Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 3 - Section 3.4 - The Chain Rule - 3.4 Exercises: 62



Work Step by Step

$$h(x)=\sqrt{4+3f(x)}$$ Apply the Chain Rule: $$h'(x)=\frac{d(\sqrt{[4+3f(x)]})}{d[4+3f(x)]}\frac{d[4+3f(x)]}{df(x)}\frac{df(x)}{dx}$$ $$h'(x)=\frac{1}{2\sqrt{4+3f(x)}}\times(0+3\times1)\times f'(x)$$ $$h'(x)=\frac{3f'(x)}{2\sqrt{4+3f(x)}}$$ Then, $$h'(1)=\frac{3f'(1)}{2\sqrt{4+3f(1)}}$$ We know $f(1)=7$ and $f'(1)=4$. Therefore, $$h'(1)=\frac{3\times4}{2\sqrt{4+3\times7}}=\frac{12}{2\sqrt{25}}=\frac{6}{5}$$
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