem is very simple as all we need to do is plug in the known speeds and
note that we will need to result from the first part here. So we have
1
2
p
x′
= −
from the problem
statement and
1
3
s
x′
= −
from the previous part.
( )
5
1
1
2
3
6
x′
= − + − = −
Because this rate is negative we can see that the tip of the shadow is moving towards the wall at a
rate of
5
6
m/s.
10. A tank of water in the shape of a cone is being filled with water at a rate of 12 m
3
/sec. The
base radius of the tank is 26 meters and the height of the tank is 8 meters. At what rate is the
depth of the water in the tank changing with the radius of the top of the water is 10 meters?
Step 1
Here is a sketch of the cross section of the tank and it is not even remotely to scale as I found it
easier to reuse an old image that I had lying around. I can be a little lazy sometimes….
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Calculus I
We want to determine
h′
when
10
r
=
and we know that
12
V ′
=
.
Step 2
We’ll need the equation for the volume of a cone.
2
1
3
V
r h
π
=
This is a problem however as it has both r and h in it and it would be best to have only h since we
need
h′
. We can use similar triangles to fix this up. Based on similar triangles we get the
following equation which can be solved for r.
13
4
26
8
r
r
h
h
=
⇒
=
Plugging this into the volume equation gives,
3
169
48
V
h
π
=
Step 3
Next, let’s differentiate this with respect to t.
2
169
16
V
h h
π
′
′
=
Step 4
To finish off this problem all we need to do is determine the value of h for the time we are
interested in. This can easily be done from the similar triangle equation and the fact that we
know
10
r
=
.
( )
40
4
4
13
13
13
10
h
r
=
=
=
The rate of change of the height of the water is then,
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Calculus I
( )
2
169
40
3
16
13
25
12
100
h
h
h
π
π
π
′
′
′
=
=
⇒
=
11. The angle of elevation is the angle formed by a horizontal line and a line joining the
observer’s eye to an object above the horizontal line. A person is 500 feet way from the launch
point of a hot air balloon. The hot air balloon is starting to come back down at a rate of 15 ft/sec.
At what rate is the angle of elevation,
θ
, changing when the hot air balloon is 200 feet above the
ground. See the (probably bad) sketch below to help visualize the angle of elevation if you are
having trouble seeing it.
Step 1
Putting variables and known quantities on the sketch from the problem statement gives,
We want to determine
θ
′
when
200
y
=
and we know that
15
y′
= −
.
Step 2
There are a variety of equations that we could use here but probably the best one that involves all
of the known and needed quantities is,
( )
tan
500
y
θ
=
Step 3
Differentiating with respect to t gives,
( )
( )
2
2
sec
cos
500
500
y
y
θ θ
θ
θ
′
′
′
′
=
⇒
=
Step 4
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Calculus I
To finish off this problem all we need to do is determine the value of
θ
for the time in question.
We can either use the original equation to do this or we could acknowledge that all we really need
is
( )
cos
θ
and we could do a little right triangle trig to determine that.
For this problem we’ll just use the original equation to find the value of
θ
.
( )
( )
1
200
2
500
5
tan
tan
0.38051radians
θ
θ
−
=
⇒
=
=
The rate of change of the angle of elevation is then,
(
)
2
15
cos
0.38051
0.02586
500
θ
−
′ =
= −
Higher Order Derivatives
1. Determine the fourth derivative of
( )
7
4
3
3
6
8
12
18
h t
t
t
t
t
=
−
+
−
+
Step 1
Not much to this problem other than to take four derivatives so each step will show each
successive derivatives until we get to the fourth. The first derivative is then,
( )
6
3
2
21
24
24
12
h t
t
t
t
′
=
−
+
−
Step 2
The second derivative is,
( )
5
2
126
72
48
h t
t
t
t
′′
=
−
+
Step 3
The third derivative is,
( )
4
630
144
48
h
t
t
t
′′′
=
−
+
Step 4
The fourth, and final derivative for this problem, is,
( )
( )
4
3
2520
144
h
t
t
=
−
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Calculus I
2. Determine the fourth derivative of
( )
3
2
1
V x
x
x
x
=
−
+ −
Step 1
Not much to this problem other than to take four derivatives so each step will show each
successive derivatives until we get to the fourth. The first derivative is then,
( )
2
3
2
1
V
x
x
x
′
=
−
+
Step 2
The second derivative is,
( )
6
2
V
x
x
′′
=
−
Step 3
The third derivative is,
( )
6
V
x
′′′
=
Step 4
The fourth, and final derivative for this problem, is,
( )
( )
4
0
V
x
=
Note that we could have just as easily used the
Fact
from the notes to arrive at this answer in one
step.
