University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 3 - Section 3.6 - The Chain Rule - Exercises - Page 159: 82


For $x=1/4$, $$(fog)'=5\pi$$

Work Step by Step

$f(u)=u+\frac{1}{\cos^2u}$, $u=g(x)=\pi x$, $x=1/4$ Applying the Chain Rule: $$(fog)'=f'(g(x))g'(x)=f'(u)u'$$ We have $$f'(u)=\Big(u+\frac{1}{\cos^2u}\Big)'=1+\Big(\frac{-1(\cos^2u)'}{\cos^4u}\Big)=1-\frac{2\cos u(\cos u)'}{\cos^4u}$$ $$f'(u)=1+\frac{2\cos u\sin u}{\cos^4u}=1+\frac{2\sin u}{\cos^3u}$$ $$u'=(\pi x)'=\pi$$ Therefore, $$(fog)'=\Big(1+\frac{2\sin u}{\cos^3 u}\Big)\pi=\Big(1+\frac{2\sin \pi x}{\cos^3 \pi x}\Big)\pi$$ For $x=1/4$, $$(fog)'=\Big(1+\frac{2\sin\frac{\pi}{4}}{\cos^3\frac{\pi}{4}}\Big)\pi=\Big(1+\frac{2\times\frac{\sqrt2}{2}}{\Big(\frac{\sqrt2}{2}\Big)^3}\Big)\pi=\Big(1+\frac{\sqrt2}{\frac{2\sqrt2}{8}}\Big)\pi$$ $$=\Big(1+\frac{8\sqrt2}{2\sqrt2}\Big)\pi=(1+4)\pi=5\pi$$
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