Contents
How You May Use This Resource Guide
22 The Derivative
Worksheet 22.1:
Worksheet 22.2:
Worksheet 22.3:
Worksheet 22.4:
Worksheet 22.5:
ii
Closing the Gap Between Secant and Tangent
The Derivative and Its Rules . . . . . . . . .
Chain Rule Exploration . . . . . . . . . . . .
Tangent/Normal Line Applications . . . . . .
Finding Derivatives–A Review Activity . . .
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i
How You May Use This Resource
Guide
This guide is divided into chapters that match the chapters in the third editions of Technical
Mathematics and Technical Mathematics with Calculus by John C. Peterson. The guide
was originally developed for the second editions of these books by Robert Kimball, Lisa
Morgan Hodge, and James A. Martin all of Wake Technical Community College, Raleigh,
North Carolina. It has been modified for the third editions by the author.
Each chapter in this Resource Guide contains the objectives for that chapter, some
teaching hints, guidelines based on NCTM and AMATYC standards, and activities. The
teaching hints are often linked to activities in the Resource Guide, but also include comments concerning the appropriate use of technology and options regarding pedagogical
strategies that may be implemented.
The guidelines provide comments from the Crossroads of the American Mathematical
Association of Two-Year Colleges (AMATYC), and the Standards of the National Council
of Teachers of Mathematics, as well as other important sources. These guidelines concern
both content and pedagogy and are meant to help you consider how you will present the
material to your students. The instructor must consider a multitude of factors in devising
classroom strategies for a particular group of students. We all know that students learn
better when they are actively involved in the learning process and know where what they
are learning is used. We all say that less lecture is better than more lecture, but each one of
us must decide on what works best for us as well as our students.
The activities provided in the resource guide are intended to supplement the excellent
problems found in the text. Some activities can be quickly used in class and some may
be assigned over an extended period to groups of students. Many of the activities built
around spreadsheets can be done just as well with programmable graphing calculators; but
we think that students should learn to use the spreadsheet as a mathematical tool. There
are obstacles to be overcome if we are to embrace this useful technology for use in our
courses, but it is worth the effort to provide meaningful experiences with spreadsheets to
people who probably will have to use them on the job.
Whether or not you use any of the activities, we hope that this guide provides you with
some thought-provoking discussion that will lead to better teaching and quality learning.
ii
Chapter 22
The Derivative
Objectives
After completing this chapter, the student will be able to:
• Find the derivative of an algebraic function using the limit method of taking a derivative;
• Find the derivative of algebraic functions using the rules for derivatives;
• Find the derivatives of implicit algebraic functions;
• Find the slope of the tangent line to a curve at a given point;
• Find the equations of tangent and normal lines to a curve at a given point;
• Find higher order derivatives of algebraic functions.
Teaching Hints
1. Activity 23.1 in Chapter 23 can be used to illustrate Chapter 22. This activity deals
with approximate rates of change with the difference quotient (or slopes of secant).
Students do not need to know how to take derivatives to do the activity.
2. Show students how the slope of the secant line between two points approaches the
slope of the tangent line as the distance between the two points gets closer to zero.
(see Activity 22.1) This will show students the relationship between secant lines,
tangent lines, limits and derivatives. This will enable students to visualize what is
meant by the slope of a tangent line to a curve and it will help them to understand
how and why derivatives can be found using the limit method.
3. Have students show all steps when they find derivatives using the limit method. Students need to understand that with this method they are not just finding the derivative
of a function, but that they are showing through the use of limits that the slope of
a tangent line can be obtained by using the slope of the secant line and making the
distance between the points get closer to zero. They also need to understand that the
slope of a tangent line to a curve at a point is the slope of the curve at that point and
is the first derivative of the function evaluated at that point.
1
Instructional Resource Guide, Chapter 22
Peterson, Technical Mathematics with Calculus, 3rd edition
2
4. Make sure that students are aware of the many different notations used to denote
taking the first derivative.
5. Students need to learn the product and quotient rules for taking derivatives. The
product rule usually does not give students much trouble, but the quotient rule
usually does. You may want to help students learn the quotient rule by using the
mnemonic “high d low minus low d high over low squared.” Some students may
u
prefer to convert to u · v −1 and use the product rule along with the chain rule.
v
Give students both options.
