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Salas, Hille, Etgen Calculus: One and Several VariablesCopyright 2007 © John Wiley & Sons, Inc. All rights reserved.Main Menu
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
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Salas, Hille, Etgen Calculus: One and Several VariablesCopyright 2007 © John Wiley & Sons, Inc. All rights reserved.Main Menu
The Derivative
Tangent line to a curve, p. 105 , figures 3.1.1
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Salas, Hille, Etgen Calculus: One and Several VariablesCopyright 2007 © John Wiley & Sons, Inc. All rights reserved.Main Menu
The Derivative
Definition – differentiable (3.1.1), p. 106
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Salas, Hille, Etgen Calculus: One and Several VariablesCopyright 2007 © John Wiley & Sons, Inc. All rights reserved.Main Menu
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)
h=
[m(x + h) + b] – [mx + b]
b=
mh
h= m
.lim)()(
lim)(00
mmh
xfhxfxf
hh
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Salas, Hille, Etgen Calculus: One and Several VariablesCopyright 2007 © John Wiley & Sons, Inc. All rights reserved.Main Menu
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)
h=
(x + h)2 – x2
h
(Figure 3.1.2)
(x + h)2 – x2
h=
(x2 + 2xh + h2) – x2
h=
2xh + h2
h= 2x + h.
Example 2, p. 106-107
.2)2(lim)()(
lim)(00
xhxh
xfhxfxf
hh
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Salas, Hille, Etgen Calculus: One and Several VariablesCopyright 2007 © John Wiley & Sons, Inc. All rights reserved.Main Menu
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
.)()(
h
xhx
h
xfhxf
xhx
.
1
)()(
xhxxhxh
xhx
xhx
xhx
h
xhx
h
xfhxf
.2
11lim
)()(lim)(
00 xxhxh
xfhxfxf
hh
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Salas, Hille, Etgen Calculus: One and Several VariablesCopyright 2007 © John Wiley & Sons, Inc. All rights reserved.Main Menu
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)
1
x + h–
1
x
h=
– x
x(x + h)
x + h
x(x + h)
h=
–h
x(x + h)
h=
–1
x(x + h).
Example 4, p. 107-108
.1
)(x
xf
.
11)()(
hxhx
h
xfhxf
.1
)(
1lim
)()(lim)(
200 xhxxh
xfhxfxf
hh
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Salas, Hille, Etgen Calculus: One and Several VariablesCopyright 2007 © John Wiley & Sons, Inc. All rights reserved.Main Menu
The Derivative
Example 5
We take f(x) = 1 – x2 and calculate f ’ (–2).
.
Example 5, p. 108
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Salas, Hille, Etgen Calculus: One and Several VariablesCopyright 2007 © John Wiley & Sons, Inc. All rights reserved.Main Menu
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
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Salas, Hille, Etgen Calculus: One and Several VariablesCopyright 2007 © John Wiley & Sons, Inc. All rights reserved.Main Menu
The Derivative
Equation of the tangent line, (3.1.2), p. 109
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Salas, Hille, Etgen Calculus: One and Several VariablesCopyright 2007 © John Wiley & Sons, Inc. All rights reserved.Main Menu
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 )(
.2
1)(
xxf
4
1)4( f
.44
12 xy
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Salas, Hille, Etgen Calculus: One and Several VariablesCopyright 2007 © John Wiley & Sons, Inc. All rights reserved.Main Menu
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)
h=
[(x + h)]3 – 12(x + h)] – [x3 – 12x]
h
=x3 + 3x2h + 3xh2 + h3 – 12x – 12h – x3 + 12x
h= 3x2 + 3xh + h2 – 12.
