## Calculus 8th Edition

Published by Cengage

# Chapter 14 - Partial Derivatives - 14.5 The Chain Rule - 14.5 Exercises: 11

#### Answer

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

#### Work Step by Step

The partial derivative with respect to $s$ is: $$\frac{\partial z}{\partial s}=\frac{\partial z}{\partial r}\frac{\partial r}{\partial s}+\frac{\partial z}{\partial \theta}\frac{\partial \theta}{\partial s}= \frac{\partial }{\partial r}(e^r\cos\theta)\frac{\partial}{\partial s}(st)+\frac{\partial}{\partial \theta}(e^r\cos\theta)\frac{\partial}{\partial s}(\sqrt{s^2+t^2})= e^r\cos\theta\cdot t+e^r\cdot(-\sin\theta)\cdot\frac{1}{2\sqrt{s^2+t^2}}\frac{\partial}{\partial s}(s^2+t^2)= e^rt\cos\theta-e^r\sin\theta\frac{1}{2\sqrt{s^2+t^2}}\cdot2s= e^rt\cos\theta-\frac{e^rs\sin\theta}{\sqrt{s^2+t^2}}$$ Expressing this in terms of $s$ and $t$ we get: $$\frac{\partial z}{\partial s}=e^rt\cos\theta-\frac{e^rs\sin\theta}{\sqrt{s^2+t^2}}=e^{st}t\cos(\sqrt{s^2+t^2})-\frac{e^{st}s\sin(\sqrt{s^2+t^2})}{\sqrt{s^2+t^2}}$$ The partial derivative with respect to $t$ is: $$\frac{\partial z}{\partial t}=\frac{\partial z}{\partial r}\frac{\partial r}{\partial t}+\frac{\partial z}{\partial \theta}\frac{\partial \theta}{\partial t}= \frac{\partial }{\partial r}(e^r\cos\theta)\frac{\partial}{\partial t}(st)+\frac{\partial}{\partial \theta}(e^r\cos\theta)\frac{\partial}{\partial t}(\sqrt{s^2+t^2})= e^r\cos\theta\cdot s+e^r(-\sin\theta)\frac{1}{2\sqrt{s^2+t^2}}\frac{\partial}{\partial t}(s^2+t^2)= e^rs\cos\theta-\frac{e^rt\sin\theta}{\sqrt{s^2+t^2}}$$ Expressing this in terms of $s$ and $t$ we get: $$\frac{\partial z}{\partial t}=e^rs\cos\theta-\frac{e^rt\sin\theta}{\sqrt{s^2+t^2}}=e^{st}s\cos(\sqrt{s^2+t^2})-\frac{e^{st}t\sin(\sqrt{s^2+t^2})}{\sqrt{s^2+t^2}}$$

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