Chapter 3 Shortcuts to Differentiation Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013...

Chapter 3 Shortcuts to Differentiation

Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013 by John Wiley & Sons, All Rights Reserved

Section 3.1 Powers and PolynomialsSection 3.2 The Exponential FunctionSection 3.3 The Product and Quotient RulesSection 3.4 The Chain RuleSection 3.5 The Trigonometric Functions

Section 3.7 Implicit Functions

Section 3.10 Theorems About Differentiable Functions

Section 3.6 The Chain Rule and Inverse Functions

Section 3.8 Hyperbolic FunctionsSection 3.9 Linear Approximation and the Derivative

Section 3.1

Powers and Polynomials


Derivative of a Constant Times a Function

Theorem 3.1: Derivative of a Constant MultipleIf f is differentiable and c is a constant

Figure 3.1 A function and its multiples: Derivative of multiple is multiple of derivative

)(')( xfcxfcdx

d


Theorem 3.2: Derivative of Sum and DifferenceIf f and g are differentiable, then

)()()()( xgdx

dxf

dx

dxgxf

dx

d

Proof using the definition of the derivative:

)(')('

)()(lim

)()(lim

)()()()(lim)()(

00

0

xgxfh

xghxg

h

xfhxfh

xgxfhxghxfxgxf

dx

d

hh

h


Example 1 Use the power rule to differentiate (a) 1/x3 , (b) x1/2 , (c)

Solution(a) For n = − 3:

(b) For n = 1/2:

(c) For n = − 1/3:

3/1 x

The Power RuleFor any constant real number n,

1 nn xnxdx

d

4

41333

333

1

xxxx

dx

d

xdx

d

xxxx

dx

d

2

1

2

1

2

1 2/112/12/1

3/4

3/413/13/1

3 3

1

3

1

3

11

xxxx

dx

d

xdx

d


Example 6If the position of a body, in meters, is given as a function of time t, in seconds, by

s = − 4.9t2 + 5t + 6,find the velocity and acceleration of the body at time t.

Solution The velocity, v, is the derivative of the position:

v = ds/dt = − 9.8t +5and the acceleration, a, is the derivative of the velocity:

a = dv/dt = − 9.8

Note that v is in meters/second and a is in meters/second2.


4 2 2 4 6 8 10

250

200

150

100

50

50

100

Finding and Visualizing Derivatives of Polynomials

Exercise 56 The graph of y(x) = x3 − 9x2 − 16x + 1 has a slope of 5 at two points. Find the coordinates of the points.

Solution y’(x) = 3x2 − 18x − 16.Setting 3x2 − 18x − 16 = 5,3(x2 − 6x − 7) = 0 .So x = 7 or -1.At red points, the slope of theblue graph, y(x), is 5.

x

Cubicy(x)

Parabolay’(x)(-1,7)

(7,-209)


Section 3.2The Exponential

Function


Derivatives of Exponential Functions

h (2h- 1)/h

-0.1 0.6697-0.01 0.6908-0.001 0.69290.001 0.69340.01 0.69560.1 0.7177

We start by calculating the derivative of g(x) = 2x, which is given by

xh

h

xh

h

xhx

h

h

h

hg'(x)

212

lim

212lim

22lim

0

0

0

Table 3.2 lists values of (2h- 1)/h as h → 0.

We might conclude that g’(x) ≈ (0.693) 2x .


Exploring the Derivative of ax

The derivative of 2x is proportional to 2x with constant of proportionality 0.693. A similar calculation shows that the derivative of f(x) = ax is

And the value of this limit is explored in Table 3.3.Is there a value of “a” between 2 and 3, such that this limit will be 1?

xh

ha

h

af'(x)

1lim

0

a

2 0.6933 1.0994 1.3865 1.6096 1.7927 1.946

h

ah

h

1lim

0

Table 3.3


The Natural Number e

If , then

for small values of h,(ah -1)/h ≈ 1 orah ≈ 1 + h , a ≈ (1 + h)1/h

We define the value of a that makes our constant of proportionality 1 as .

This limit is explored in the table on the right.