Example 8, p. 109-110
.1231233lim)()(
lim)( 222
00
xhxhx
h
xfhxfxf
hh
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Salas, Hille, Etgen Calculus: One and Several VariablesCopyright 2007 © John Wiley & Sons, Inc. All rights reserved.Main Menu
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
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Salas, Hille, Etgen Calculus: One and Several VariablesCopyright 2007 © John Wiley & Sons, Inc. All rights reserved.Main Menu
The Derivative
Figure 3.1.5-10, p. 110-111
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Salas, Hille, Etgen Calculus: One and Several VariablesCopyright 2007 © John Wiley & Sons, Inc. All rights reserved.Main Menu
The Derivative
If f is differentiable at x, then f is continuous at x, p. 111
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Salas, Hille, Etgen Calculus: One and Several VariablesCopyright 2007 © John Wiley & Sons, Inc. All rights reserved.Main Menu
Differentiation Formulas
Derivatives of sums, differences and scalar multiples, (3.2.3), (3.2.4), p. 115, 116
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Salas, Hille, Etgen Calculus: One and Several VariablesCopyright 2007 © John Wiley & Sons, Inc. All rights reserved.Main Menu
Differentiation Formulas
The product rule, p. 117
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Differentiation Formulas
Derivatives of powers and polynomials, (3.2.7), (3.2.8), pp. 117, 118
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Salas, Hille, Etgen Calculus: One and Several VariablesCopyright 2007 © John Wiley & Sons, Inc. All rights reserved.Main Menu
Differentiation Formulas
Example 1, p. 118
Example 1
Differentiate F(x) = (x3 – 2x + 3)(4x2 + 1) and find F ’(–1).
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Differentiation Formulas
Example 2, p. 118
Example 2
Differentiate F(x) = (ax + b)(cx + d), where a, b, c, d are constants.
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Differentiation Formulas
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.
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Salas, Hille, Etgen Calculus: One and Several VariablesCopyright 2007 © John Wiley & Sons, Inc. All rights reserved.Main Menu
Differentiation Formulas
The reciprocal rule, p. 119-120
5
x2
6
x
1
2
Example 4
Differentiate f(x) = – and find f ’( ).
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Differentiation Formulas
Example 5, p. 120
Example 5
Differentiate f(x) = , where a, b, c are constants. 1
ax2 + bx + c
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Differentiation Formulas
The quotient rule, p. 121
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Differentiation Formulas
Example 6, p. 121
Example 6
Differentiate F(x) = .6x2 – 1
x4 + 5x + 1
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Differentiation Formulas
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).
3x
1 – 2x
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Differentiation Formulas
Example 8, p. 122
Example 8
Find the point on the graph of
f(x) =
where the tangent line is horizontal.
4x
x2 + 4
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Derivatives of Higher Order
The d/dx notation, p. 124–125
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Derivatives of Higher Order
Example 1, p. 125
Example 1
Find for y = .dy
dx
3x – 1
5x + 2
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Derivatives of Higher Order
Example 2, p. 125
Example 2
Find for y = (x3 + 1)(3x5 + 2x – 1).dy
dx
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Derivatives of Higher Order
Example 3, p. 126
Example 3
Find .12
3
t
tt
dt
d
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Derivatives of Higher Order
Example 4, p. 126
Example 4
Find for u = x(x + 1)(x + 2).dx
du
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Derivatives of Higher Order
Example 5, p. 126
Example 5
Find dy/dx at x = 0 and x = 1 given that y = . x2
x2 – 4
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Derivatives of Higher Order
Derivatives of higher order, p. 127
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Derivatives of Higher Order
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.
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Derivatives of Higher Order
Example 7, p. 127
Example 7
.2460)2420()74(
,2420)7125()74(
,7125)74(
23353
3
32452
2
2435
xxxdx
dxxx
dx
d
xxxdx
dxxx
dx
d
xxxxxdx
d
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Derivatives of Higher Order
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.
dy
dx
d2y
dx2
d3y
dx3
d4y
dx4
dny
dxn
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The Derivative as a Rate of Change
The derivative as a rate of change, p. 130
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The Derivative as a Rate of Change
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.
dA
dx
d
dx
1
4
1
2
1
2
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The Derivative as a Rate of Change
Figure 3.4.3, p. 131
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The Derivative as a Rate of Change
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.
.34
1 2xA
32x32
.32
1x
dx
dA
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The Derivative as a Rate of Change
Example 3, p. 131
Example 3
Set
(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.
.2
2x
xy
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The Derivative as a Rate of Change
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
dVdh
= πr2.
dVdh
= 2πrh.
