Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 3 - Differentiation - 3.7 The Chain Rule - Exercises - Page 147: 98

Answer

$-11\sqrt 2$

Work Step by Step

Applying the chain rule, we have $\frac{d}{dx}[\sin(g(x))]=\cos (g(x))\times g'(x)$ $\frac{d^{2}}{dx^{2}}[\sin(g(x))]_{x=2}=$ $\frac{d}{dx}[\cos (g(x))\times g'(x)]_{x=2}=$ $[-\sin(g(2))g'(2)\times g'(2)]+[\cos(g(2))\times g''(2)]$ (where we applied the chain rule and the product rule) $=(-\sin \frac{\pi}{4}\times5\times5)+(\cos\frac{\pi}{4}\times3)$ $=-\frac{1}{\sqrt 2}\times25+\frac{1}{\sqrt 2}\times3$ $=-\frac{22}{\sqrt 2}=-11\sqrt 2$
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