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14.33 Find the shortest distance between points of the curve y =
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14.34 Find the dimensions of the rectangle of maximum area that can be inscribed in a right triangle whose sides are 3, 4, and 5, if one side of the rectangle lies on the side of the triangle of length 3 and the other side lies on the side of the triangle of length 4. 14.35 GC Find the relative extrema of f (x) = x 4 + 2x 2 5x 2 in two ways. (a) Trace the graph of f to nd the relative maximum and relative minimum directly. (b) Trace the graph of f to nd the critical numbers of f . 14.36 GC Find the relative extrema of f (x) = x 3 3x 2 + 4x 1 on ( 5, 3).
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15.1 COMPOSITE FUNCTIONS There are still many functions whose derivatives we do not know how to calculate; for example, (i) x3 x + 2 (ii) 3 x+4 (iii) (x 2 + 3x 1)23
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In case (iii), we could, of course, multiply x 2 + 3x 1 by itself 22 times and then differentiate the resulting polynomial. But without a computer, this would be extremely arduous. The above three functions have the common feature that they are combinations of simpler functions: (i) x 3 x + 2 is the result of starting with the function f (x) = x 3 x + 2 and then applying the function g(x) = x to the result. Thus, x 3 x + 2 = g( f (x)) (ii) 3 x + 4 is the result of starting with the function F(x) = x + 4 and then applying the function G(x) = 3 x. Thus, 3 x + 4 = G(F(x)) (iii) (x 2 + 3x 1)23 is the result of beginning with the function H(x) = x 2 + 3x 1 and then applying the function K(x) = x 23 . Thus, (x 2 + 3x 1)23 = K(H(x)) Functions that are put together this way out of simpler functions are called composite functions. De nition: If f and g are any functions, then the composition g f of f and g is the function such that (g f )(x) = g( f (x)) The process of composition is diagrammed in Fig. 15-1.
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Fig. 15-1 EXAMPLES
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(a) Let f (x) = x 1 and g(x) = x 2 . Then, (g f )(x) = g( f (x)) = g(x 1) = (x 1)2 On the other hand, ( f g)(x) = f (g(x)) = f (x 2 ) = (x 2 1) Thus, f g and g f are not necessarily the same function (and usually they are not the same). (b) Let f (x) = x 2 + 2x and g(x) = x. Then, (g f )(x) = g( f (x)) = g(x 2 + 2x) = x 2 + 2x
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( f g)(x) = f (g(x)) = f ( x) = ( x)2 + 2( x) = x + 2 x Again, g f and f g are different.
A composite function g f is de ned only for those x for which f (x) is de ned and g( f (x)) is de ned. In other words, the domain of g f consists of those x in the domain of f for which f (x) is in the domain of g. Theorem 15.1: The composition of continuous functions is a continuous function. If f is continuous at a, and g is continuous at f (a), then g f is continuous at a. For a proof, see Problem 15.25.
DIFFERENTIATION OF COMPOSITE FUNCTIONS
First, let us treat an important special case. The function [ f (x)]n is the composition g f of f and the function g(x) = x n . We have: Theorem 15.2 (Power Chain Rule): Let f be differentiable and let n be any integer. Then, Dx (( f (x))n ) = n( f (x))n 1 Dx ( f (x)) EXAMPLES
(a) Dx ((x 2 5)3 ) = 3(x 2 5)2 Dx (x 2 5) = 3(x 2 5)2 (2x) = 6x(x 2 5)2 (b) Dx ((x 3 2x 2 + 3x 1)7 ) = 7(x 3 2x 2 + 3x 1)6 Dx (x 3 2x 2 + 3x 1) = 7(x 3 2x 2 + 3x 1)6 (3x 2 4x + 3) (c) Dx 1 (3x 5)4 = Dx ((3x 5) 4 ) = 4(3x 5) 5 Dx (3x 5) = 12 4 (3) = 5 (3x 5) (3x 5)5
(15.1)
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