1 2x 3 (2x 3) = 2 (x 2 3x + 1)1/2 2 x 2 3x + 1 in .NET

Create QR Code in .NET 1 2x 3 (2x 3) = 2 (x 2 3x + 1)1/2 2 x 2 3x + 1

1 1 2x 3 (2x 3) = 2 (x 2 3x + 1)1/2 2 x 2 3x + 1
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[CHAP. 15
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(b) Dx ((3x 2 1)5/4 ) = = (c) Dx 3 1 7x + 2
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1 = Dx ((7x + 2) 1/3 ) = (7x + 2) 4/3 Dx (7x + 2) 3
1 7 1 (7) = 3 (7x + 2)4/3 3( 3 7x + 2)4
Solved Problems
15.1 For each pair of functions f and g, nd formulas for g f and f g, and determine the domains of g f and f g. (b) g(x) = x 2 and f (x) = x 1 (a) g(x) = x and f (x) = x + 1
(a) Because (g f )(x) = g( f (x)) = g(x + 1) = x+1
x + 1 is de ned if and only if x 1, the domain of g f is ( 1, ). ( f g)(x) = f (g(x)) = f ( x) = x + 1
Because (b)
x + 1 is de ned if and only if x 0, the domain of f g is (0, ). (g f )(x) = g( f (x)) = g(x 1) = (x 1)2 ( f g)(x) = f (g(x)) = f (x 2 ) = x 2 1
Both composite functions are polynomials, and so the domain of each is the set of all real numbers.
15.2 Calculate the derivatives of: (a) (x 4 3x 2 + 5x 2)3 (b)
7x 3 2x 2 + 5
(5x 2
1 + 4)3
The power chain rule is used in each case. (a) Dx ((x 4 3x 2 + 5x 2)3 ) = 3(x 4 3x 2 + 5x 2)2 Dx (x 4 3x 2 + 5x 2) = 3(x 4 3x 2 + 5x 2)2 (4x 3 6x + 5) (b) Dx ( 7x 3 2x 2 + 5) = Dx ((7x 3 2x 2 + 5)1/2 ) = = 1 (5x 2 + 4)3 1 3 (7x 2x 2 + 5) 1/2 Dx (7x 3 2x 2 + 5) 2
1 1 x(21x 4) (21x 2 4x) = 2 (7x 3 2x 2 + 5)1/2 2 7x 3 2x 2 + 5
(c) Dx
= Dx ((5x 2 + 4) 3 ) = 3(5x 2 + 4) 4 Dx (5x 2 + 4) = 3 (5x 2 + 4)4 (10x) = 30x (5x 2 + 4)4
CHAP. 15]
THE CHAIN RULE
15.3 Find the derivative of the function f (x) =
By the power chain rule, used twice, f (x) = = = = =
(x + 1) = [1 + (x + 1)1/2 ]1/2 .
1 (1 + (x + 1)1/2 ) 1/2 Dx (1 + (x + 1)1/2 ) 2 1 1 (1 + (x + 1)1/2 ) 1/2 (x + 1) 1/2 Dx (x + 1) 2 2 1 (1 + (x + 1)1/2 ) 1/2 (x + 1) 1/2 (1) 4 1 {(1 + (x + 1)1/2 )(x + 1)} 1/2 4 1 1 1 = 4 {(1 + (x + 1)1/2 )(x + 1)}1/2 4 (1 + 1 x + 1)(x + 1)
15.4 Find the absolute extrema of f (x) = x 1 x 2 on [0, 1].
Dx (x 1 x 2 ) = xDx ( 1 x 2 ) + 1 x 2 Dx (x) 1 x2 1 x2 x 2 1 x2 by + 1 x2 [by the power chain rule] [by the product rule]
= xDx ((1 x 2 )1/2 ) + =x = =
1 (1 x 2 ) 1/2 Dx (1 x 2 ) + 2 1 x2 =
x 1 ( 2x) + 2 (1 x 2 )1/2 x 2 + (1 x 2 ) 1 x2 =
1 2x 2 1 x2
a + bc a +b= c c
The right-hand side is not de ned when the denominator is 0; that is, when x 2 = 1. Hence, 1 and 1 are critical numbers. The right-hand side is 0 when the numerator is 0; that is, when 2x 2 = 1 Thus,
1 and 2
x2 =
1, 2
1 are also critical numbers. The only critical number in (0, 1) is 2
algebra
1 = 2
2 2 = 0.707 4 2
1 = 2
1 1 = 2 2
1 ) and 0 is the absolute minimum 2
At the endpoints, f (0) = f (1) = 0. Hence, 1 is the absolute maximum (achieved at x = 2 (achieved at x = 0 and x = 1).
15.5 A spy on a submarine S, 6 kilometers off a straight shore, has to reach a point B, which is 9 kilometers down the shore from the point A opposite S (see Fig. 15-2). The spy must row a boat to some point C on the shore and then walk the rest of the way to B. If he rows at 4 kilometers per hour and walks at 5 kilometers per hour, at what point C should he land in order to reach B as soon as possible
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