Calculus 10th Edition

Published by Brooks Cole
ISBN 10: 1-28505-709-0
ISBN 13: 978-1-28505-709-5

Chapter 13 - Functions of Several Variables - 13.3 Exercises - Page 896: 63

Answer

$$\frac{\partial f(0,\frac{\pi}{2},-4)}{\partial x}=0$$ $$\frac{\partial f(0,\frac{\pi}{2},-4)}{\partial y}=0$$ $$\frac{\partial f(0,\frac{\pi}{2},-4)}{\partial z}=1$$

Work Step by Step

The partial derivative with respect to $x$ is: $$\frac{\partial f(x,y,z)}{\partial x}=\frac{\partial}{\partial x}(z\sin(x+y))=z\cos(x+y)\frac{\partial}{\partial x}(x+y)=z\cos(x+y)\cdot1=z\cos(x+y)$$ The partial derivative with respect to $y$ is: $$\frac{\partial f(x,y,z)}{\partial y}=\frac{\partial}{\partial y}(z\sin(x+y))=z\cos(x+y)\frac{\partial}{\partial y}(x+y)=z\cos(x+y)\cdot1=z\cos(x+y)$$ The partial derivative with respect to $z$ is: $$\frac{\partial f(x,y,z)}{\partial z}=\frac{\partial}{\partial z}(z\sin(x+y))=\sin(x+y)$$ Now we will evaluate these partial derivatives at the given point $(x,y,z)=(0,\frac{\pi}{2},-4)$: $$\frac{\partial f(0,\frac{\pi}{2},-4)}{\partial x}=\left.z\cos(x+y)\right|_{(x,y,z)=(0,\pi/2,-4)}=-4\cos(0+\frac{\pi}{2})=-4\cdot0=0$$ $$\frac{\partial f(0,\frac{\pi}{2},-4)}{\partial y}=\left.z\cos(x+y)\right|_{(x,y,z)=(0,\pi/2,-4)}=-4\cos(0+\frac{\pi}{2})=-4\cdot0=0$$ $$\frac{\partial f(0,\frac{\pi}{2},-4)}{\partial z}=\left.\sin(x+y)\right|_{(x,y,z)=(0,\pi/2,-4)}=\sin(0+\frac{\pi}{2})=1$$
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