6. Before teaching Section 22.4 make sure that you review the basics of composite
functions. Have students take a function and split it into its two composite functions.
For example, if h(x) = f (g(x)), find f (x) and g(x), when h(x) = (x2 + x)3 .
7. Use a graphing utility that will find derivatives, such as the TI-86, to check a derivative found algebraically. Graph the algebraic derivative and compare the graph of it
to the graph of the derivative computed by the graphing utility. If both graphs are
the same, then the algebraic derivative is correct.
8. Introduce Section 22.5 by giving the class an equation such as x2 y 3 + 6xy − 5x =
xy 2 +4 and asking them to find the derivative by first solving the equation for y. Stop
them after a few minutes of frustration and introduce implicit differentiation. This
will help students to understand the importance of implicit differentiation. Many
students try to avoid using implicit differentiation because they do not understand
why they may need to use it. Students need to be shown that they may not always be
able to solve for one variable in an equation before differentiating.
Guidelines
The emphasis of calculus should be on the fundamental concepts of the subject, not symbolic manipulation. (Preparing for a New Calculus Conference
Proceedings, Mathematical Association of America, 1994, page 55)
It is very important for students to learn the fundamental concepts in calculus. Students
need to understand what a derivative is and why the algebraic rules for finding derivatives
work. Once they understand the fundamental concepts of a derivative, students are ready
to use the technology that is available. “Appropriate technology should be available at
all times for graphing, numerical computations, and symbolic manipulations.” (page 55)
Students should view technology as a tool for understanding concepts and as a tool for
solving complex problems. “In this data-driven, technologically advanced environment,
technicians must be proficient in the application of technology, in mathematics at the level
of the Foundation and beyond, and in critical thinking.” (Crossroads in Mathematics, 1995,
page 31)
Guidelines for Content (Crossroads)
Increased Attention
Realistic applications and modeling
Visual representation of concepts
Pattern recognition, drawing inferences
Decreased Attention
Verification of complex identities
Rote memorization and use of formulas
Rote application of formulas
Instructional Resource Guide, Chapter 22
Peterson, Technical Mathematics with Calculus, 3rd edition
3
Guidelines for Pedagogy (Crossroads)
Increased Attention
active involvement of students
technology to aid in concept development
mathematical reasoning
Decreased Attention
Passive listening
Paper-and-pencil drills
Memorization of facts and procedures
Activities
1. Closing the Gap Between Secant and Tangent
Students will use numerical analysis and graphs to understand the relationship between secant lines, tangent lines, limits, and derivatives.
2. The Derivative and Its Rules
Problems designed for students to gain better understanding of the derivative and the
algebraic rules used to find derivatives.
3. A Chain Rule Exploration
Students will explore why the chain rule works to take derivatives of certain functions.
4. Tangent/Normal Line Applications
Students will write equations of tangent and normal lines to curves using given information.
5. Finding Derivatives—A Review Activity
Students will compare derivatives found both algebraically and with the use of a
graphing utility. They will also examine graphical situations where derivatives are
undefined
Instructional Resource Guide, Chapter 22
Peterson, Technical Mathematics with Calculus, 3rd edition
4
Student Worksheet 22.1
Closing the Gap Between Secant and Tangent
You learned in Chapter 4 that the slope of a line through two points (x1 , y1 ) and (x2 , y2 ) can be obtained through the
y2 − y 1
∆y
formula m =
=
. Now we are going to use this formula to find slopes of secant lines and learn how we can
x2 − x1
∆x
use the slope of a secant line to approximate the slope of a tangent line and find the exact slope of the tangent line.
Numerical
Graph the function f (x) = x2 − 9. We want to find the slope of the tangent line at the point (2, −5). First, look at the
slopes of some secant lines that pass through the point (2, −5) and some second point (x2 , y2 ).
1. Using the function f (x) = x2 − 9 fill out the table below.
x1 = 2
x2
3
2.5
2.25
2.1
2.01
2.001
2.0001
y1 = −5
y2
Slope of secant line
2. What appears to be the limit of the slope of the secant lines as the x-values get closer together?
3. On your graph draw lines through the points (x1 , y1 ) and (x2 , y2 ). Name at least one thing similar about all the lines.
