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Chapter 3: Differentiation Topics
The Derivative
a. Tangent line to a curve, p.105, figures 3.1–3
b. Definition – differentiable (3.1.1), p. 106
c. Equation of the tangent line, (3.1.2), p. 109
d. The derivative as a function, (3.1.3), p. xxx
e. Tangent lines and normal lines, p. xxx, figure xxx
f. If f is differentiable at x, then f is continuous at
x, p.111
Differentiating the Trigonometric Functions
Basic formulas, (3.6.1), (3.6.2), (3.6.3), (3.6.4), pp. 142, 143
b.
The chain rule and the trig functions, (3.6.5), p. 144
c.
My table of differentiation formulas
a.
Implicit differentiation; Rational Powers
Example 1, p.147, Figures 1.71–2
b.
The derivative of rational powers, (3.7.1), p. 149
c.
Chain-rule version, (3.7.2), p. 150
a.
Differentiation Formulas
a. Derivatives of sums, differences and scalar
multiples, (3.2.3), (3.2.4), p. 115
b. The product rule, p. 117
c. The reciprocal rule, p. 119
d. Derivatives of powers and polynomials, (3.2.7),
(3.2.8), pp. 117, 118
e. The quotient rule, p. 121
Derivatives of higher Order
a. The d/dx notation, p. 124, 125
b. Derivatives of higher order, p. 127
The Derivative as a Rate of Change
The Chain Rule
a. Leibnitz form of the chain rule, p. 133
b. The chain-rule theorem (3.5.6), p. 138
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The Derivative
Tangent line to a curve, p. 105 , figures 3.1.1
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The Derivative
Definition – differentiable (3.1.1), p. 106
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The Derivative
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,
f(x + h) – f(x)
h
[m(x + h) + b] – [mx + b]
=
=
b
mh
=m
h
and therefore
f ( x)  lim
h 0
f ( x  h)  f ( x )
 lim m  m.
h 0
h
Example 1, p. 106
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The Derivative
Example 2
Now we look at the squaring function
f(x) = x2.
(Figure 3.1.2)
To find f ’ (x), we form the difference quotient
(x + h)2 – x2
f(x + h) – f(x)
=
h
h
and take the limit as h → 0. Since
(x2 + 2xh + h2) – x2
(x + h)2 – x2
h
=
h
2xh + h2
=
h
= 2x + h.
Therefore
f ( x)  lim
h 0
f ( x  h)  f ( x )
 lim (2 x  h)  2 x.
h 0
h
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.
Example 2, p. 106-107
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The Derivative
Example 3
Here we look for f ’ (x) for the square-root function
f ( x)  x ,
x  0.
(Figure 3.1.3)
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
f ( x  h)  f ( x )
xh  x

.
h
h
We simplify this expression by multiplying both numerator and denominator by
gives us
f ( x  h)  f ( x)  x  h  x  x  h  x 


 



h
h

 x  h  x 
x  h   x 
1

.
h
x

h

x
x

h

x
It follows that

f ( x)  lim
h 0
x  h  x This

f ( x  h)  f ( x )
1
1
 lim

.
h 0
h
xh  x 2 x
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.
Example 3, p. 107
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The Derivative
Example 4
Let’s differentiate the reciprocal function
1
f ( x)  .
x
We begin by forming the difference quotient
1
1

f ( x  h)  f ( x ) x  h x

.
h
h
Now we simplify:
1
–
x+h
(Figure 3.1.4)
1
x
x
x(x + h)
–
x+h
–h
x(x + h)
x(x + h)
=
=
h
h
h
–1
=
.
x(x + h)
It follows that
f ( x)  lim
h 0
f ( x  h)  f ( x )
1
1
 lim
 2.
h 0 x ( x  h)
h
x
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).
Example 4, p. 107-108
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The Derivative
Example 5
We take f(x) = 1 – x2 and calculate f ’ (–2).
.
Example 5, p. 108
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The Derivative
Example 6
Let’s find f ’ (–3) and f ’ (1) given that
f(x) =
x2,
2x – 1,
x≦1
.
x > 1.
Example 6, p. 109
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The Derivative
Equation of the tangent line, (3.1.2), p. 109
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The Derivative
Example 7
We go back to the square-root function
f ( x)  x
and write an equation for the tangent line at the point (4, 2).
As we showed in Example 3, for x > 0
f ( x) 
Thus f (4) 
1
2 x
.
1
. The equation for the tangent line at the point (4, 2) can be written
4
y2
1
x  4.
4
Example 7, p. 109
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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:
f(x + h) – f(x)
h
=
[(x + h)]3 – 12(x + h)] – [x3 – 12x]
h
x3 + 3x2h + 3xh2 + h3 – 12x – 12h – x3 + 12x
=
= 3x2 + 3xh + h2 – 12.
h
Now we take the limit as h → 0:
f ( x)  lim
h 0
f ( x  h)  f ( x )
 lim 3x 2  3xh  h 2  12  3x 2  12.
h 0
h

