Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.5 The Chain Rule - 14.5 Exercises: 10

Answer

$$\frac{\partial z}{\partial s}=\frac{e^{(1+st)(s^2-t^2)}(t+2(1+st)(2s+3s^2t-t^3)}{2\sqrt{1+st}}$$ $$\frac{\partial z}{\partial t}=\frac{e^{(1+st)(s^2-t^2)}(s+2(1+st)(s^3-3st^2-2t))}{2\sqrt{1+st}}$$

Work Step by Step

The partial derivative with respect to $s$ is: $$\frac{\partial z}{\partial s}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial s}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial s}= \frac{\partial }{\partial x}(\sqrt xe^{xy})\frac{\partial}{\partial s}(1+st)+\frac{\partial}{\partial y}(\sqrt xe^{xy})\frac{\partial}{\partial s}(s^2-t^2)= \Big(\frac{\partial}{\partial x}(\sqrt x)\cdot e^{xy}+\sqrt x\frac{\partial}{\partial x}e^{xy}\Big)\cdot t+\sqrt xe^{xy}\frac{\partial}{\partial y}(xy)\cdot2s= \Big(\frac{1}{2\sqrt x}e^{xy}+\sqrt xe^{xy}\frac{\partial}{\partial x}(xy)\Big)\cdot t+\sqrt xe^{xy}\cdot x\cdot2s= \frac{te^{xy}}{2\sqrt x}+t\sqrt xye^{xy}+2sx\sqrt xe^{xy}= \frac{te^{xy}+2txye^{xy}+4sx^2e^{xy}}{2\sqrt x}= \frac{e^{xy}(t+2txy+4sx^2)}{2\sqrt x}$$ Now we will express solution in terms of $s$ and $t$: $$\frac{\partial z}{\partial s}=\frac{e^{xy}(t+2txy+4sx^2)}{2\sqrt x}= \frac{e^{(1+st)(s^2-t^2)}(t+2t(1+st)(s^2-t^2)+4s(1+st)^2)}{2\sqrt{1+st}}=\frac{e^{(1+st)(s^2-t^2)}(t+2(1+st)(2s+3s^2t-t^3)}{2\sqrt{1+st}}$$ Using that $\frac{\partial}{\partial x}(\sqrt xe^{xy})=\frac{e^{xy}(1+2xy)}{2\sqrt x}$ and $\frac{\partial}{\partial y}(\sqrt xe^{xy})=\sqrt x xe^{xy}$ which we have already found, we will evaluate the partial derivative with respect to $t$: $$\frac{\partial z}{\partial t}=\frac{\partial z}{\partial x}\frac{\partial x}{\partial t}+\frac{\partial z}{\partial y}\frac{\partial y}{\partial t}= \frac{\partial }{\partial x}(\sqrt xe^{xy})\frac{\partial}{\partial t}(1+st)+\frac{\partial}{\partial y}(\sqrt xe^{xy})\frac{\partial}{\partial t}(s^2-t^2)= \frac{e^{xy}(1+2xy)}{2\sqrt x}\cdot s+\sqrt xxe^{xy}\cdot(-2t)= \frac{e^{xy}(1+2xy)s-4tx^2e^{xy}}{2\sqrt x}= \frac{e^{xy}(s+2sxy-4tx^2)}{2\sqrt x}$$ Expressing this in terms of $s$ and $t$ we get: $$\frac{\partial z}{\partial t}=\frac{e^{xy}(s+2sxy-4tx^2)}{2\sqrt x}= \frac{e^{(1+st)(s^2-t^2)}(s+2s(1+st)(s^2-t^2)-4t(1+st)^2)}{2\sqrt{1+st}}= \frac{e^{(1+st)(s^2-t^2)}(s+2(1+st)(s^3-st^2-2t-2st^2))}{2\sqrt{1+st}}= \frac{e^{(1+st)(s^2-t^2)}(s+2(1+st)(s^3-3st^2-2t))}{2\sqrt{1+st}}$$
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