1

1lim

0

h

ah

hh (1+h)1/h

−0.001 2.719642

−0.0001 2.718418

−0.00001 2.718295

0.00001 2.718268

0.0001 2.718146

0.001 2.716924

h

hhe /1

01lim

Table 3.4


Formula for the Derivative of ax

Based on Table 3.4, the number e ≈ 2.718… and this irrational number is the base of the natural logarithms. Using this fact, a = eln a and

aaat

e

h

e

ht hath

e

h

e

h

a

t

t

ha

h

ha

h

a

h

h

h

h

lnln1)(ln1

lim1

lim

and 0 as0,)ln( Letting

1lim

1lim

1lim

0

)(ln

0

)(ln

0

ln

00

xxxx eedx

daaa

dx

d and )(ln


Exercise 41In 2009, the population of Mexico was 111 million and growing 1.13% annually, while the population of the US was 307 million and growing 0.975% annually. If we measure growth rates in people per year, which population was growing faster in 2009?Solution Country Mexico United StatesPopulation Function

111 (1.0113)t 307 (1.00975)t

Population Rate of Change

111 ln(1.0113) (1.0113)t

307 ln(1.00975) (1.00975)t

Rate of Growth in 2009 (t = 0)

1.24727 million/year

2.97875 million/year


Section 3.3

The Product and Quotient Rules


Difference of a Productf(x + h)g(x + h) − f(x)g(x) = (Area of whole rectangle) − (Unshaded area) = Area of the three shaded rectangles

= Δ f ・ g(x) + f(x) ・ Δ g + Δ f ・ Δ g.

Figure 3.13: Illustration for the product rule (with Δf, Δg positive)


Theorem 3.3: The Product RuleIf u = f(x) and v = g(x) are differentiable, then

(f g) = f g + f g .′ ′ ′The product rule can also be written

In words:The derivative of a product is the derivative of the first times the second plus the first times the derivative of the second.

dx

dvuv

dx

duuv

dx

d)(


Exercise 6Find the derivative of y = (t2 + 3) et

Solutiondy/dt = (2t) et + (t2 + 3) et = (t2 + 2t +3) et

Exercise 54Let f(3) = 6, g(3) = 12, f′(3) = 1/2 , and g (3) = 4/3 .′Evaluate the following when x = 3.

(f(x)g(x)) − (′ g(x) − 4f′(x)) SolutionAt x = 3, (fg)’= f’(3)g(3) + f(3)g’(3) = (1/2)(12)+6(4/3) = 6+8 =14So at x = 3, (f(x)g(x)) − (′ g(x) − 4f′(x)) = 14 – (12-4·(1/2))

= 14 – 10 = 4


Theorem 3.4: The Quotient RuleIf u = f(x) and v = g(x) are differentiable, then

(f /g) = (f g - f g )/g′ ′ ′ 2

or equivalently,

In words:The derivative of a quotient is the derivative of the numerator times the denominator minus the numerator times the derivative of the denominator, all over the denominator squared.

2vdx

dvuv

dx

du

v

u

dx

d


Example 2aDifferentiate 5x2 /(x3 + 1)

Solution

23

4

23

44

23

223

23

3232

3

2

1

105

1

151010

1

35110

1

15)1(5

1

5

x

xx

x

xxx

x

xxxx

x

xdx

dxxx

dx

d

x

x

dx

d


Section 3.4

The Chain Rule


Intuition Behind the Chain RuleImagine we are moving straight upward in a hot air balloon. Let y be our distance from the ground. The temperature, H, is changing as a function of altitude, so H = f(y). How does our temperature change with time? Since temperature is a function of height, H = f(y), and height is a function of time, y = g(t), we can think of temperature as a composite function of time, H = f(g(t)), with f as the outside function and g as the inside function. The example suggests the following result, which turns out to be true:

Rate of change of composite function = Rate of change of

outside function × Rate of change of inside function

dt

dg

dg

df

dt

df

t

g

g

f

t

f

or

The Derivative of a Composition of FunctionsSymbolically, for our H = f(g(t)):


Theorem 3.5: The Chain RuleIf f and g are differentiable functions, then

In words:The derivative of a composite function is the product of the derivatives of the outside and inside functions. The derivative of the outside function must be evaluated at the inside function.

).('))(('))(( xgxgfxgfdx

d

Example 2aFind the derivative of (x2 + 1)100 .Solution

9929921002 1200211001 xxxxxdx

d


Example 4 Differentiate

Solution The chain rule is needed four times:

231 xe

2

2

2

2

2

32

3

2

3

3

3

132

232

1

12

11

x

x

x

x

x

ex

xe

xx

ee

edx

d


Section 3.5

The Trigonometric Functions


The Sine Function and Its Derivative Graphical Perspective

Figure 3.22: The sine function

First we might ask where the derivative is zero. Then ask where the derivative is positive and where it is negative.In exploring these answers, we get something like the following graph.

