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: 53

Answer

$$\frac{\partial H(x,y,z)}{\partial x}=\cos(x+2y+3z)$$ $$\frac{\partial H(x,y,z)}{\partial y}=2\cos(x+2y+3z)$$ $$\frac{\partial H(x,y,z)}{\partial z}=3\cos(x+2y+3z)$$

Work Step by Step

We will use chain rule to find these partial derivatives. The partial derivative with respect to $x$ is: $$\frac{\partial H(x,y,z)}{\partial x}=\frac{\partial }{\partial x}(\sin(x+2y+3z))=\cos(x+2y+3z)\frac{\partial}{\partial x}(x+2y+3z)=\cos(x+2y+3z)\cdot1=\cos(x+2y+3z)$$ Because $2y+3z$ is treated as a constant its' derivative is $0.$ The partial derivative with respect to $y$ is: $$\frac{\partial H(x,y,z)}{\partial y}=\frac{\partial }{\partial x}(\sin(x+2y+3z))=\cos(x+2y+3z)\frac{\partial}{\partial y}(x+2y+3z)=\cos(x+2y+3z)\cdot2=2\cos(x+2y+3z)$$ Here $x+3z$ is treated as a constant, so its' derivative is $0.$ The partial derivative with respect to $z$ is: $$\frac{\partial H(x,y,z)}{\partial z}=\frac{\partial }{\partial z}(\sin(x+2y+3z))=\cos(x+2y+3z)\frac{\partial }{\partial z}(x+2y+3z)=\cos(x+2y+3z)\cdot3=3\cos(x+2y+3z)$$ $x+2y$ is treated as a constant, so its' derivative is $0.$
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