lecture 14: the derivative as a function

Lecture 14: The Derivative as aFunction

Objective:

� Present the concept of the derivative as a function.

� Suggested problems: Section 2.8 #1 − 61 (odd)

Introduction

Recall that given a function, f(x), the derivative of a function at x = a can be found usingone of two formulas

f ′(a) = limx→a

f(x)− f(a)

x− a

f ′(a) = limh→0

f(a + h)− f(a)

h

Geometrically, f ′(a) is the slope of the tangent line to the curve of y = f(x) at (a, f(a)). Italso represents the instantaneous rate of change of y with respect to x.

In Lecture 13, the value at which the derivative was calculated at, was fixed. In this set ofnotes, we will examine what happens when a is allowed to vary.

Derivative of as a Function

Suppose that we have a function, f(x), and wanted to calculate the derivative of f at x. To doso, simply replace a with x in the definition of the derivative to generate

f ′(x) = limh→0

f(x + h)− f(x)

h

Observe the f ′(x) is a function of x. Thus, it has a domain and a range. Moreover, f ′(x)provides us with a convenient means for calculating the derivative of the function y = f(x) atany point in its domain provided that the limit exists.

1

August 20, 2021 201-NYA-05: Calculus I Sect. 00013

Since f ′(x) represents the rate of change of y with respect to x. The value of f ′(x) tells us ifthe function is increasing, decreasing, or flat at a specific value of x, or on an interval. Thus is,

f ′(x) > 0 ⇒ f is increasing.

f ′(x) < 0 ⇒ f is decreasing.

f ′(x) = 0 ⇒ f is neither increasing nor decreasing; so it is flat.

The graphs below show how the derivative of a function, is the slope of the tangent line to f(x)at a. Observe that whenever the f(x) is increasing, f ′(x) is above the x−axis. This is becausewhenever f(x) is increasing, the slopes of the tangent lines will be positive; and hence, f ′(x)will be positive. Similarly, whenever f(x) is decreasing; f ′(x) is below the x−axis.

154 CHAPTER 2 Limits and Derivatives

SOLUTION We can estimate the value of the derivative at any value of x by drawing the tangent at the point sx, f sxdd and estimating its slope. For instance, for x − 3 we draw a tangent at P in Figure 2 and estimate its slope to be about 22

3. (We have drawn a triangle to help estimate the slope.) Thus f 9s3d < 22

3 < 20.67 and this allows us to plot the point P9s3, 20.67d on the graph of f 9 directly beneath P. (The slope of the graph of f becomes the y-value on the graph of f 9.)

3 51

m=0

P ª(3, _0.67)

y

B

A

D

P

x

0

y=ƒ

y

Aª

Bª

Cª

0

y=fª(x)

mÅ_23

3

5

1

x

m=_1m=0

m=1

C

Dª

1

1

The slope of the tangent drawn at A appears to be about 21, so we plot the point A9 with a y-value of 21 on the graph of f 9 (directly beneath A). The tangents at B and D are horizontal, so the derivative is 0 there and the graph of f 9 crosses the x-axis (where y − 0) at the points B9 and D9, directly beneath B and D. Between B and D, the graph of f is steepest at C and the tangent line there appears to have slope 1, so the largest value of f 9sxd between B9 and D9 is 1 (at C9).

Notice that between B and D the tangents have positive slope, so f 9sxd is positive there. (The graph of f 9 is above the x-axis.) But to the right of D the tangents have negative slope, so f 9sxd is negative there. (The graph of f 9 is below the x-axis.) ■

EXAMPLE 2 (a) If f sxd − x 3 2 x, find a formula for f 9sxd.(b) Illustrate this formula by comparing the graphs of f and f 9.

SOLUTION(a) When using Equation 2 to compute a derivative, we must remember that the variable

FIGURE 2

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Example 1

Below is a graph of f(x). Sketch a graph of f ′(x).

f(x)

x

y

2


Solution 1

It is best two sketch f ′(x) directly under f(x) using some key features from the graph of f .

f(x)

x

y

f ′(x)

x

y

Example 2


f(x)

x

y

3


Solution 2

f(x)

x

y

f ′(x)

x

y

Example 3


f(x)

x

y

4


Solution 3

f(x)

x

y

f ′(x)

x

y

The next set of problems illustrate how to find the derivative of a function using the definitionof the derivative.

Definition 1

The derivative of f(x) with respect to x is the function f ′(x) and is defined as

f ′(x) = limh→0

f(x + h)− f(x)

h

We say that f itself is differentiable at x, and that f has a derivative whenever the limitexists.