Figure 3.23: Derivative of f(x) = sin x

The graph of the derivative in Figure 3.23 looks suspiciously like the graph of the cosine function. This might lead us to conjecture, quite correctly, that the derivative of the sine is the cosine.


The Cosine Function and Its Derivative Graphical Perspective

Example 2 Starting with the graph of the cosine function, sketch a graph of its derivative.Solution Looking at the graph of g(x) = cos x its derivative is 0 at x = 0, ±π, ± 2π, …It’s derivative (slope) is positive for (- π,0), (π, 2π), (3π, 4π), …It’s derivative (slope) is negative for (- 2π, π), (0, π), (2π, 3π), …

Figure 3.24: g(x) = cos x and its derivative, g (′ x)

The derivative of the cosine in Figure 3.24(b) looks exactly like the graph of sine, except reflected across the x-axis.


Examples

Solution


Derivative of the Tangent FunctionSince tan x = sin x/cos x, we differentiate tan x using the quotient rule. Writing (sin x) for d(sin x)/dx, we have:′

xx

x

xx

x

xxxx

x

xxxx

x

x

dx

dx

dx

d

22

2

22

2

2

seccos

1cos

sincos

cos

sinsincoscos

cos

'cossincos'sin

cos

sintan

For x in radians,x

xx

dx

d 22

seccos

1tan

Exercise 30 Find the derivative of g(z) = tan ez .Solution g’(z) = ez/(cos2 ez)


Section 3.6

The Chain Rule and Inverse Functions


We use the chain rule to differentiate an identity involving ln x. Since eln x = x, on the one hand we have

However, we can also use the chain rule to get

Equating these to one another, we have our desired result.

Derivative of ln x

.1ln xdx

de

dx

d x

.lnlnlnln xdx

dxx

dx

dee

dx

d xx

xx

dx

d 1ln


Derivative of ax RevisitedIn Section 3.2, we saw that the derivative of ax is proportional to ax. Now we see another way of calculating the constant of proportionality. We use the identity

ln(ax) = x ln a.Differentiating both sides, using the chain rule, and remembering that ln a is a constant, we obtain:

So we have the result from Section 3.2.

aadx

d

aa

dx

d xx

x ln1

ln

xx aaadx

dln


Derivatives of Inverse Trigonometric FunctionsTo find d/dx (arctan x) we use the identity tan(arctan x) = x. Differentiating both sides and using the chain rule gives

So

Using the identity 1 + tan2θ = 1/cos2θ, and replacing θ by arctan x,

Thus we have

1arctanarctancos

1arctantan

2 x

dx

d

xx

dx

d

xxdx

darctancosarctan 2

222

1

1

arctantan1

1arctancos

xxx

21

1arctan

xx

dx

d


Derivative of the Arcsine and Examples

By a similar argument, we obtain the result:

21

1arcsin

xx

dx

d

Example 2 Differentiate (a) arctan(t2) (b) arcsin(tan θ).Solution Use the chain rule:(a)

(b)

4

222

2

1

2

1

1arctan

t

tt

dt

d

tt

dt

d

222 cos

1

tan1

1tan

tan1

1tanarcsin

d

d

d

d


Derivative of a General Inverse FunctionIn general, if a function f has a differentiable inverse, f−1, we find its derivative by differentiating f(f−1(x)) = x by the chain rule, yielding the following result:

)('

1)(

11

xffxf

dx

d

x f(x) f (x′ )3 1 76 2 109 3 5

Exercise 65 Use the table and the fact that f(x) is invertible and differentiable everywhere to find (f−1) (3).′

SolutionBegin by observing that f−1(3)=9, since f(9) = 3Then (f−1) (3) = 1/′ f’(9) = 1/5


Section 3.7

Implicit Functions


What Is an Implicit Function?In earlier chapters, most functions were written in the form y = f(x); here y is said to be an explicit function of x. An equation such as

x2 + y2 = 4is said to give y as an implicit function of x. Its graph is the circle below. Since there are x-values which correspond to two y-values, y is not a function of x on the whole circle.

