Calculus (3rd Edition)

Published by W. H. Freeman
ISBN 10: 1464125260
ISBN 13: 978-1-46412-526-3

Chapter 15 - Differentiation in Several Variables - 15.6 The Chain Rule - Exercises - Page 808: 3


$$\frac{\partial f}{\partial s}=6rs^2, \quad \frac{\partial f}{\partial r}=2s^3+4r^3.$$

Work Step by Step

We have $\frac{\partial f}{\partial x}=y$, $\frac{\partial f}{\partial y}=x$, $\frac{\partial f}{\partial z}=2z$; then \begin{aligned} \frac{\partial f}{\partial s}&=\frac{\partial f}{\partial x} \frac{\partial x}{\partial s}+\frac{\partial f}{\partial y} \frac{\partial y}{\partial s}+\frac{\partial f}{\partial z} \frac{\partial z}{\partial s}\\ &=y \frac{\partial}{\partial s}\left(s^{2}\right)+x \frac{\partial}{\partial s}(2rs )+2z\frac{\partial}{\partial s}\left(r^{2}\right) \\ &=(y)(2 s)+(x)(2r)+0 \\ &=2 s y+2r x \\ &=4rs^2+2r s^2=6rs^2 . \end{aligned} \begin{aligned} \frac{\partial f}{\partial r}&=\frac{\partial f}{\partial x} \frac{\partial x}{\partial r}+\frac{\partial f}{\partial y} \frac{\partial y}{\partial r}+\frac{\partial f}{\partial z} \frac{\partial z}{\partial r}\\ &=y \frac{\partial}{\partial r}\left(s^{2}\right)+x \frac{\partial}{\partial r}(2rs )+2z\frac{\partial}{\partial r}\left(r^{2}\right) \\ &=(y)(0)+(x)(2s)+2z(2r) \\ &= 2s x+4rz\\ &=2s^3+4r^3 . \end{aligned}
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