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: 33

Answer

$$\frac{\partial z}{\partial x}=ye^y\cos xy$$ $$\frac{\partial z}{\partial y}=e^y(\sin xy+x\cos xy)$$

Work Step by Step

The partial derivative with respect to $x$ is: $$\frac{\partial z}{\partial x}=\frac{\partial }{\partial x}(e^y\sin xy)=e^y\frac{\partial }{\partial x}(\sin xy)=e^y\cos xy\frac{\partial }{\partial x}(xy)=e^y\cos xy\cdot y=ye^y\cos xy$$ Here we treated $y$ as a constant, as we did with $e^y$. We used chain rule to find partial derivative $\frac{\partial }{\partial x}(\sin xy).$ We will use product rule to find partial derivative with respect to $y$: $$\frac{\partial z}{\partial y}=\frac{\partial }{\partial y}(e^y\sin xy)=\sin xy\frac{\partial }{\partial y}(e^y)+e^y\frac{\partial }{\partial y}(\sin xy)=\sin xy\cdot e^y+e^y\cdot \cos xy\frac{\partial }{\partial y}(xy)=e^y\sin xy+e^y\cos xy\cdot x=e^y(\sin xy+x\cos xy)$$ Here is $x$ treated as a constant. We used chain rule to find partial derivative $\frac{\partial }{\partial y}(\sin xy).$
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