Figure 3.35: Graph of x2 + y2 = 4

Note that y is a function of x on the top half, and y is a different function of x on the bottom half.


Differentiating ImplicitlyLet us consider the circle as a whole. The equation does represent a curve which has a tangent line at each point. The slope of this tangent can be found by differentiating the equation of the circle with respect to x:

d/dx (x2) + d/dx (y2) = d/dx (4).If we think of y as a function of x and use the chain rule, we get

2x + 2y dy/dx = 0.Solving gives dy/dx = − x/y.The derivative here depends on both x and y (instead of just on x). Differentiating the equation of the circle has given us the slope of the curve at all points except (2, 0) and (−2, 0), where the tangent is vertical. In general, this process of implicit differentiation leads to a derivative whenever the expression for the derivative does not have a zero in the denominator.


Example 2 Find all points where the tangent line to y3 − x y = −6 is either horizontal or vertical.Solution Differentiating implicitly, 3y2·dy/dx–y–x·dy/dx=0 So dy/dx = y/(3y2 − x). The tangent is horizontal when the numerator of dy/dx equals 0, so y = 0. Since we also must satisfy y3 − x y = −6, we get 0=-6, which is impossible. We conclude that there are no points on the curve where the tangent line is horizontal.The tangent is vertical when the denominator of dy/dx is 0, giving 3y2−x = 0. Thus, x = 3y2 at any point with a vertical tangent line. Again, we must also satisfy y3 − x y = −6, so y3=3. Solving for x, with this value of y, we conclude there is a vertical tangent at the point (6.240,1.442).

1 0 2 0 3 0 4 0 5 0x

2

4

6

8y

y3 − x y = −6

·(6.240,1.442)

The figure below verifies graphically what was determined analytically.


Section 3.8

Hyperbolic Functions


Hyperbolic Functions

2sinh

2cosh

xxxx eex

eex

Graphs of Hyperbolic Cosine and Sine

Figure 3.37: Graph of y = cosh x Figure 3.38: Graph of y = sinh x


Properties of Hyperbolic Functions

cosh 0 = 1 sinh 0 = 0

cosh(−x) = cosh x sinh(−x) = −sinh x

Example 2Describe and explain the behavior of cosh x as x → ∞ and as x → −∞.Solution From Figure 3.37, it appears that as x → ∞, the graph of cosh x resembles the graph of ½ ex. Similarly, as x → −∞, the graph of cosh x resembles the graph of ½ e-x. This behavior is explained by using the formula for cosh x and the facts that e−x → 0 as x → ∞ and ex → 0 as x → −∞:


Identity Involving cosh x and sinh xcosh2 x − sinh2 x = 1

This identity shows us how the hyperbolic functions got their name. Suppose (x, y) is a point in the plane and x = cosh t and y = sinh t for some t. Then the point (x, y) lies on the hyperbola x2 − y2 = 1.Extending the analogy to the trigonometric functions, we define the hyperbolic tangent.

The Hyperbolic Tangent

xx

xx

ee

ee

x

xx

cosh

sinhtanh

4 2 2 4

1.0

0.5

0.5

1.0

x

y =tanh x

y


Derivatives of Hyperbolic FunctionsWe calculate the derivatives using the fact that d/dx (ex) = ex. The results are again reminiscent of the trigonometric functions.

xxdx

dxx

dx

dcoshsinhsinhcosh

Example 3 Compute the derivative of tanh x.Solution Using the quotient rule gives

22

22

cosh

1

cosh

sinhcosh

cosh

sinhtanh

xx

xx

x

x

dx

dx

dx

d


Section 3.9

Linear Approximation

and the Derivative


The Tangent Line Approximation

Suppose f is differentiable at a. Then, for values of x near a, the tangent line approximation

to f(x) is

f(x) ≈ f(a) + f (′ a)(x − a).

The expression f(a)+ f (′ a)(x −a) is called the local linearization of f near x = a. We are thinking of a as fixed, so that f(a) and f (′ a) are constant.

The error, E(x), in the approximation is defined by

E(x) = f(x) − f(a) − f (′ a)(x − a).


It can be shown that the tangent line approximation is the best linear approximation to f near a.

