## Calculus 8th Edition

Published by Cengage

# Chapter 14 - Partial Derivatives - 14.5 The Chain Rule - 14.5 Exercises: 4

#### Answer

$$\frac{dz}{dt}=\frac{1}{2\sqrt{1+\tan t\arctan t}}\Big(\frac{\arctan t}{\cos^2t}+\frac{\tan t}{1+t^2}\Big)$$

#### Work Step by Step

$$\frac{dz}{dt}=\frac{\partial z}{\partial x}\frac{dx}{dt}+\frac{\partial z}{\partial y}\frac{dy}{dt}=\frac{\partial}{\partial x}(\sqrt{1+xy})\frac{d}{dt}(\tan t)+\frac{\partial}{\partial y}(\sqrt{1+xy})\frac{d}{dt}(\arctan t)= \frac{1}{2\sqrt{1+xy}}\frac{\partial}{\partial x}(1+xy)\frac{1}{\cos^2t}+ \frac{1}{2\sqrt{1+xy}}\frac{\partial}{\partial y}(1+xy)\frac{1}{1+t^2}= \frac{1}{2\sqrt{1+xy}}\cdot y\cdot\frac{1}{\cos^2t}+\frac{1}{2\sqrt{1+xy}}\cdot x\cdot\frac{1}{1+t^2}= \frac{1}{2\sqrt{1+xy}}\Big(\frac{y}{\cos^2t}+\frac{x}{1+t^2}\Big)$$ Now we have to express solution in terms of $t$: $$\frac{dz}{dt}=\frac{1}{2\sqrt{1+xy}}\Big(\frac{y}{\cos^2t}+\frac{x}{1+t^2}\Big)=\frac{1}{2\sqrt{1+\tan t\arctan t}}\Big(\frac{\arctan t}{\cos^2t}+\frac{\tan t}{1+t^2}\Big)$$

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