Example 8, p. 109-110
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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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The Derivative
Figure 3.1.5-10, p. 110-111
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The Derivative
If f is differentiable at x, then f is continuous at x, p. 111
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Differentiation Formulas
Derivatives of sums, differences and scalar multiples, (3.2.3), (3.2.4), p. 115, 116
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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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Differentiation Formulas
Example 1
Differentiate F(x) = (x3 – 2x + 3)(4x2 + 1) and find F ’(–1).
Example 1, p. 118
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Differentiation Formulas
Example 2
Differentiate F(x) = (ax + b)(cx + d), where a, b, c, d are constants.
Example 2, p. 118
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Differentiation Formulas
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.
Example 3, p. 119
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Differentiation Formulas
Example 4
Differentiate f(x) =
5
x2
–
6
x
and find f ’(
1
).
2
The reciprocal rule, p. 119-120
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Differentiation Formulas
Example 5
Differentiate f(x) =
1
, where a, b, c are constants.
ax2 + bx + c
Example 5, p. 120
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Differentiation Formulas
The quotient rule, p. 121
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Differentiation Formulas
Example 6
Differentiate F(x) =
6x2 – 1
x4
.
+ 5x + 1
Example 6, p. 121
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Differentiation Formulas
Example 7
Find equations for the tangent and normal lines to the graph of
3x
f(x) =
1 – 2x
at the point (2, f(2)) = (2, –2).
Example 7, p. 121
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Differentiation Formulas
Example 8
Find the point on the graph of
4x
f(x) =
x2 + 4
where the tangent line is horizontal.
Example 8, p. 122
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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
Find
dy
dx
for y =
3x – 1
.
5x + 2
Example 1, p. 125
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Derivatives of Higher Order
Example 2
Find
dy
for y = (x3 + 1)(3x5 + 2x – 1).
dx
Example 2, p. 125
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Derivatives of Higher Order
Example 3
d  3
t 
t


.
Find
2
dt 
t 1 
Example 3, p. 126
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Derivatives of Higher Order
Example 4
Find
du
for u = x(x + 1)(x + 2).
dx
Example 4, p. 126
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Derivatives of Higher Order
Example 5
x2
Find dy/dx at x = 0 and x = 1 given that y =
x2
–4
.
Example 5, p. 126
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Derivatives of Higher Order
Derivatives of higher order, p. 127
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Derivatives of Higher Order
Example 6
If f(x) = x4 – 3x–1 + 5, then
f ’(x) = 4x3 + 3x–2
and
f ”(x) = 12x2 – 6x–3.
Example 6, p. 127
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Derivatives of Higher Order
Example 7
d 5
( x  4 x 3  7 x)  5 x 4  12 x 2  7,
dx
d2 5
d
4
2
3
(
x

4
x

7
x
)

(
5
x

12
x

7
)

20
x
 24,
2
dx
dx
d3 5
d
3
3
2
(
x

4
x

7
x
)

(
20
x

24
x
)

60
x
 24.
3
dx
dx
Example 7, p. 127
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Derivatives of Higher Order
Example 8
Finally, we consider y = x–1. In the Leibniz notation.
dy
=
–x–2,
dx
d2y
= 2x–3,
dx2
d3y
=
–6x–4,
dx3
d4y
= 24x–5, … .
dx4
On the basis of these calculations, we are led to the general result
dny
dxn
= (–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.
Example 8, p. 127-128
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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
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
dA
dx
When x =
1
4
=
d
(x2) = 2x.
dx
, this rate of change is
1
: the area is changing at half the rate of x.
2
1
When x =
2
, 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.
Example 1, p. 130
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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
An equilateral triangle of side x has area
A
1
3x 2 .
4
The rate of change of A with respect to x is the derivative
dA 1

3 x.
dx 2
When x  2 3, the rate of change of A with respect to x is 3. In other words, when
the side has length 2 3 , the area is changing three time as fast as the length of the
side.
Example 2, p. 131
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The Derivative as a Rate of Change
Example 3
Set
y
x2
.
2
x
(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 3, p. 131
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The Derivative as a Rate of Change
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.
V  r 2 h.
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
dV
= πr2.
dh
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
dV
= 2πrh.
dh
Example 4, p. 131-132
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The Derivative as a Rate of Change
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
h=
V
πr2
V
=
π
r –2.
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
2h
2(πr2h)
2V
dh
–3
–3
=
r =–
r =–
dr
r
π
π
.
Example 4, p. 131-132
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The Chain Rule
Leibnitz form of the chain rule, p. 133
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The Chain Rule
Example 1
Find dy/dx by the chain rule given that
y=
u–1
u+2
and
u = x2.
Example 1, p. 133
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The Chain Rule
Example 2
d 
1
x



dx 
x
3
4
4

1
d
1
1
1





 


3
x

x



3
x

1

.





