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

Answer

$$h'(1)=\frac{6}{5}$$

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, $$h'(1)=\frac{3\times4}{2\sqrt{4+3\times7}}=\frac{12}{2\sqrt{25}}=\frac{6}{5}$$
Update this answer!

You can help us out by revising, improving and updating this answer.

Update this answer

After you claim an answer you’ll have 24 hours to send in a draft. An editor will review the submission and either publish your submission or provide feedback.