3. Determine the fourth derivative of
( )
5
3
2
1
4
8
f x
x
x
x
=
−
−
Step 1
Not much to this problem other than to take four derivatives so each step will show each
successive derivatives until we get to the fourth. After a quick rewrite of the function to help
with the differentiation the first derivative is,
( )
( )
3
2
1
1
2
3
5
5
2
2
1
12
1
1
4
8
5
4
2
f x
x
x
x
f
x
x
x
x
−
−
−
−
′
=
−
−
⇒
=
+
−
Step 2
The second derivative is,
( )
7
3
4
5
2
24
3
1
25
4
4
f
x
x
x
x
−
−
−
′′
= −
−
+
Step 3
The third derivative is,
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Calculus I
( )
12
5
5
5
2
168
3
3
125
8
f
x
x
x
x
−
−
−
′′′
=
+
−
Step 4
The fourth, and final derivative for this problem, is,
( )
( )
17
7
4
6
5
2
2016
15
15
625
16
f
x
x
x
x
−
−
−
= −
−
+
4. Determine the fourth derivative of
( )
( )
(
)
3
7 sin
cos 1 2
w
f w
w
=
+
−
Step 1
Not much to this problem other than to take four derivatives so each step will show each
successive derivatives until we get to the fourth. The first derivative is then,
( )
( )
(
)
3
7
cos
2 sin 1 2
3
w
f
w
w
′
=
+
−
Step 2
The second derivative is,
( )
( )
(
)
3
7
sin
4 cos 1 2
9
w
f
w
w
′′
= −
−
−
Step 3
The third derivative is,
( )
( )
(
)
3
7
cos
8sin 1 2
27
w
f
w
w
′′′
= −
−
−
Step 4
The fourth, and final derivative for this problem, is,
( )
( )
( )
(
)
4
3
7
sin
16 cos 1 2
81
w
f
w
w
=
+
−
5. Determine the fourth derivative of
( )
5
4
8 ln 2
z
y
z
−
=
+
e
Step 1
Not much to this problem other than to take four derivatives so each step will show each
successive derivatives until we get to the fourth. The first derivative is then,
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Calculus I
3
5
5
1
4
8
5
8
5
32
2
z
z
dy
z
z
dx
z
−
−
−
= −
+
= −
+
e
e
Step 2
The second derivative is,
2
5
2
2
25
32
z
d y
z
dx
−
−
=
−
e
Step 3
The third derivative is,
3
5
3
3
125
64
z
d y
z
dx
−
−
= −
+
e
Step 4
The fourth, and final derivative for this problem, is,
4
5
4
4
625
192
z
d y
z
dx
−
−
=
−
e
6. Determine the second derivative of
( )
(
)
3
sin 2
9
g x
x
x
=
−
Step 1
Not much to this problem other than to take two derivatives so each step will show each
successive derivatives until we get to the second. The first derivative is then,
( )
(
) (
)
2
3
6
9 cos 2
9
g x
x
x
x
′
=
−
−
Step 2
Do not forget that often we will end up needing to do a product rule in the second derivative even
though we did not need to do that in the first derivative. The second derivative is then,
( )
(
) (
) (
)
2
3
2
3
12 cos 2
9
6
9 sin 2
9
g
x
x
x
x
x
x
x
′′
=
−
−
−
−
7. Determine the second derivative of
(
)
3
ln 7
z
x
=
−
Step 1
Not much to this problem other than to take two derivatives so each step will show each
successive derivatives until we get to the second. The first derivative is then,
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Calculus I
2
3
3
7
dz
x
dx
x
−
=
−
Step 2
Do not forget that often we will end up needing to do a quotient rule in the second derivative even
though we did not need to do that in the first derivative. The second derivative is then,
(
) (
)(
)
(
)
(
)
3
2
2
2
4
2
2
2
3
3
6
7
3
3
42
3
7
7
x
x
x
x
d z
x
x
dx
x
x
−
−
− −
−
−
−
=
=
−
−
8. Determine the second derivative of
( )
(
)
4
2
2
6 2
Q v
v v
=
+
−
Step 1
Not much to this problem other than to take two derivatives so each step will show each
successive derivatives until we get to the second. We’ll do a quick rewrite of the function to help
with the derivatives and then the first derivative is,
( )
(
)
( )
(
)
(
)
4
2
5
2
2 6 2
8 2 2
6 2
Q v
v v
Q v
v
v v
−
−
=
+
−
′
= −
−
+
−
Step 2
Do not forget that often we will end up needing to do a product rule in the second derivative even
though we did not need to do that in the first derivative. The second derivative is then,
