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The Derivativea. Tangent line to a curve, p.105, figures 3.1–3b. Definition – differentiable (3.1.1), p. 106c. Equation of the tangent line, (3.1.2), p. 109d. The derivative as a function, (3.1.3), p. xxxe. Tangent lines and normal lines, p. xxx, figure xxxf. If f is differentiable at x, then f is continuous at

x, p.111

Differentiation Formulasa. Derivatives of sums, differences and scalar multiple

s, (3.2.3), (3.2.4), p. 115b. The product rule, p. 117c. The reciprocal rule, p. 119d. Derivatives of powers and polynomials, (3.2.7), (3.

2.8), pp. 117, 118e. The quotient rule, p. 121

Derivatives of higher Ordera. The d/dx notation, p. 124, 125b. Derivatives of higher order, p. 127

The Derivative as a Rate of Change

The Chain Rulea. Leibnitz form of the chain rule, p. 133b. The chain-rule theorem (3.5.6), p. 138

Chapter 3: Differentiation Topics Differentiating the Trigonometric Functionsa. Basic formulas, (3.6.1), (3.6.2), (3.6.3), (3.6.4), pp. 142, 143b. The chain rule and the trig functions, (3.6.5), p. 144c. My table of differentiation formulas

Implicit differentiation; Rational Powersa. Example 1, p.147, Figures 1.71–2b. The derivative of rational powers, (3.7.1), p. 149c. Chain-rule version, (3.7.2), p. 150

The Derivative

Tangent line to a curve, p. 105 , figures 3.1.1

The Derivative

Definition – differentiable (3.1.1), p. 106

The Derivative

Example 1, p. 106

Example 1

We begin with a linear function

f(x) = mx + b.

The graph of this function is the line y = mx + b, a line with constant slope m. We therefore expect f ’ (x) to be constantly m. Indeed it is: for h ≠ 0,

and therefore

f(x + h) – f(x)

[m(x + h) + b] – [mx + b]

.lim)()(

lim)(00

xfhxfxf

The Derivative

Example 2

Now we look at the squaring function

f(x) = x2.

To find f ’ (x), we form the difference quotient

and take the limit as h → 0. Since

Therefore

The slope of the graph changes with x. For x ＜ 0, the slope is negative and the curve tends down; at x = 0, the slope is 0 and the tangent line is horizontal; for x ＞ 0, the slope is positive and the curve tends up.

f(x + h) – f(x)

(x + h)2 – x2

(Figure 3.1.2)

(x + h)2 – x2

(x2 + 2xh + h2) – x2

2xh + h2

h= 2x + h.

Example 2, p. 106-107

.2)2(lim)()(

lim)(00

xfhxfxf

The Derivative

Example 3

Here we look for f ’ (x) for the square-root function

Since f ’ (x) is a two-side limit, we can expect a derivative at most for x ＞ 0.

We take x ＞ 0 and form the difference quotient

We simplify this expression by multiplying both numerator and denominator by This gives us

It follows that

At each point of the graph to the right of the origin the slope is positive. As x increase, the slope diminishes and the graph flattens out.

(Figure 3.1.3)

Example 3, p. 107

.0 ,)( xxxf

xhxxhxh

)()(lim)(

00 xxhxh

xfhxfxf

The Derivative

Example 4

Let’s differentiate the reciprocal function

We begin by forming the difference quotient

Now we simplify:

It follows that

The graph of the function consists of two curves. On each curve the slope, –1/x2, is negative: large negative for x close to 0 (each curve steepens as x approaches 0 and tends toward the vertical) and small negative for x far from 0 (each curve flattens out as x moves away from 0 and tends toward the horizontal).

(Figure 3.1.4)

x + h–

x(x + h)

x(x + h).

Example 4, p. 107-108

11)()(

)()(lim)(

200 xhxxh

xfhxfxf

The Derivative

Example 5

We take f(x) = 1 – x2 and calculate f ’ (–2).

Example 5, p. 108

The Derivative

Example 6

Let’s find f ’ (–3) and f ’ (1) given that

f(x) = . x2, x 1≦

2x – 1, x ＞ 1.

Example 6, p. 109

The Derivative

Equation of the tangent line, (3.1.2), p. 109

The Derivative

Example 7, p. 109

Example 7

We go back to the square-root function

and write an equation for the tangent line at the point (4, 2).

As we showed in Example 3, for x ＞ 0

Thus . The equation for the tangent line at the point (4, 2) can be written

xxf )(

1)4( f

The Derivative

Example 8

We differentiate the function

f(x) = x3 – 12x

and seek the point of the graph where the tangent line is horizontal. Then we write an equation for the tangent line at the point of the graph where x = 3.

First we calculate the difference quotient:

Now we take the limit as h → 0:

f(x + h) – f(x)

[(x + h)]3 – 12(x + h)] – [x3 – 12x]

=x3 + 3x2h + 3xh2 + h3 – 12x – 12h – x3 + 12x

h= 3x2 + 3xh + h2 – 12.

