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: 59

Answer

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

Work Step by Step

The partial derivative with respect to $x$ is: $$\frac{\partial f(x,y,z)}{\partial x}=\frac{\partial }{\partial x}(x^3yz^2)=yz^2\frac{\partial }{\partial x}(x^3)=yz^2\cdot3x^2=3x^2yz^2$$ because $y$ and $z$ are treated as a constant, $yz^2$ is treated as a constant as well. The partial derivative with respect to $y$ is: $$\frac{\partial f(x,y,z)}{\partial y}=\frac{\partial }{\partial y}(x^3yz^2)=x^3z^2\frac{\partial }{\partial y}(y)=x^3z^3\cdot1=x^3z^2$$ because $x$ and $z$ are treated as a constant, $x^3z^2$ is treated as a constant as well. The partial derivative with respect to $z$ is: $$\frac{\partial f(x,y,z)}{\partial z}=\frac{\partial}{\partial z}(x^3yz^2)=x^3y\frac{\partial }{\partial z}(z^2)=x^3y\cdot2z=2x^3yz$$ because $x$ and $y$ are treated as a constant, $x^2y$ is treated as a constant as well. Now we will evaluate these partial derivatives in the given point $(x,y,z)=(1,1,1)$: $$\frac{\partial f(1,1,1)}{\partial x}=\left.3x^2yz^2\right|_{(x,y,z)=(1,1,1)}=3\cdot1^2\cdot1\cdot1^2=3$$ $$\frac{\partial f(1,1,1)}{\partial y}=\left.x^3z^2\right|_{(x,y,z)=(1,1,1)}=1^3\cdot1^2=1$$ $$\frac{\partial f(1,1,1)}{\partial z}=\left.2x^3yz\right|_{(x,y,z)=(1,1,1)}=2\cdot1^3\cdot1\cdot1=2$$
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