( )
(
)
(
)
(
)
5
6
2
2
2
16 6 2
40 2 2
6 2
Q v
v v
v
v v
−
−
′′
=
+
−
+
−
+
−
9. Determine the second derivative of
( )
( )
2
cos
7
H t
t
=
Step 1
Not much to this problem other than to take two derivatives so each step will show each
successive derivatives until we get to the second. The first derivative is then,
( )
( ) ( )
14 cos 7 sin 7
H t
t
t
′
= −
Step 2
Do not forget that often we will end up needing to do a product rule in the second derivative even
though we did not need to do that in the first derivative. The second derivative is then,
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Calculus I
( )
( ) ( )
( ) ( )
( )
( )
2
2
98sin 7 sin 7
98 cos 7 cos 7
98sin
7
98 cos
7
H
t
t
t
t
t
t
t
′′
=
−
=
−
Note that, in this case, if we recall our trig formulas we could have reduced the product in the first
derivative to a single trig function which would have then allowed us to avoid the product rule for
the second derivative. Can you figure out what the formula is?
10. Determine the second derivative of
3
2
2
1 4
x
y
y
+
= −
Step 1
Not much to this problem other than to take two derivatives so each step will show each
successive derivatives until we get to the second. Note however that we are going to have to do
implicit differentiation
to do each derivative.
Here is the work for the first derivative. If you need a refresher on implicit differentiation go
back to that section and check some of the problems in that section.
(
)
2
2
2
2
6
2
4
6
3
2
4
6
2
4
2
x
y y
y
x
x
y
y
x
y
y
y
′
′
+
= −
−
−
′
′
+
= −
⇒
=
=
+
+
Step 2
Now, the second derivative will also need implicit differentiation. Note as well that we can work
with the first derivative in its present form which will require the quotient rule or we can rewrite
it as,
(
)
1
2
3
2
y
x
y
−
′ = −
+
and use the product rule.
These get messy enough as it is so we’ll go with the product rule to try and keep the “mess” down
a little. Using implicit differentiation to take the derivative of first derivative gives,
( )
(
)
(
)
1
2
2
6
2
3
2
d
y
y
x y
x
y
y
dx
−
−
′′
′
′
=
= −
+
+
+
Step 3
Finally, recall that we don’t want a
y′
in the second derivative so to finish this out we need to
plug in the formula for
y′
(which we know…) and do a little simplifying to get the final answer.
(
)
(
)
(
)
(
)
(
)
(
)
1
2
1
1
3
2
2
4
6
2
3
2
3
2
6
2
9
2
y
x y
x
y
x
y
x y
x
y
−
−
−
−
−
′′ = −
+
+
+
−
+
= −
+
−
+
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Calculus I
11. Determine the second derivative of
2
6
1
y
xy
−
=
Step 1
Not much to this problem other than to take two derivatives so each step will show each
successive derivatives until we get to the second. Note however that we are going to have to do
implicit differentiation
to do each derivative.
Here is the work for the first derivative. If you need a refresher on implicit differentiation go
back to that section and check some of the problems in that section.
(
)
2
2
2
6
2
0
6 2
6 2
y
y
xy y
y
xy y
y
y
xy
′
′
−
−
=
′
′
−
=
⇒
=
−
Step 2
Now, the second derivative will also need implicit differentiation. Note as well that we can work
with the first derivative in its present form which will require the quotient rule or we can rewrite
it as,
(
)
1
2
6 2
y
y
xy
−
′ =
−
and use the product rule.
These get messy enough as it is so we’ll go with the product rule to try and keep the “mess” down
a little. Using implicit differentiation to take the derivative of first derivative gives,
( )
(
)
(
) (
)
1
2
2
2
6 2
6 2
2
2
d
y
y
y y
xy
y
xy
y
xy
dx
−
−
′′
′
′
′
=
=
−
−
−
−
−
Step 3
Finally, recall that we don’t want a
y′
in the second derivative. So, to finish this out let’s do a
little “simplifying” of the to make it “easier” to plug in the formula for
y′
.
