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: 48

Answer

The partial derivatives $$\frac{\partial f(x,y)}{\partial x}$$ and $$\frac{\partial f(x,y)}{\partial y}$$ are not defined in the given point $(x,y)=(1,1).$

Work Step by Step

To solve both partial derivative we will use chain rule. The partial derivative with respcet to $x$ is: $$\frac{\partial f(x,y)}{\partial x}=\frac{\partial }{\partial x}(\arccos xy)=-\frac{1}{\sqrt{1-(xy)^2}}\cdot\frac{\partial }{\partial x}(xy)=-\frac{y}{\sqrt{1-(xy)^2}}$$ because $y$ is treated as a constant. The partial derivative with the respect to $y$ is: $$\frac{\partial f(x,y)}{\partial y}=\frac{\partial }{\partial y}(\arccos xy)=-\frac{1}{\sqrt{1-(xy)^2}}\frac{\partial }{\partial y}(xy)=-\frac{x}{\sqrt{1-(xy)^2}}$$ because here $x$ is treated as a constant. Now we will evaluate these partial derivatives in the given point $(x,y)=(1,1)$: $$\frac{\partial f(1,1)}{\partial x}=\left.-\frac{y}{\sqrt{1-(xy)^2}}\right|_{(x,y)=(1,1)}=-\frac{1}{\sqrt{1-1\cdot1}}=-\frac{1}{0}$$ We see that $0$ is in the denominator, so because we cannot divide with $0$ the partial derivative with respect to $x$ is not defined at this point. $$\frac{\partial f(1,1)}{\partial y}=\left.-\frac{x}{\sqrt{1-(xy)^2}}\right|_{(x,y)=(1,1)}=-\frac{1}{\sqrt{1-1\cdot1}}=-\frac{1}{0}$$ We see that $0$ is in the denominator, so because we cannot divide with $0$ the partial derivative with respect to $y$ is not defined at this point.
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