.2hrV
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The Derivative as a Rate of Change
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
πr2
V
π
dhdr
= r –3 = – r –3 = – .2V
π
2(πr2h)
π
2h
r
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The Chain Rule
Leibnitz form of the chain rule, p. 133
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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
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The Chain Rule
Example 2, p. 135
Example 2
.1
11
311
31
2
443
xxx
xx
dx
d
xx
xx
dx
d
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The Chain Rule
Example 3, p. 135
Example 3
Since
We have
d
dx
d
dx
d
dx[1 + (2 + 3x)5]3 = 3[1 + (2 + 3x)5]2 [1 + (2 + 3x)5].
d
dx
[1 + (2 + 3x)5]3 = 5(2 + 3x)4 (2 + 3x) = 5(2 + 3x)4(3) = 15(2 + 3x)4.d
dx
[1 + (2 + 3x)5]3 = 3[1 + (2 + 3x)5]2 [15(2 + 3x)4]
= 45(2 + 3x)4[1 + (2 + 3x)5]2.
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The Chain Rule
Example 4, p. 135
Example 4
Calculate the derivative of f(x) = 2x3(x2 – 3)4.
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The Chain Rule
Example 5, p. 135
Example 5
Find dy/ds given that y = 3u + 1, u = x–2, x = 1 – s.
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The Chain Rule
Example 6, p. 136
Example 6
Find dy/dt at t = 9 given that
. ,)73( ,
1
2 2 tssuu
uy
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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?
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The Chain Rule
The chain-rule theorem (3.5.6), p. 138
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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
dx
d
dx
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Differentiating the Trigonometric Functions
Basic formulas (3.6.3), (3.6.4), p.143
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Differentiating the Trigonometric Functions
Example 2, p.143
Example 2
Find f ’(π/4) for f(x) = x cot x.
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Differentiating the Trigonometric Functions
Example 3, p.143
Example 3
Find . tan
sec1
x
x
dx
d
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Differentiating the Trigonometric Functions
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
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Differentiating the Trigonometric Functions
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.
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Differentiating the Trigonometric Functions
The chain rule and the trig functions, (3.6.5), p. 144
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Differentiating the Trigonometric Functions
Example 6, p. 144
Example 6
d
dx(cos 2x) = –sin 2x (2x) = –2sin 2x.
d
dx
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Differentiating the Trigonometric Functions
Example 7, p. 144
Example 7
d
dx[sec(x2 + 1)] = sec(x2 + 1)tan(x2 + 1) (x2 + 1)
= 2x sec(x2 + 1)tan(x2 + 1).
d
dx
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Differentiating the Trigonometric Functions
Example 8, p. 145
Example 8
. cos sin 3)(cos)3(sin
)( cos)3(sin
)sin ()3(sin
)sin () sin(
22
2
2
33
xxxx
xdx
dxx
xdx
dx
xdx
dx
dx
d
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Differentiating the Trigonometric Functions
Example 9, p. 145
Example 9
Find (sin x°).d
dx
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Implicit Differentiation; Rational Powers
Example 1, p. 147, Figures 3.71–2
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Implicit Differentiation; Rational Powers
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.
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Implicit Differentiation; Rational Powers
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?
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Implicit Differentiation; Rational Powers
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.
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Implicit Differentiation; Rational Powers
The derivative of rational powers, (3.7.1), p. 149
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Implicit Differentiation; Rational Powers
Chain-rule version, (3.7.2), p. 150
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Implicit Differentiation; Rational Powers
Example 5, p. 150
Example 5
d
dx(a) [(1 + x2)1/5] = (1 + x2)–4/5(2x) = x(1 + x2)–4/5.
1
5
2
5
d
dx(b) [(1 + x2)2/3] = (1 – x2)–1/3(–2x) = – x(1 – x2)–1/3.
2
3
4
3
d
dx(c) [(1 – x2)1/4] = (1 – x2)–3/4(–2x) = – x(1 – x2)–3/4.
1
4
1
2
The first statement holds for all real x, the second for all x ≠ ±1, and the third only for
x ∊ (–1, 1).
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Implicit Differentiation; Rational Powers
Example 6, p. 150
Example 6
The result holds for all x > 0.
.)1(2
1
)1(
11
2
1
)1(
)2()1)(1(
12
1
112
1
1
2/322/1
2
22
22/12
22
22/1
2
2
2/1
2
2/1
2
xx
x
x
x
x
x
x
xxx
x
x
x
x
dx
d
x
x
x
x
dx
d