## Calculus 8th Edition

Published by Cengage

# Chapter 14 - Partial Derivatives - 14.5 The Chain Rule - 14.5 Exercises - Page 983: 3

#### Answer

$$\frac{dz}{dt}=\frac{\cos\sqrt t\cos\frac{1}{t}}{2\sqrt t}+\frac{\sin\sqrt t\sin\frac{1}{t}}{t^2}$$

#### Work Step by Step

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

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