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: 44

Answer

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

Work Step by Step

The partial derivative with the respect to $x$ is: $$\frac{\partial f(x,y)}{\partial x}=\frac{\partial }{\partial x}(e^{-x}\cos y)=\cos y\frac{\partial}{\partial x}(e^{-x})=\cos y\cdot e^{-x}\frac{\partial}{\partial x}(-x)=-e^{-x}\cos y$$ because $y$ is treated as a constant. In the last two steps we have used the chain rule to find the derivative of $e^{-x}.$ The partial derivative with the respect to $y$ is: $$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y}(e^{-x}\cos y)=e^{-x}\frac{\partial }{\partial y}(\cos y)=-e^{-x}\sin y$$ because now $x$ is treated as a constant. Now we will evaluate the partial derivatives in the given point $(x,y)=(0,0)$: $$\frac{\partial f(0,0)}{\partial x}=\left.-e^{-x}\cos y\right|_{(x,y)=(0,0)}=-e^{-0}\cos 0=-1\cdot1=-1$$ $$\frac{\partial f(0,0)}{\partial y}=\left.-e^{-x}\sin y\right|_{(x,y)=(0,0)}=-e^{-0}\sin0=-1\cdot0=0$$
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