(
)
(
)
(
)
(
)
(
)
(
)
(
)
1
2
2
3
2
1
1
2
3
2
6 2
2
6 2
2
6 2
2
6 2
1
6 2
2
6 2
y
y y
xy
y
xy
xy y
xy
y y
xy
xy
xy
y
xy
−
−
−
−
−
−
′′
′
′
=
−
+
−
+
−
′
=
−
+
−
+
−
The point of all of this was to get down to a single
y′
in the formula for the second derivative,
which won’t always be possible to do, and a little factoring to try and make things a little easier to
deal with.
Finally all we need to do is plug in the formula for
y′
to get the final answer.
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Calculus I
(
) (
)
(
)
(
)
(
)
(
)
(
)
(
)
(
)
1
1
1
2
2
3
2
1
2
3
3
2
6 2
6 2
1
6 2
2
6 2
2
6 2
1
6 2
2
6 2
y
y y
xy
xy
xy
xy
y
xy
y
xy
xy
xy
y
xy
−
−
−
−
−
−
−
′′ =
−
−
+
−
+
−
=
−
+
−
+
−
Note that for a further simplification step, if we wanted to go further, we could factor a
(
)
2
3
2
6 2
y
xy
−
−
out of both terms to get,
(
)
(
)
(
)
2
1
3
2
6 2
2
6 2
y
y
xy
xy
xy
−
−
′′ =
−
+
−
Logarithmic Differentiation
1. Use logarithmic differentiation to find the first derivative of
( )
(
)
7
2
2
5 3
6
8
12
f x
x
x
x
= −
+
−
.
Step 1
Take the logarithm of both sides and do a little simplifying.
( )
(
)
(
)
(
)
(
)
(
)
1
2
7
2
2
7
2
2
2
2
1
2
ln
ln
5 3
6
8
12
ln
5 3
ln
6
8
12
7 ln 5 3
ln 6
8
12
f x
x
x
x
x
x
x
x
x
x
=
−
+
−
=
−
+
+
−
=
−
+
+
−
Step 2
Use implicit differentiation to differentiate both sides with respect to x.
( )
( )
2
2
2
2
6
1
12
8
42
6
4
7
5 3
2 6
8
12
5 3
6
8
12
f
x
x
x
x
x
f x
x
x
x
x
x
x
′
−
+
−
+
=
+
=
+
−
+
−
−
+
−
Step 3
Finally, solve for the derivative and plug in the equation for
( )
f x
.
© 2007 Paul Dawkins
108
http://tutorial.math.lamar.edu/terms.aspx
Calculus I
( )
( )
(
)
2
2
7
2
2
2
2
42
6
4
5 3
6
8
12
42
6
4
5 3
6
8
12
5 3
6
8
12
x
x
f
x
f x
x
x
x
x
x
x
x
x
x
x
x
−
+
′
=
+
−
+
−
−
+
=
−
+
−
+
−
+
−
2. Use logarithmic differentiation to find the first derivative of
(
)
(
)
2
3
4
sin 3
6
z
z
y
z
+
=
−
.
Step 1
Take the logarithm of both sides and do a little simplifying.
( )
(
)
(
)
(
)
(
)
(
)
2
3
2
4
3
4
2
4
sin 3
ln
ln
ln sin 3
ln
6
6
ln sin 3
3ln 6
z
z
y
z
z
z
z
z
z
z
+
=
=
+
−
−
−
=
+
−
−
Step 2
Use implicit differentiation to differentiate both sides with respect to z.
(
)
(
)
(
)
(
)
(
)
2
3
3
2
4
4
2
3 2
cos 3
4
12
3
3 2
cot 3
6
6
sin 3
z
z
z
y
z
z
z
z
z
y
z
z
z
z
+
+
′
−
=
−
= +
+
+
−
−
+
Step 3
Finally, solve for the derivative and plug in the equation for y .
(
)
(
)
(
)
(
)
(
)
(
)
2
3
3
2
2
3
4
4
4
sin 3
12
12
3 2
cot 3
3 2
cot 3
6
6
6
z
z
z
z
y
y
z
z
z
z
z
z
z
z
z
+
′ =
+
+
+
=
+
+
+
−
−
−
3. Use logarithmic differentiation to find the first derivative of
( )
( )
3
2
4
5
8
1 9 cos 4
10
t
t
h t
t
t
+
−
=
+
.
