Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.3 Partial Derivatives - 14.3 Exercises - Page 964: 34

Answer

Regard $y$ and $z$ and differentiate with respect to $x$: $$\frac{\partial}{\partial x}w=\frac{y}{\cos^2(x+2z)}.$$ Regard $x$ and $z$ and differentiate with respect to $y$: $$\frac{\partial}{\partial y}w=\tan(x+2z).$$ Regard $x$ and $y$ and differentiate with respect to $z$: $$\frac{\partial}{\partial z}w=\frac{2y}{\cos^2(x+2z)}.$$

Work Step by Step

Regard $y$ and $z$ and differentiate with respect to $x$: $$\frac{\partial}{\partial x}w=\frac{\partial}{\partial x}(y\tan(x+2z))=y\frac{\partial}{\partial x}\tan(x+2z)=y\frac{1}{\cos^2(x+2z)}\frac{\partial}{\partial x}(x+2z)=\frac{y}{\cos^2(x+2z)}.$$ Regard $x$ and $z$ and differentiate with respect to $y$: $$\frac{\partial}{\partial y}w=\frac{\partial}{\partial y}(y\tan(x+2z))=\tan(x+2z)\frac{\partial}{\partial y}y=\tan(x+2z).$$ Regard $x$ and $y$ and differentiate with respect to $z$: $$\frac{\partial}{\partial z}w=\frac{\partial}{\partial z}(y\tan(x+2z))=y\frac{\partial}{\partial z}\tan(x+2z)=y\frac{1}{\cos^2(x+2z)}\frac{\partial}{\partial z}(x+2z)=\frac{2y}{\cos^2(x+2z)}.$$
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