Thomas' Calculus 13th Edition

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

Chapter 3: Derivatives - Practice Exercises - Page 178: 56


a. $-\frac{13}{10}$ b. $-\frac{1}{3}$ c. $\frac{1}{10}$ d. $-1$ e. $-\frac{2}{3}$ f. $-12$

Work Step by Step

a. Let $g(x)=\sqrt xf(x)$, we have $g'(x)=\frac{1}{2\sqrt x}f(x)+\sqrt xf'(x)$ and $g'(1)=\frac{1}{2\sqrt 1}f(1)+\sqrt 1f'(1)=\frac{1}{2}(-3)+(1/5)=-\frac{13}{10}$ b. Let $g(x)=\sqrt {f(x)}$, we have $g'(x)=\frac{f'(x)}{2\sqrt {f(x)}}$ and $g'(0)=\frac{f'(0)}{2\sqrt {f(0)}}=\frac{-2}{2\sqrt {9}}=-\frac{1}{3}$ c. Let $g(x)=f(\sqrt x)$, we have $g'(x)=f'(\sqrt x)\frac{1}{2\sqrt x}$ and $g'(1)=f'(\sqrt 1)\frac{1}{2\sqrt 1}=(1/5)/2=\frac{1}{10}$ d. Let $g(x)=f(1-5tan(x))$, we have $g'(x)=f'(1-5tan(x))(-5sec^2x)$ and $g'(0)=f'(1-5tan(0))(-5sec^20)=f'(1)(-5)=(1/5)(-5)=-1$ e. Let $g(x)=\frac{f(x)}{2+cos(x)}$, we have $g'(x)=\frac{(2+cos(x))f'(x)-f(x)(-sin(x))}{(2+cos(x))^2}$ and $g'(0)=\frac{(2+cos(0))f'(0)+f(0)sin(0)}{(2+cos(0))^2}=\frac{(2+1)(-2)}{(2+1)^2}=-\frac{2}{3}$ f. Let $g(x)=10sin(\frac{\pi x}{2})f^2(x)$, we have $g'(x)=10(\pi/2)cos(\frac{\pi x}{2})f^2(x)+10sin(\frac{\pi x}{2})(2f(x)f'(x))$ and $g'(1)=5\pi cos(\frac{\pi}{2})f^2(1)+20sin(\frac{\pi}{2})f(1)f'(1)=20(-3)(1/5)=-12$
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