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: 54

Answer

$$\frac{\partial f(x,y,z)}{\partial x}=6xy-5yz$$ $$\frac{\partial f(x,y,z)}{\partial y}=3x^2-5xz+10z^2$$ $$\frac{\partial f(x,y,z)}{\partial z}=20yz-5xy$$

Work Step by Step

The partial derivative with respect to $x$ is: $$\frac{\partial f(x,y,z)}{\partial x}=\frac{\partial }{\partial x}(3x^2y-5xyz+10yz^2)=3y\frac{\partial }{\partial x}(x^2)-5yz\frac{\partial f}{\partial x}(x)+\frac{\partial }{\partial x}(10yz^2)=3y\cdot2x-5yz\cdot1+0=6xy-5yz$$ Here are $y$ and $z$ treated as a constant. The partial derivative with respect to $y$ is: $$\frac{\partial f(x,y,z)}{\partial y}=\frac{\partial }{\partial y}(3x^2y-5xyz+10yz^2)=3x^2\frac{\partial }{\partial y}(y)-5xz\frac{\partial }{\partial y}(y)+10z^2\frac{\partial }{\partial y}(y)=3x^2\cdot1-5xz\cdot1+10z^2\cdot1=3x^2-5xz+10z^2$$ Here are $x$ and $z$ treated as a constant. The partial derivative with respect to $z$ is: $$\frac{\partial f(x,y,z)}{\partial z}=\frac{\partial }{\partial z}(3x^2y-5xyz+10yz^2)=\frac{\partial }{\partial z}(3x^2y)-5xy\frac{\partial }{\partial z}(z)+10y\frac{\partial }{\partial z}(z^2)=0-5xy\cdot1+10y\cdot2z=20yz-5xy$$ Here are $x$ and $y$ treated as a constant.
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