Example 8, p. 109-110

.1231233lim)()(

lim)( 222

xfhxfxf

The Derivative

The function has a horizontal tangent at the points (x, f(x)) where f ’(x) = 0. In this case

f ’(x) = 0 iff 3x2 – 12 = 0 iff x = ±2.

The graph has a horizontal tangent at the points

(–2, f(–2) ) = (–2, 16) and (2, f(2)) = (2, –16).

The graph of and the horizontal tangents are shown in Figure 3.1.5.

The point on the graph where x = 3 is the point (3, f(3)) = (3, –9). The slope at this point is f ’(3) = 15, and the equation of the tangent line at this point can be written

y + 9 = 15(x – 3).

Example 8, p. 109-110

The Derivative

Figure 3.1.5-10, p. 110-111

The Derivative

If f is differentiable at x, then f is continuous at x, p. 111

Differentiation Formulas

Derivatives of sums, differences and scalar multiples, (3.2.3), (3.2.4), p. 115, 116

The product rule, p. 117

Derivatives of powers and polynomials, (3.2.7), (3.2.8), pp. 117, 118

Example 1, p. 118

Example 1

Differentiate F(x) = (x3 – 2x + 3)(4x2 + 1) and find F ’(–1).

Example 2, p. 118

Example 2

Differentiate F(x) = (ax + b)(cx + d), where a, b, c, d are constants.

Example 3, p. 119

Example 3

Suppose that g is differentiable at each x and that F(x) = (x3 – 5x)g(x). Find F ’(2) given that g(2) = 3 and g’(2) = –1.

The reciprocal rule, p. 119-120

Example 4

Differentiate f(x) = – and find f ’( ).

Example 5, p. 120

Example 5

Differentiate f(x) = , where a, b, c are constants. 1

ax2 + bx + c

The quotient rule, p. 121

Example 6, p. 121

Example 6

Differentiate F(x) = .6x2 – 1

x4 + 5x + 1

Example 7, p. 121

Example 7

Find equations for the tangent and normal lines to the graph of

f(x) =

at the point (2, f(2)) = (2, –2).

1 – 2x

Example 8, p. 122

Example 8

Find the point on the graph of

f(x) =

where the tangent line is horizontal.

x2 + 4

Derivatives of Higher Order

The d/dx notation, p. 124–125

Example 1, p. 125

Example 1

Find for y = .dy

3x – 1

5x + 2

Example 2, p. 125

Example 2

Find for y = (x3 + 1)(3x5 + 2x – 1).dy

Example 3, p. 126

Example 3

Find .12

Example 4, p. 126

Example 4

Find for u = x(x + 1)(x + 2).dx

Example 5, p. 126

Example 5

Find dy/dx at x = 0 and x = 1 given that y = . x2

x2 – 4

Derivatives of higher order, p. 127

Example 6, p. 127

Example 6

If f(x) = x4 – 3x–1 + 5, then

f ’(x) = 4x3 + 3x–2 and f ”(x) = 12x2 – 6x–3.

Example 7, p. 127

Example 7

.2460)2420()74(

,2420)7125()74(

,7125)74(

xxxxxdx

Example 8, p. 127-128

Example 8

Finally, we consider y = x–1. In the Leibniz notation.

= –x–2, = 2x–3, = –6x–4, = 24x–5, … .

On the basis of these calculations, we are led to the general result

= (–1)nn!x–n – 1. [Recall that n! = n(n – 1)(n – 2)…3 ． 2 ． 1.]

In Exercise 61 you are asked to prove this result. In the prime notation we have

y’ = –x–2, y” = 2x–3, y’” = –6x–4, y(4) = 24x–5, … .

In general

y(n) = (–1)nn!x–n – 1.

The derivative as a rate of change, p. 130

Example 1, p. 130

Example 1

The area of a square is given by the formula A = x2 where x is the length of a side. As x changes, A changes. The rate of change of A with respect to x is the derivative

= (x2) = 2x.

When x = , this rate of change is : the area is changing at half the rate of x.

When x = , the rate of change of A with respect to x is 1: the area is changing at the

same rate as x. When x = 1, the rate of change of A with respect to x is 2: the area is changing at twice the rate of x .

In Figure 3.4.3 we have plotted A against x. The rate of change of A with respect to x at each of the indicated point appears as the slope as the slope of the tangent line.

Figure 3.4.3, p. 131

Example 2, p. 131

Example 2

An equilateral triangle of side x has area

The rate of change of A with respect to x is the derivative

When , the rate of change of A with respect to x is 3. In other words, when the side has length , the area is changing three time as fast as the length of the side.

Example 3, p. 131

Example 3

(a) Find the rate of change of y with respect to x at x = 2.

(b) Find the value(s) of x at which the rate of change of y with respect to x is 0.

Example 4, p. 131-132

Example 4

Suppose that we have a right circular cylinder of changing dimensions. (Figure 3.4.4)

When the base radius is r and the height is h, the cylinder has volume.

If r remains constant while h changes, then V can be viewed as a

function of h. The rate of change of V with respect to h is the derivative

If h remains constant while r changes, then V can be viewed as a function of r.