Example 4

Find the derivative of the function using the definition of derivative. State the domain ofthe function and the domain of its derivative.

f(x) = mx + b

5


Solution 4

The definition of the derivative is f ′(x) = limh→0

f(x + h)− f(x)

h

f ′(x) = limh→0

f(x + h)− f(x)

h

= limh→0

[ m(x + h) + b ]− (mx + b)

h

= limh→0

��mx + mh + ��b−��mx− ��bh

= limh→0

m��h

��h= m

Domains: Df : (−∞,∞)Df ′ : (−∞,∞)

Example 5


f(x) = 4 + 8x− 5x2

Solution 5

f ′(x) = limh→0

f(x + h)− f(x)

h

= limh→0

[ 4 + 8(x + h)− 5(x + h)2 ]− (4 + 8x− 5x2)

h

= limh→0

�4 +��8x + 8h−��5x2 − 10xh− h2 − �4−��8x +��5x2

h

= limh→0

��h(8− 10x− h)

��h

= limh→0

(8− 10x− ��0

h )

= 8− 10x

Domains: Df : (−∞,∞)Df ′ : (−∞,∞)

6


Example 6


f(v) =v

v + 2

Solution 6

f ′(v) = limh→0

f(v + h)− f(v)

h

= limh→0

v+h(v+h)+2

− vv+2

h

= limh→0

1

h

[v + h

v + h + 2− v

v + 2

]= lim

h→0

1

h

[(v + 2)(v + h)− (v + h + 2)(v)

(v + h + 2)(v + 2)

]= lim

h→0

1

h

[��v2 +��vh +��2v + 2h−��v2 −��vh−��2v

(v + h + 2)(v + 2)

]

= limh→0

1

��h

[2��h

(v + h + 2)(v + 2)

]= lim

h→0

2

(v + ��0

h + 2)(v + 2)

=2

(v + 2)2

Domains: Df : (−∞,−2) ∪ (−2,∞)Df ′ : (−∞,−2) ∪ (−2,∞)

7


Example 7


f(x) =√

9− x

Solution 7

f ′(x) = limh→0

f(x + h)− f(x)

h

= limh→0

√9− (x + h)−

√9− x

h

= limh→0

(√

9− (x + h)−√

9− x)

h·

(√

9− (x + h) +√

9− x)

(√

9− (x + h) +√

9− x)

= limh→0

9− (x + h)− (9− x)

h(√

9− (x + h) +√

9− x)

= limh→0

�9−�x−��h− �9 +�x

��h(√

9− (x + h) +√

9− x)

= limh→0

−1√9− x− ��

0

h +√

9− x

= − 1

2√

9− x

Domains: Df : (−∞, 9]Df ′ : (−∞, 9)

Example 8

a. Sketch the graph of f(x) = 1 +√x + 3 by starting with the graph of y =

√x and

using transformations.

b. Use the graph from part (a) to sketch the graph of f ′(x).

c. Use the definition of a derivative to find f ′(x).

d. What are the domains of f(x) and f ′(x)?

8


Solution 8

a. Starting from y =√x we move y three units to the left, and then shift the resulting

graph up by one unit.

f(x)

(−3, 1)x

y

b. Below is the graph of f ′(x)

f ′(x)−3x

y

c. Using the definition of the derivative, we have

f ′(x) = limh→0

f(x + h)− f(x)

h= lim

h→0

1 +√

(x + h) + 3− (1 +√x + 3)

h

= limh→0

�1 +√x + h + 3− �1−

√x + 3

h

= limh→0

√x + h + 3−

√x + 3

h· (√x + h + 3 +

√x + 3)

(√x + h + 3 +

√x + 3)

= limh→0

x + h + 3− (x + 3)

h(√x + h + 3 +

√x + 3)

= limh→0

�x +��h + �3−�x− �3��h(√x + h + 3 +

√x + 3)

= limh→0

1√x + ��

0

h + 3 +√x + 3

=1

2√x + 3

d. Domains Df : [−3,∞) and Df ′ : (−3,∞)

9


Higher Derivatives

If f is differentiable, the its derivative, f ′ is also a function; so f ′ too may have a derivativeof its own (provided that certain conditions are met). The derivative of f ′ is denoted as f ′′

and is called the second derivative of f . The same could be said about f ′′ having its ownderivative, f ′′′, and so on.