Step 1
Take the logarithm of both sides and do a little simplifying.
© 2007 Paul Dawkins
109
http://tutorial.math.lamar.edu/terms.aspx
Calculus I
( )
( )
( )
(
)
( )
(
)
(
)
(
)
( )
(
)
(
)
3
2
4
2
4
3
1
1
1
2
4
3
2
2
1
1
1
2
3
4
5
8
1 9 cos 4
ln
ln
10
ln
5
8
1 9 cos 4
ln
10
ln
5
8
ln 1 9 cos 4
ln
10
ln 5
8
ln 1 9 cos 4
10
t
t
h t
t
t
t
t
t
t
t
t
t
t
t
t
t
t
+
−
=
+
=
+
−
−
+
=
+
+
−
−
+
=
+ +
−
−
+
Note that the logarithm simplification work was a little complicated for this problem, but if you
know your logarithm properties you should be okay with that.
Step 2
Use implicit differentiation to differentiate both sides with respect to t.
( )
( )
( )
( )
1
1
1
2
3
4
2
36 sin 4
5
2
10
5
8
1 9 cos 4
10
h t
t
t
h t
t
t
t
t
′
+
=
+
−
+
−
+
Step 3
Finally, solve for the derivative and plug in the equation for
( )
h t
.
( ) ( )
( )
( )
( )
( )
( )
5
5
1
2
2
2
2
3
5
5
1
2
2
2
2
2
4
12 sin 4
5
8
1 9 cos 4
10
5
8
1 9 cos 4
12 sin 4
5
8
1 9 cos 4
10
10
t
t
h t
h t
t
t
t
t
t
t
t
t
t
t
t
t
t
t
+
′
=
+
−
+
−
+
+
−
+
=
+
−
+
−
+
+
4. Find the first derivative of
( ) (
)
4
3
7
w
g w
w
=
−
.
Step 1
We just need to do some logarithmic differentiation so take the logarithm of both sides and do a
little simplifying.
( )
(
)
(
)
4
ln
ln
3
7
4
ln 3
7
w
g w
w
w
w
=
−
=
−
Step 2
Use implicit differentiation to differentiate both sides with respect to w. Don’t forget to product
rule the right side.
( )
( )
(
)
(
)
3
12
4 ln 3
7
4
4 ln 3
7
3
7
3
7
g w
w
w
w
w
g w
w
w
′
=
− +
=
− +
−
−
© 2007 Paul Dawkins
110
http://tutorial.math.lamar.edu/terms.aspx
Calculus I
Step 3
Finally, solve for the derivative and plug in the equation for
( )
g w
.
( )
( )
(
)
(
)
(
)
4
12
4 ln 3
7
3
7
12
3
7
4 ln 3
7
3
7
w
w
g w
g w
w
w
w
w
w
w
′
=
− +
−
=
−
− +
−
5. Find the first derivative of
( )
(
)
( )
sin 2
8
2
x
x
f x
x
=
− e
.
Step 1
We just need to do some logarithmic differentiation so take the logarithm of both sides and do a
little simplifying.
( )
(
)
( )
( )
(
)
sin 2
8
8
ln
ln
2
sin 2
ln 2
x
x
x
f x
x
x
x
=
−
=
−
e
e
Step 2
Use implicit differentiation to differentiate both sides with respect to x. Don’t forget to product
rule the right side.
( )
( )
( )
(
)
( )
( )
(
)
( )
8
8
8
8
8
8
2 8
2 cos 2
ln 2
sin 2
2
2 8
2 cos 2
ln 2
sin 2
2
x
x
x
x
x
x
f
x
x
x
x
f x
x
x
x
x
x
′
−
=
−
+
−
−
=
−
+
−
e
e
e
e
e
e
Step 3
Finally, solve for the derivative and plug in the equation for
( )
f x
.
( )
( )
( )
(
)
( )
(
)
( )
( )
(
)
( )
8
8
8
8
sin 2
8
8
8
2 8
2 cos 2
ln 2
sin 2
2
2 8
2
2 cos 2
ln 2
sin 2
2
x
x
x
x
x
x
x
x
f
x
f x
x
x
x
x
x
x
x
x
x
−
′
=
−
+
−
−
=
−
−
+
−
e
e
e
e
e
e
e
© 2007 Paul Dawkins
111
http://tutorial.math.lamar.edu/terms.aspx