Calculus with Applications (10th Edition)

Published by Pearson
ISBN 10: 0321749006
ISBN 13: 978-0-32174-900-0

Chapter 4 - Calculating the Derivative - 4.3 The Chain Rule - 4.3 Exercises - Page 225: 43


$a.\quad-2$ $b.\quad-\displaystyle \frac{24}{7}$

Work Step by Step

$D_{x}(f[g(x)]) =f^{\prime}[g(x)]\cdot g^{\prime}(x)$ (a) $D_{x}(f[g(x)])\quad $ at $x=1$ $=f^{\prime}[g(1)]\cdot g^{\prime}(1)$ ... read the table: $g(1)=2,\displaystyle \quad g^{\prime}(1)=\frac{2}{7}$ $=f^{\prime}(2)\displaystyle \cdot(\frac{2}{7})$ ... read the table: $f^{\prime}(2)=-7$ $=-7(\displaystyle \frac{2}{7}]$ $=\color{red}{-2}$ (b) $D_{x}(f[g(x)])\quad$ at $x=2$ $=f^{\prime}[g(2)]\cdot g^{\prime}(2)$ ... read the table: $g(2)=3,\displaystyle \quad g^{\prime}(2)=\frac{3}{7}$ $=f^{\prime}(3)\displaystyle \cdot(\frac{3}{7})$ ... read the table: $f^{\prime}(3)=-8$ $=-8(\displaystyle \frac{3}{7}]$ $=\displaystyle \color{red}{-\frac{24}{7}}$
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.