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: 45

Answer

$\frac{\partial f}{\partial x}(\frac{\pi}{4},\frac{\pi}{3})=-1$ $\frac{\partial f}{\partial y}(\frac{\pi}{4},\frac{\pi}{3})=\frac{1}{2}$

Work Step by Step

To solve these partial derivatives we will use chain rule. The partial derivative with the respect to $x$ is: $$\frac{\partial f(x,y)}{\partial x}=\frac{\partial }{\partial x}(\cos(2x-y))=-\sin(2x-y)\frac{\partial}{\partial x}(2x-y)=-2\sin(2x-y),$$ because $y$ is treated as a constant. The partial derivative with the respect to $y$ is: $$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y}(\cos(2x-y))=-\sin(2x-y)\frac{\partial }{\partial y}(2x-y)=-\sin(2x-y)\cdot(-1)=\sin(2x-y),$$ because now $x$ is treated as a constant. Now we will evaluate these partial derivatives in the given point $(x,y)=(\frac{\pi}{4},\frac{\pi}{3})$: $\frac{\partial f}{\partial x}(\frac{\pi}{4},\frac{\pi}{3})=\left.-2\sin(2x-y)\right|_{(x,y)=(\frac{\pi}{4},\frac{\pi}{3})}=-2\sin(2\cdot\frac{\pi}{4}-\frac{\pi}{3})=-2\sin\frac{\pi}{6}$ $\frac{\partial f}{\partial y}(\frac{\pi}{4},\frac{\pi}{3})=\left.\sin(2x-y)\right|_{(x,y)=(\frac{\pi}{4},\frac{\pi}{3})}=\sin(2\cdot \frac{\pi}{4}-\frac{\pi}{3})=\sin\frac{\pi}{6}$ $\sin\frac{\pi}{6}=\frac{1}{2}$ $\frac{\partial f}{\partial x}(\frac{\pi}{4},\frac{\pi}{3})=-2\sin\frac{\pi}{6}=(-2)(\frac{1}{2})=-1$ $\frac{\partial f}{\partial y}(\frac{\pi}{4},\frac{\pi}{3})=\sin\frac{\pi}{6}=\frac{1}{2}$
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