The rate of change of V with respect to r is the derivative

= πr2.

= 2πrh.

Example 4, p. 131-132

Suppose now that r changes but V is kept constant. How does h change with respect to r? To answer this, we express h in term of r and V

Since V is held constant, h is now a function of r. There rate of change of h with respect to r is the derivative

h = = r –2.V

= r –3 = – r –3 = – .2V

2(πr2h)

The Chain Rule

Leibnitz form of the chain rule, p. 133

The Chain Rule

Example 1, p. 133

Example 1

Find dy/dx by the chain rule given that

y = and u = x2.u – 1u + 2

The Chain Rule

Example 2, p. 135

Example 2

The Chain Rule

Example 3, p. 135

Example 3

We have

dx[1 + (2 + 3x)5]3 = 3[1 + (2 + 3x)5]2 [1 + (2 + 3x)5].

[1 + (2 + 3x)5]3 = 5(2 + 3x)4 (2 + 3x) = 5(2 + 3x)4(3) = 15(2 + 3x)4.d

[1 + (2 + 3x)5]3 = 3[1 + (2 + 3x)5]2 [15(2 + 3x)4]

= 45(2 + 3x)4[1 + (2 + 3x)5]2.

The Chain Rule

Example 4, p. 135

Example 4

Calculate the derivative of f(x) = 2x3(x2 – 3)4.

The Chain Rule

Example 5, p. 135

Example 5

Find dy/ds given that y = 3u + 1, u = x–2, x = 1 – s.

The Chain Rule

Example 6, p. 136

Example 6

Find dy/dt at t = 9 given that

. ,)73( ,

2 2 tssuu

The Chain Rule

Example 7, p. 136

Example 7

Gravel is being poured by a conveyor onto a conical pile at the constant rate of 60π cubic feet per minute. Frictional forces within the pile are such that the height is always two-thirds of the radius. How fast is the radius of the pile changing at the instant the radius is 5 feet?

The Chain Rule

The chain-rule theorem (3.5.6), p. 138

Differentiating the Trigonometric Functions

Basic formulas, (3.6.1), (3.6.2), Example 1, p. 142

Example 1

To differentiate f(x) = cos x sin x, we use the product rule:

f ’(x) = cos x (sin x) + sin x (cos x)d

Basic formulas (3.6.3), (3.6.4), p.143

Example 2, p.143

Example 2

Find f ’(π/4) for f(x) = x cot x.

Example 3, p.143

Example 3

Find . tan

Example 4, p.143

Example 4

Find an equation for the line tangent to the curve y = cos x at the point where x = π/3

Example 5, p.144

Example 5

Set f(x) = x + 2 sin x. Find the numbers x in the open interval (0, 2π) at which (a) f ’(x) = 0, (b) f ’(x) ＞ 0, (c) f ’(x) ＜ 0.

The chain rule and the trig functions, (3.6.5), p. 144

Example 6, p. 144

Example 6

dx(cos 2x) = –sin 2x (2x) = –2sin 2x.

Example 7, p. 144

Example 7

dx[sec(x2 + 1)] = sec(x2 + 1)tan(x2 + 1) (x2 + 1)

= 2x sec(x2 + 1)tan(x2 + 1).

Example 8, p. 145

Example 8

. cos sin 3)(cos)3(sin

)( cos)3(sin

)sin ()3(sin

)sin () sin(

Example 9, p. 145

Example 9

Find (sin x°).d

Implicit Differentiation; Rational Powers

Example 1, p. 147, Figures 3.71–2

Example 2, p. 147

Example 2

Assume that y is a differentiable function of x which satisfies the given equation. Use

implicit differentiation to express dy/dx in terms of x and y.

(a) 2x2y – y3 + 1 = x + 2y. (b) cos(x – y) = (2x + 1)3y.

Example 3, p. 147-148, Figures 3.7.3

Example 3

Figure 3.7.3 show the curve 2x3 + 2y3 = 9xy and the tangent line at the point (1, 2).

What is the slope of the tangent line at that point?

Example 4, p. 148

Example 4

The function y = (4 + x2)1/3 satisfies differentiation

y3 – x2 = 4.

Use implicit differentiation to express d2y/dx2 in terms of x and y.

The derivative of rational powers, (3.7.1), p. 149

Chain-rule version, (3.7.2), p. 150

Example 5, p. 150

Example 5

dx(a) [(1 + x2)1/5] = (1 + x2)–4/5(2x) = x(1 + x2)–4/5.

dx(b) [(1 + x2)2/3] = (1 – x2)–1/3(–2x) = – x(1 – x2)–1/3.

dx(c) [(1 – x2)1/4] = (1 – x2)–3/4(–2x) = – x(1 – x2)–3/4.

The first statement holds for all real x, the second for all x ≠ ±1, and the third only for

x ∊ (–1, 1).

Example 6, p. 150

Example 6

The result holds for all x ＞ 0.

)2()1)(1(

2/322/1

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