Symbolic
Letting h represent the distance from x1 = 2 to x2 results is x2 =
2 + h and y2 = f (2 + h). The slope of the secant line using the
f (2 + h) − f (2)
.
points (2, −5) and (2 + h, f (2 + h)) is
(2 + h) − 2
4. Find f (2 + h)
f (x) = x2 − 9
y
f (2 + h)
2
5. Using f (2 + h), simplify the secant line slope above.
6. Find the slope of the tangent by finding lim
x→0
f (2 + h) − f (2)
.
(2 + h) − 2
2+h
x
f (2) = −5
ant
7. Compare your limit with the numerical guess from above.
sec
A graphing utility can also be used to see that the slope of the secant
line will approximate the slope of the tangent line as the distance between the x coordinates approaches zero. Graph the function f (x) = x3 + 6x2 + 3x − 10 for −6 ≤ x ≤ 5. Zoom in on
the point (1, 0) until the graph in the viewing window looks like a straight line. You may have to zoom in several times.
If your graphing utility has a box feature, box in around the point (1, 0) until the graph looks like a straight line.
Instructional Resource Guide, Chapter 22
Peterson, Technical Mathematics with Calculus, 3rd edition
5
Pick another point on your curve that is in your viewing window. Pick an x value within your range and do one of the
following to find the y value for your point. If your graphing utility has an evaluate function use it to find the y value at
your new point. If your graphing utility does not have an evaluation function use the original function to find the y value
that corresponds with the x you have chosen.
Calculate the slope of the line through these two points. The slope that you just calculated is an approximation of
the slope of the function at the point (1, 0). This slope is called the derivative f (x) of your function f (x) at the point
(1, 0). You can now say that for your function f (x) = x3 + 6x2 + 3x − 10, f 0 (x) =
. This process works
because when you zoom in to make the function look like a straight line you are creating a situation where the distance
between your x values is becoming smaller. So you are causing the slope of the secant line to approximate the slope of
the tangent line to the curve at a certain point.
Numerical
Now it is your turn. Using the function g(x) = 2x2 − 8, find the slope of the tangent lint to the curve at the point (2, 0).
8. Fill out the table below.
x1 = 2
x2
3
2.5
2.25
2.1
2.01
2.001
2.0001
y1 = 0
y2
Slope of secant line
9. What appears to be the limit of the slope of the secant lines as the x values get closer together?
Symbolic
g(2 + h) − g(2)
.
x→0
(2 + h) − 2
10. Now find the slope of the tangent line by finding lim
11. Compare your limit with your numerical guess from above.
It is time to do some on your own. In Exercises 12–13, use the given function and point to find: (a) The limit of the
slope of the secant lines as the x values of the points get closer together (set up a table). (b) The slope of the tangent by
f (x + h) − f (x)
finding lim
.
x→0
(x + h) − x
12. f (x) = 3x − 4 at (2, 2)
13. f (x) = x2 − 3x + 1 at (3, 1)
Instructional Resource Guide, Chapter 22
Peterson, Technical Mathematics with Calculus, 3rd edition
6
Student Worksheet 22.2
The Derivative and Its Rules
1. Air is pumped into a spherical balloon at a constant rate. Let S(t) be the surface area of the balloon and V (t) be the
volume of gas in the balloon.
(a) In your own words, explain what is meant by
dS
dV
and
.
dt
dt
dS
positive, negative, or zero, when the balloon is half full?
dt
dV
(c) Is
positive, negative, or zero, when the balloon is half full?
dt
dS
dV
(d) Which of
or
is constant? Explain your answer.
dt
dt
(e) If gas was escaping from the balloon, would your answers to (b), (c), and (d) change? If so, what would they
change to and why?
(b) Is
2. Suppose f (3) = 4, g(3) = −2, f 0 (3) = −3, g 0 (3) = 1, f 0 (−2) = 2, and g 0 (4) = 5. Find the derivative at 3 for each
of the following.
(a) h(x) = f (x) + g(x)
(b) j(x) = f (x)g(x)
(c) k(x) = f (x)/g(x)
(d) m(x) = f (g(x))
(e) p(x) = g(f (x))
3. A train is moving away from the station at a rate of 35 mph. A person on the train is walking towards the front of the
train at a rate of 4 mph.
(a) How fast is the person moving away from the station?
(b) Addition
4. The length of a rectangle is increasing at a rate of 5 cm/sec, and its width is increasing at a rate of 4 cm/sec. At what
rate is the area of the rectangle increasing when its length is 8 cm and its width is 6 cm?