 

2 
x  dx 
x
x  x 



Example 2, p. 135
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The Chain Rule
Example 3
d
[1 + (2 + 3x)5]3 = 3[1 + (2 + 3x)5]2
dx
d
[1 + (2 + 3x)5].
dx
Since
d
dx
[1 + (2 +
3x)5]3 =
5(2 +
3x)4
d
(2 + 3x) = 5(2 + 3x)4(3) = 15(2 + 3x)4.
dx
We have
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.
Example 3, p. 135
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The Chain Rule
Example 4
Calculate the derivative of f(x) = 2x3(x2 – 3)4.
Example 4, p. 135
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The Chain Rule
Example 5
Find dy/ds given that y = 3u + 1, u = x–2, x = 1 – s.
Example 5, p. 135
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The Chain Rule
Example 6
Find dy/dt at t = 9 given that
y
u2
,
u 1
u  (3s  7) 2 ,
s  t.
Example 6, p. 136
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The Chain Rule
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?
Example 7, p. 136
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The Chain Rule
The chain-rule theorem (3.5.6), p. 138
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Differentiating the Trigonometric Functions
Example 1
To differentiate f(x) = cos x sin x, we use the product rule:
f ’(x) = cos x
d
dx
(sin x) + sin x
d
(cos x)
dx
Basic formulas, (3.6.1), (3.6.2), Example 1, p. 142
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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
Find f ’(π/4) for f(x) = x cot x.
Example 2, p.143
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Differentiating the Trigonometric Functions
Example 3
Find
d 1  sec x 
.


dx  tan x 
Example 3, p.143
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Differentiating the Trigonometric Functions
Example 4
Find an equation for the line tangent to the curve y = cos x at the point
where x = π/3
Example 4, p.143
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Differentiating the Trigonometric Functions
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.
Example 5, p.144
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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
d
dx
(cos 2x) = –sin 2x
d
(2x) = –2sin 2x.
dx
Example 6, p. 144
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Differentiating the Trigonometric Functions
Example 7
d
dx
[sec(x2
+ 1)] =
sec(x2
= 2x
+
1)tan(x2
sec(x2
+
+ 1)
1)tan(x2
d
(x2 + 1)
dx
+ 1).
Example 7, p. 144
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Differentiating the Trigonometric Functions
Example 8
d
d
(sin 3 x)  (sin x)3
dx
dx
d
 3(sin x) 2 (sin x)
dx
d
 3(sin x) 2 cos x (x)
dx
 3(sin x) 2 cosx( )  3 sin 2x cos x.
Example 8, p. 145
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Differentiating the Trigonometric Functions
Example 9
Find
d
(sin x°).
dx
Example 9, p. 145
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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
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 2, p. 147
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Implicit Differentiation; Rational Powers
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 3, p. 147-148, Figures 3.7.3
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Implicit Differentiation; Rational Powers
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.
Example 4, p. 148
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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
1
d
(a)
[(1 +
x2)1/5]
=
5
dx
(b)
d
2
[(1 +
x2)2/3]
[(1 –
x2)1/4]
=
dx
(c)
d
dx
3
1
=
4
(1 +
x2)–4/5(2x)
2
=
(1 –
x2)–1/3(–2x)
(1 –
x2)–3/4(–2x)
5
=–
=–
x(1 + x2)–4/5.
4
3
1
2
x(1 – x2)–1/3.
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 5, p. 150
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Implicit Differentiation; Rational Powers
Example 6
d  x 


dx  1  x 2 
1/ 2
 1  x  1/ 2 d  x 

 
2 
2 
2
1

x
dx
1

x





1 x 
 
2 
2 1 x 
1 / 2
1/ 2
1  1 x2 

 
2  x 
(1  x 2 )(1)  x(2 x)
(1  x 2 ) 2
1 x2
(1  x 2 ) 2
1 x2
 1/ 2
.
2 x (1  x 2 ) 3 / 2
The result holds for all x > 0.
Example 6, p. 150
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