Notation

There are several ways to represent derivative. The two frequently used notational systemsused in calculus are: Lagrange (prime) notation, and Leibniz notation.

� Prime notation

y = f(x) y′ = f ′(x) y′′ = f ′′(x) y′′′ = f ′′′(x) y(4) = f (4)(x) . . .

� Leibniz notation

y = f(x) y′ =dy

dxy′′ =

d2y

dx2y′′′ =

d3y

dx3y(4) =

d4y

dx4. . .

Towards the end of the course, we use second order derivatives to help us sketch complicatedfunctions. Just like the first derivatives which tells us if the function is increasing or decreasingon a certain interval, the second derivative tells us if the graph of the function on that intervalis concave (opening) up or concave (opening) down.In application involving the movement of an object, suppose that s(t) represents the displace-ment of the object at time t, then the

� velocity is: v(t) = s′(t) =ds

dt

� acceleration is: a(t) = v′(t) = s′′(t) =d2s

dt2

� jerk is: j(t) = a′(t) = v′′s′′′(t) =d3s

dt3

10


Example 9

Let f(x) = x3−3x. Use the definition of a derivative to find f ′(x), f ′′(x), and f ′′(x). Thengraph on the same axes to see if your answers are reasonable.

Solution 9

We are given f(x) = x3 − 3x

f ′(x) = limh→0

f(x + h)− f(x)

h

= limh→0

(x + h)3 − 3(x + h)− (x3 − 3x)

h

= limh→0

(��x3 + 3x2h + 3xh2 + h3 −��3x− 3h−��x3 +��3x

h

= limh→0

��h(3x2 + 3xh + h2 − 3)

��h

= limh→0

(3x2 +��*0

3xh +��0

h2 − 3)

= 3x2 − 3

To get f ′′(x), simply feed f ′(x) into the definition of the derivative.

f ′′(x) = limh→0

f ′(x + h)− f ′(x)

h

= limh→0

[ 3(x + h)2 − 3 ]− (3x2 − 3)

h

= limh→0

��3x2 + 6xh + 3h2 − �3−��3x2 + �3

h

= limh→0

��h(6x + 3h)

��h

= limh→0

(6x−��>0

3h )

= 6x

To get f ′′′(x), simply feed f ′′(x) into the definition of the derivative.

f ′′′(x) = limh→0

f ′′(x + h)− f ′′(x)

h

= limh→0

[ 6(x + h) ]− (6x)

h

= limh→0

��6x + 6��h−��6x��h

= 6

11


f(x)

x

y

f ′′(x)

x

y

f ′(x)

x

y

f ′′′(x)

x

y

When is a Function Not Differentiable

Not all functions are differentiable. A function is differentiable at x = a provided that

f ′(a) = limh→0

f(a + h)− f(a)

h

exists. In other words, the (general) limit has to exist. Functions can fail to be differentiablefor several reasons: they can have a cusp/corner at x = a, be discontinuous at x = a, or havea vertical tangent line at x = a

SECTION 2.8 The Derivative as a Function 159

A third possibility is that the curve has a vertical tangent line when x − a; that is, f is continuous at a and

limx l a | f 9sxd | − `

This means that the tangent lines become steeper and steeper as x l a. Figure 6 shows one way that this can happen; Figure 7(c) shows another. Figure 7 illustrates the three possibilities that we have discussed.

(a) A corner (c) A vertical tangent(b) A discontinuity

x

y

a0 x

y

a0x

y

a0

A graphing calculator or computer provides another way of looking at differentiability. If f is differentiable at a, then when we zoom in toward the point sa, f sadd the graph straightens out and appears more and more like a line. (See Figure 8. We saw a specific example of this in Figure 2.7.2.) But no matter how much we zoom in toward a point like the ones in Figures 6 and 7(a), we can’t eliminate the sharp point or corner (see Figure 9).

x

y

a0x

y

a0

FIGURE 8f is differentiable at a.

FIGURE 9f is not differentiable at a.

■ Higher DerivativesIf f is a differentiable function, then its derivative f 9 is also a function, so f 9 may have a derivative of its own, denoted by s f 9d9 − f 0. This new function f 0 is called the second derivative of f because it is the derivative of the derivative of f. Using Leibniz notation, we write the second derivative of y − f sxd as

d

dx S dy

dxD − d 2y

dx 2

derivative of

firstderivative

second derivative

FIGURE 6

vertical tangent line

x

y

a0

FIGURE 7Three ways for f not to be

differentiable at a



Definition 2

A function f is differentiable at a of f ′(a) exists.