5. If t = x2 − 3x and y = (4t − 6)4 , find
dy
.
dx
6. Consider the curve x2 + y 2 = 25.
dy
(a) Sketch the graph of this curve and use the graph to predict in what quadrants
will be positive and in what
dx
quadrants it will be negative.
dy
(b) Find
and confirm the predictions you made in part (a).
dx
(c) Find the equations of two different lines that are tangent to the curve at x = 3.
Instructional Resource Guide, Chapter 22
Peterson, Technical Mathematics with Calculus, 3rd edition
7
Student Worksheet 22.3
Chain Rule Exploration
1. If f (x) = (g(x))2 , use the product rule to find f 0 (x).
2. If h(x) = (g(x))3 , write h(x) as (g(x))2 g(x); use the product rule along with your answer to Exercise 1 to find h(x).
3. Suppose the pattern you see in parts Exercises 1 and 2 continues to hold true for k 0 (x), where k(x) = (g(x))n−1 .
Write an expression for k 0 (x).
4. Now let p(x) = (g(x))n , and find p0 (x) by writing p(x) as (g(x))n−1 g(x). Use the product rule and the result of
Exercise 3 to find p0 (x).
5. Do you think this pattern will hold for all natural numbers? Explain your answer.
Instructional Resource Guide, Chapter 22
Peterson, Technical Mathematics with Calculus, 3rd edition
8
Student Worksheet 22.4
Tangent/Normal Line Applications
1. Find an equation for the tangent line to the graph of f (x) =
3x + 5
at the point where x = 1.
7x − 3
2. Find an equation for the line normal to the graph of y = (x2 − 3)(5x − x3 ) at the point where x = 1.
3. Find the coordinates of the points, if any, where the graph of f (x) = x2 − 3x has a horizontal tangent line.
4. Find the coordinates of the points, if any, where the graph of f (x) =
5x
has a horizontal tangent line.
+1
x2
5. Find the points on the graph of y = (2x − 1)3 at which the normal line is parallel to the line x + 12y = 36.
6. Find the points where the tangent to f (x) = x3 − x2 is perpendicular to the line 5y + x + 2 = 0.
7. Find an equation of a line with slope 4 that is normal to the graph of y =
1
.
(8x + 3)2
8. Find the area of the triangle formed by the x-axis and the lines tangent and normal to the curve f (x) = 6x − x2 at the
point where x = 5.
9. Find the value of c, if any, so that the tangent line to f (x) =
(1, f (1)) and (3, f (3)).
x
at (c, f (c)) is parallel to the secant line through
x+1
Instructional Resource Guide, Chapter 22
Peterson, Technical Mathematics with Calculus, 3rd edition
9
Student Worksheet 22.5
Finding Derivatives–A Review Activity
For each function below:
I. Algebraically
(a) Find the derivative at any x.
(b) Evaluate the derivative at the specified points.
(c) Specify all x-values where the derivative does not exist.
II. Geometrically
(a) Sketch a graph of the function on the grid provided.
(b) Draw the tangent line at the specified points.
(c) Evaluate the derivative at the specified points and compare these values with the ones you found in Part I.
√
3
2. g(x) = 2 − 4 x
x
x = 2 and x = 0
1. f (x) = 3x3 − 5x2 + 2x − 1
x = −1 and x = 1
y
y
x
2x − 1
x2 + 1
x = −2 and x = 0
x
4. j(x) = 19 2x2 − 3
x = 1.7 and x = 0
3. h(x) =
3
y
y
x
x
Instructional Resource Guide, Chapter 22
Peterson, Technical Mathematics with Calculus, 3rd edition
√
5. f (x) = (1 − x2 ) 2x + 4
x = 0 and x = −2
10
3
1
6. g(x) = 2 −
+1
4x
2x
x = −2 and x = 0
y
y
x
x
2−x
x = 2 and x = 0
x
3x − 9
9 − x2
x = 3 and x = 0
7. h(x) = √
4
8. j(x) =
y
x
y
x
9. As you have noticed, the derivative is sometimes undefined. Give three different graphical situations that result in
f 0 (x) being undefined.
(a)
(b)
(c)
Answers
Student Worksheet 22.1
1.
x1 = 2
x2
3
2.5
2.25
2.1
2.01
2.001
2.0001
y1 = −5
y2
0
−2.75
−3.9375
−4.59
−4.9599
−4.995999
−4.99959999
8.
