Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 3: Derivatives - Section 3.6 - The Chain Rule - Exercises 3.6 - Page 149: 67

Answer

$-\frac{\pi}{4}$

Work Step by Step

$(fog)'(x)$ = $f'(g(x))g'(x)$ $g(x)$ = $5\sqrt x$ $g'(x)$ = $\frac{5}{2\sqrt x}$ $g'(1)$ = $\frac{5}{2\sqrt 1}$ = $\frac{5}{2}$ $f(u)$ = $cot(\frac{\pi(u)}{10})$ $f'(u)$ = $-csc^{2}(\frac{\pi(u)}{10})[\frac{\pi}{10}]$ $f'(u)$ = $-\frac{\pi}{10}csc^{2}(\frac{\pi(u)}{10})$ $f'(u)$ = $f'(g(x))$ = $-\frac{\pi}{10}csc^{2}(\frac{5\pi\sqrt x}{10})$ = $-\frac{\pi}{10}csc^{2}(\frac{\pi\sqrt x}{2})$ $f'(g(1))$ = $-\frac{\pi}{10}csc^{2}(\frac{\pi\sqrt 1}{2})$ = $-\frac{\pi}{10}csc^{2}(\frac{\pi}{2})$ = $-\frac{\pi}{10}(1)$ = $-\frac{\pi}{10}$ $(fog)'(1)$ = $f'(g(1))g'(1)$ = $(-\frac{\pi}{10})$$(\frac{5}{2})$ = $-\frac{\pi}{4}$
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