Calculus, 10th Edition (Anton)

Published by Wiley
ISBN 10: 0-47064-772-8
ISBN 13: 978-0-47064-772-1

Chapter 2 - The Derivative - 2.6 The Chain Rule - Exercises Set 2.6: 4

Answer

$(a) (f\circ g)(x)=5\sqrt{4+\cos x},(f\circ g)'(x)=-\frac{5\sin x}{2\sqrt{4+\cos x}}$ $(b) (g\circ f)(x)=4+\cos(5\sqrt x),(g\circ f)'(x)=-\frac{5\sin(5\sqrt x)}{2\sqrt x}$

Work Step by Step

$f(x)=5\sqrt x,g(x)=4+\cos x$ $(a) (f\circ g)(x)=f(g(x))=f(4+\cos x)=5\sqrt{4+\cos x}$ $(f\circ g)'(x)=f'(g(x))\cdot g'(x)$ $$(f\circ g)'(x)=(5\sqrt{4+\cos x})'=5\frac{1}{2\sqrt{4+\cos x}}\cdot(4+\cos x)'=\frac{5}{2\sqrt{4+\cos x}}\cdot(-\sin x)=-\frac{5\sin x}{2\sqrt{4+\cos x}}$$ $(b) (g\circ f)(x)=g(f(x))=g(5\sqrt x)=4+\cos (5\sqrt x)$ $(g\circ f)'(x)=g'(f(x))\cdot f'(x)$ $$(g\circ f)'(x)=(4+\cos(5\sqrt x))'=-\sin(5\sqrt x)\cdot(5\sqrt x)'=-\sin (5\sqrt x)\cdot5\frac{1}{2\sqrt x}=-\frac{5\sin(5\sqrt x)}{2\sqrt x}$$
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