Figure 3.40: The tangent line approximation and its error

Visualization of the Tangent Line Approximation


Example 1 What is the tangent line approximation for f(x) = sin x near x = 0?Solution The tangent line approximation of f near x = 0 is

f(x) ≈ f(0) + f (0)(′ x − 0).If f(x) = sin x, then f (′ x) = cos x, so f(0) = sin 0 = 0 and f (0) = cos 0 = 1, and ′the approximation is

sin x ≈ x.This means that, near x = 0, the function f(x) = sin x is well approximated by the function y = x. If we zoom in on the graphs of the functions sin x and x near the origin, we won’t be able to tell them apart.

Figure 3.41: Tangent line approximation to y = sin x


Estimating the Error in Linear ApproximationTheorem 3.6: Differentiability and Local LinearitySuppose f is differentiable at x = a and E(x) is the error in the tangent line approximation, that is:

E(x) = f(x) − f(a) − f (′ a)(x − a).Then

0)(

lim ax

xEax

Proof Using the definition of E(x), we have

Taking the limit as x → a and using the definition of the derivative, we see that

An error estimate that will be developed later is

)(')()())((')()()(

afax

afxf

ax

axafafxf

ax

xE

0)(')(')(')()(

lim)(

lim

afafafax

afxf

ax

xE

axax

2

2

)('')( ax

afxE


Example 3 Let E(x) be the error in the tangent line approximation to f(x) = x3 − 5x + 3 for x near 2.(a) What does a table of values for E(x)/(x − 2) suggest about the limit of

this ratio as x→2?(b) Make another table to see that E(x) ≈ k(x − 2) 2. Estimate the value of

k. Check that a possible value is k = f (2)/2.′′

Solution Since f(x) = x3 − 5x + 3 , we have f (′ x) = 3x2 − 5, and f (′′ x) = 6x. Thus, f(2) = 1 and f (2) = 3·2′ 2 − 5 = 7, so the tangent line approximation for x near 2 is

f(x) ≈ f(2) + f (2)(′ x − 2) ≈ 1 + 7(x − 2).E(x) = True value − Approximation = (x3 − 5x + 3) − (1 + 7(x − 2)).

x E(x)/(x − 2) E(x)/(x − 2)2

2.1 0.61 6.1

2.01 0.0601 6.01

2.001 0.006001 6.001

2.0001 0.0006 6.0001

These values suggest that E(x)/(x − 2) → 0 as x → 2 and E(x)/(x − 2)2 → 6 as x → 2. So E(x) ≈ 6(x − 2) 2

Also note that f (2)/2 = 12/2=6′′


Section 3.10

Theorems About Differentiable

Functions


A Relationship Between Local and Global: The Mean Value Theorem

Theorem 3.7: The Mean Value TheoremIf f is continuous on a ≤ x ≤ b and differentiable on a < x < b, then there exists a number c, with a < c < b, such that

f (c) = (′ f(b) − f(a))/(b − a).In other words, f(b) − f(a) = f (c)(′ b − a).

To understand this theorem geometrically, look at Figure 3.44. Join the points on the curve where x = a and x = b with a secant line and observe that the slope of secant line = (f(b) − f(a))/(b − a).

Figure 3.44

There appears to be at least one point between a and b where the slope of the tangent line to the curve is precisely the same as the slope of the secant line.


Theorem 3.8: The Increasing Function TheoremSuppose that f is continuous on a ≤ x ≤ b and differentiable on a < x < b.• If f (′ x) > 0 on a < x < b, then f is increasing on a ≤ x ≤ b.• If f (′ x) ≥ 0 on a < x < b, then f is nondecreasing on a ≤ x ≤ b.

Theorem 3.9: The Constant Function TheoremSuppose that f is continuous on a ≤ x ≤ b and differentiable on a < x < b. If f (′ x) = 0 on a < x < b, then f is constant on a ≤ x ≤ b.

Theorem 3.10: The Racetrack PrincipleSuppose that g and h are continuous on a ≤ x ≤ b and differentiable on a < x < b, and that g (′ x) ≤ h (′ x) for a < x < b.• If g(a) = h(a), then g(x) ≤ h(x) for a ≤ x ≤ b.• If g(b) = h(b), then g(x) ≥ h(x) for a ≤ x ≤ b.


Chapter 3 Shortcuts to Differentiation Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013...

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Transcript of Chapter 3 Shortcuts to Differentiation Calculus, 6th edition, Hughes-Hallett et. al., Copyright 2013...