It is differentiable on an open interval (a, b) [or (a,∞), or (−∞, b), or(−∞,∞)] if it isdifferentiable at every number in the interval.

12


Theorem 1

If f is differentiable at a then f is continuous at a.

Remark

� The converse is not true. Continuity does not guarantee differentiability.

Example 10

The graph of f is given. State, with reasons, the numbers at which f is not differentiable.


37. The table gives the height as time passes of a typical pine tree grown for lumber at a managed site.

Tree age (years) 14 21 28 35 42 49

Height (feet) 41 54 64 72 78 83

Source: Arkansas Forestry Commission

If Hstd is the height of the tree after t years, construct a table of estimated values for H9 and sketch its graph.

38. Water temperature affects the growth rate of brook trout. The table shows the amount of weight gained by brook trout after 24 days in various water temperatures.

Temperature (°C) 15.5 17.7 20.0 22.4 24.4

Weight gained (g) 37.2 31.0 19.8 9.7 29.8

If Wsxd is the weight gain at temperature x, construct a table of estimated values for W9 and sketch its graph. What are the units for W9sxd?Source: Adapted from J. Chadwick Jr., “Temperature Effects on Growth and Stress Physiology of Brook Trout: Implications for Climate Change Impacts on an Iconic Cold-Water Fish.” Masters Theses. Paper 897. 2012. scholarworks.umass.edu/theses/897.

39. Let P represent the percentage of a city’s electrical power that is produced by solar panels t years after January 1, 2020.

(a) What does dPydt represent in this context? (b) Interpret the statement

dP

dt Zt −2

− 3.5

40. Suppose N is the number of people in the United States who travel by car to another state for a vacation in a year when the average price of gasoline is p dollars per gallon. Do you expect dNydp to be positive or negative? Explain.

41–44 The graph of f is given. State, with reasons, the numbers at which f is not differentiable.

41.

_2 2 x

y

0

42.

2 4 x

y

_2

43.

2 4 6 x

y

0

44.

_2 2 4 x

y

0

45. Graph the function f sxd − x 1 s| x | . Zoom in repeatedly, first toward the point (21, 0) and then toward the origin. What is different about the behavior of f in the vicinity of these two points? What do you conclude about the differen-tiability of f ?

46. Zoom in toward the points (1, 0), (0, 1), and (21, 0) on the graph of the function tsxd − sx 2 2 1d2y3. What do you notice? Account for what you see in terms of the differen-tiability of t.

47–48 The graphs of a function f and its derivative f 9 are shown. Which is bigger, f 9s21d or f 99s1d ?

47.

1 x

y

0

48.

x

y

0 1

49. The figure shows the graphs of f , f 9, and f 0. Identify each curve, and explain your choices.

x

ya

b

c

;

;



Solution 10

The function, f , is not differentiable at:

x = −1 because the function is discontinuous at that point

x = 2 because there is a cusp

Example 11

The graph of f is given. State, with reasons, the numbers at which f is not differentiable.



Tree age (years) 14 21 28 35 42 49

Height (feet) 41 54 64 72 78 83




Temperature (°C) 15.5 17.7 20.0 22.4 24.4

Weight gained (g) 37.2 31.0 19.8 9.7 29.8




dP

dt Zt −2

− 3.5



41.

_2 2 x

y

0

42.

2 4 x

y

_2

43.

2 4 6 x

y

0

44.

_2 2 4 x

y

0




47.

1 x

y

0

48.

x

y

0 1


x

ya

b

c

;

;



13


Solution 11

The function, f , is not differentiable at:

x = −4 because there is a cusp

x = 0 because the function is discontinuous at that point

Example 12

The graphs of a function f and its derivative f ′(x) are shown. Which is bigger, f(1) orf ′(1) ?



Tree age (years) 14 21 28 35 42 49

Height (feet) 41 54 64 72 78 83




Temperature (°C) 15.5 17.7 20.0 22.4 24.4

Weight gained (g) 37.2 31.0 19.8 9.7 29.8




dP

dt Zt −2

− 3.5



41.

_2 2 x

y

0

42.

2 4 x

y

_2

43.

2 4 6 x

y

0

44.

_2 2 4 x

y

0




47.

1 x

y

0

48.

x

y

0 1


x

ya

b

c

;

;



Solution 12

By inspection, f(x) is the blue graph, and f ′(x) is the red graph. f(1) is bigger.

Example 13

The graphs of a function f and its derivative f ′(x) are shown. Which is bigger, f(1) orf ′(1) ?