Slope of secant line
5
4.5
4.25
4.1
4.01
4.001
4.0001
2. 4
y1 = 0
y2
10
4.5
2.125
0.82
0.0802
0.008002
0.00080002
Slope of secant line
10
9
8.5
8.2
8.02
8.002
8.0002
9. 8
4. f (2 + h) = (2 + h)2 − 9 = h2 + 4h − 5
5.
x1 = 2
x2
3
2.5
2.25
2.1
2.01
2.001
2.0001
f (2 + h) − f (2)
(2 + h) − 2
h2 + 4h − 5 − (−5)
(2 + h) − 2
h2 + 4h
=
h
= h+4
g(2 + h) − g(2)
2(2 + h)2 − 8 − (2(22 ) − 8)
=
=
(2 + h) − 2
(2 + h) − 2
2h2 + 8h − (0)
2h2 + 8h
=
= 2h + 8 which has a limit
(2 + h) − 2
h
of 8
10.
=
11. They are the same
6. 4
12. 3
7. They are the same.
13. 3
Student Worksheet 22.2
dS
represents the instantaneous change in the surface is what dV means.
dt
dt
dV
area at time t and
represents the instantaneous change
dt
dS
dV
in the volume at time t.
1e. Both
and
would be negative but we cannot tell
dt
dt
if either would be constant.
1b. Positive
1a.
1c. Positive
2a. h0 (3) = f 0 (3) + g 0 (3) = −3 + 1 = −2
dV
1d.
because air being pumped in at a constant rate 2b. j 0 (3) = f (3)g 0 (3) + g(3)f 0 (3) = 4(1) + (−2)(−3) =
dt
means that the volume is changing at a contant rate, which 10
11
Instructional Resource Guide, Answers
Peterson, Technical Mathematics with Calculus, 3rd edition
2c. k 0 (3) =
12
dy
g(3)f 0 (3) − f (3)g 0 (3)
(−2)(−3) − (4)(2)
should be positive in quadrants II and IV and nega=
=
2
2
dx
[g(3)]
[−2]
tive in quadrants I and III.
6−8
= 0.5
4
2d. m0 (3) = f 0 (g(3))g 0 (3) = f 0 (−2)g 0 (3) = 2(1) = 2
6b. Think of this as two functions y1 = (25−x2 )0.5 graphs
the top half of the circle and y2 = −(25 − x2 )0.5 graphs
0
0
0
0
0
2e. p (3) = g (f (3))f (3) = g (4)f (3) = 5(−3) = −15 the bottom half. Then y10 = 0.5(25 − x2 )−0.5 (−2x) =
−x
and y20 = −0.5(25 −
−x(25 − x2 )−0.5 = √
3a. 39 mph
25 − x2
x
4. 62 cm2 /s
x2 )−0.5 (−2x) = x(25 − x2 )−0.5 = √
. y10 is
2
25
−
x
dy
dy dt
5.
=
= 3(4t − 6)3 (2x − 3) = 3(4(x2 − 3x) − positive in quadrant II and negative in quadrant I and y20
dx
dt dx
is positive in quadrant IV and negative in quadrant III.
6)3 (2x − 3) = 3(4x2 − 12x − 6)3 (2x − 3)
y
6c. When x = 3, the points on the curve are (3, 4) and
(3, −4). The line tangent at (3, 4) is y − 4 = 0.75(x − 3)
or y = 0.75x + 1.75. The line tangent at (3, −4) is
y + 4 = 0.75(x − 3) or y = 0.75x − 6.25.
6a.
x
Student Worksheet 22.3
1. Let f (x) = (g(x))2 = g(x) · g(x). Then, by the product 3. Let k 0 (x) = (n − 1)(g(x))n−2 · g 0 (x).
rule, f 0 (x) = g(x) · g 0 (x) + g(x) · g 0 (x) = 2g(x) · g 0 (x).
4. Let p(x) = (g(x))n = (g(x))n−1 · g(x) = k(x) ·
2. Let h(x) = (g(x))3 = (g(x))2 ·g(x). Then, by the prod- g(x). Then, by the product rule and Exercise 3, p0 (x) =
uct rule and Exercise 1, h0 (x) = (g(x))2 · g 0 (x) + g(x) · k(x) · g 0 (x) + g(x) · k 0 (x) = (g(x))n−1 · g 0 (x) + g(x) ·
[2g(x) · g 0 (x)] = 3[g(x)]2 · g 0 (x).
(n − 1)(g(x))n−2 · g 0 (x) = (n)(g(x))n−1 · g 0 (x).
Student Worksheet 22.4
1. y = −2.75x + 4.75
7. y −
1
16
=4 x−
1
8
or y = 4x −
7
16
2. y = −0.25x − 7.75
3. (1.5, −2.25)
8. 53.125 units2
√ !
4. (−1, −2.5) and (1, 2.5)
√ −1 − 2 2
√
√
≈ (−3.8284, 1.3536) and
9. −1 − 2 2,
5. (−0.5 +√0.5 2, −2.0101) ≈ (−0.2071, −2.0101) and
−2 2
!
√
(0.5 + 0.5 2, 2.8284) ≈ (1.2071, 2.8284)
√ −1 + 2 2
√
−1
+
2
2,
≈ (1.8284, 0.6464)
6. (−1, −2) and 53 , 50
2 2
27
Instructional Resource Guide, Answers
Peterson, Technical Mathematics with Calculus, 3rd edition
13
Student Worksheet 22.5
y
1. I. (a) f 0 (x) = 9x2 − 10x + 2, (b) f 0 (−1) = 21, II.
f 0 (1) = 1, (c) the derivative exists at all values of x.
II. In the graphs for Exercises 1–8, the graph of the given
function is in blue and the graphs of any tangent lines are
in red and green.
5
y
–1
1
x
2
10
5
±1
1
–5
5. I. (a) f 0 (x)
x
2
√
1 − x2
−2 x 2 x + 4 + √
2x + 4
=
=
2
±10
1 − 8x − 5 x
√
, (b) f 0 (0) ≈ 0.1890, f 0 (−2) does not exist
2x + 4
(c) the derivative does not exist when x ≤ −2.
±15
II.
±5
y
5
2. I. (a) g 0 (x) = −6 x−3 − 0.25 x−0.75 , (b) g 0 (2) ≈
−0.8987, g 0 (0) does not exist, (c) the derivative does not
exist for values of x ≤ 0.
–1
1
x
2
y
II.
10
–5
5
±1
1
2
x
3
0.5x − 1.5
6. I. (a) g 0 (x) =
, (b) g 0 (−2) ≈ 0.3305, g 0 (0)
x3
does not exist, (c) the derivative does not exist for values of
x = 0.
y
II.
5
±5
−1
(2 x − 1) x
3. I. (a) h0 (x) = 2 x2 + 10
− 2
2 =
(x2 + 10)
2 + 2x − 2x2
–3
–2
–1
, (b) h0 (−2) ≈ −0.2890, h0 (0) = 1, (c) the
(x2 + 1)2
derivative exists for all values of x.
II.
7. I. (a) h0 (x) =
2
x
1
0.25x
+
, (b) h0 (2) does
sqrt[4]2 − x 2 − x1.25
not exist, h0 (0) ≈ 0.8409, (c) the derivative exists for all
values of x < 2.
y
1
– 5 – 4 – 3 – 2 –1
1
y
1
2
3
4
5
6
x
II.
5
–1
–2
2
4x(2x2 −3)
4. I. (a) j 0 (x) =
, (b) j 0 (1.7) ≈ 17.7228,
3
j 0 (0) = 0, (c) the derivative exists for all values of x.
–3
–2
–1
1
2
x
Instructional Resource Guide, Answers
Peterson, Technical Mathematics with Calculus, 3rd edition
14
3
, (b) j 0 (3) is not 9. Answers will vary. Some possible answers:
(x + 3)2
defined, j 0 (0) = 19 , (c) the derivative exists for all values of (a) When the function is undefined, such as f (x) = 1
x−1
x 6= ±3.
is undefined when x = 1, thus the derivative does not
y
exist at x = 1,
II.
5
−2
8. I. (a) j 0 (x) = 3 (x + 3)
=
(b) When the function has a sharp point, for example,
f (x) = |x| does not have a derivative at x = 0,
±5
5
±5
x
(c) When there
( is a “break” in the function, such as for
x,
if x ≤ 1
does not have a derivative
f (x) =
5 − x, if x > 1
at x = 1.