Tree age (years) 14 21 28 35 42 49

Height (feet) 41 54 64 72 78 83




Temperature (°C) 15.5 17.7 20.0 22.4 24.4

Weight gained (g) 37.2 31.0 19.8 9.7 29.8




dP

dt Zt −2

− 3.5



41.

_2 2 x

y

0

42.

2 4 x

y

_2

43.

2 4 6 x

y

0

44.

_2 2 4 x

y

0




47.

1 x

y

0

48.

x

y

0 1


x

ya

b

c

;

;



Solution 13

By inspection, f(x) is the red graph, and f ′(x) is the blue graph. f ′(1) is bigger.

14


Example 14

a. If g(x) = x2/3 show that g′(0) does not exist.

b. If a 6= 0 find g′(a)

c. Show that g(x) = x2/3 has a vertical tangent line at (0, 0)

d. Illustrate part(c) by sketching a graph of g(x) = x2/3

Solution 14

a. We are given g(x) = x2/3 and a = 0. Therefore,

g′(a) = limx→a

g(x)− g(a)

x− a

g′(0) = limx→0

g(x)− g(0)

x− 0= lim

x→0

x2/3 − 02/3

x= lim

x→0

x2/3

x= lim

x→0

1

x1/3= dne

b. If a 6= 0. Then,

g′(a) = limx→a

g(x)− g(a)

x− a= lim

x→a

x2/3 − a2/3

x− a

Let u = x1/3 v = a1/3

=u2 − v2

u3 − v3

=(u− v)(u + v)

(u− v)(u2 + uv + v2)

=��

�(u− v)(u + v)

��(u− v)(u2 + uv + v2)

=u + v

u2 + uv + v2

= limx→a

x1/3 + a1/3

x2/3 + x1/3a1/3 + a2/3

=a1/3 + a1/3

a2/3 + a1/3a1/3 + a2/3

=2a1/3

3a2/3

=2

3a1/3

c. g(x) = x2/3 is continuous at x = 0, but

limx→0|g′(x)| = lim

x→0

2

3|x|1/3=∞

This shows that g(x) has a vertical tangent line at x = 0

15


d . Below is the graph of g(x) = x2/3

f ′′′(x)

x

y

Example 15

Determine if f(x) = |x| is differentiable at x = 0. Sketch both f(x) and f ′(x)

Solution 15

For f(x) to be differentiable at x = a the limit of the expression f ′(a) = limh→0

f(a + h)− f(a)

hmust exist.

Also recall that |x| =

−x ; x < 0

0 ; x = 0

+x ; x > 0

When x < 0

f ′(0) = limh→0−

f(0 + h)− f(0)

h= lim

h→0−

|h| − 0

h= lim

h→0−

−��h��h

= −1

when x > 0

f ′(0) = limh→0+

f(0 + h)− f(0)

h= lim

h→0+

|h| − 0

h= lim

h→0+

+��h

��h= +1

∵ these limits are different, f ′(0) = dne⇒ f(x) = |x| is not differentiable at x = 0.

The graphs of f(x) = |x| and f ′(x) =

{−1 ; x < 0

+1 ; x > 0are shown below

f(x)

x

y

f(x)

1

−1

x

y

16


Example 16

a. Sketch the graph of the function f(x) = x + |x|.

b. For what values of x is f(x) differentiable?

c. Find a formula for f ′.

Solution 16

The function |x| =

−x ; x < 0

0 ; x = 0

+x ; x > 0

can be rewritten as |x| =

{−x ; x < 0

+x ; x ≥ 0

This means that we can rewrite f(x) = x + |x| as as piecewise defined function

f(x) =

{x− x ; x < 0

x + x ; x ≥ 0⇒ f(x) =

{0 ; x < 0

2x ; x ≥ 0

a. Below is the graph of the function f(x) = x + |x|.

0x

y

b. The function is differentiable at ∀x in its domain except for when x = 0. That is, fis differentiable on (−∞, 0) ∪ (0,∞)

c. Using the definition of the derivative on the f we have When x < 0

f ′(x) = limh→0

f(x + h)− f(x)

h= lim

h→0

0− 0

h= 0

When x ≥ 0

f ′(x) = limh→0

f(x + h)− f(x)

h= lim

h→0

2(x + h)− 2x

h= lim

h→0

��2x + 2��h−��2x��h

= 2

Therefore

f ′(x) =

{0 ; x < 0

2 ; x ≥ 0

17

lecture 14: the derivative as a function

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