Calculus 8th Edition

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

Chapter 14 - Partial Derivatives - 14.3 Partial Derivatives - 14.3 Exercises: 36

Answer

$$\frac{\partial}{\partial x}u=\frac{y}{z}x^\frac{y-z}{z}.$$ $$\frac{\partial}{\partial y}u=\frac{\ln x}{z}x^\frac{y}{z}.$$ $$\frac{\partial}{\partial z}u=-\frac{x^\frac{y}{z}y\ln x}{z^2}.$$

Work Step by Step

Regard $y$ and $z$ as constant and differentiate with respect to $x$: $$\frac{\partial}{\partial x}u=\frac{\partial}{\partial x}x^{y/z}=\frac{y}{z}x^{\frac{y}{z}-1}=\frac{y}{z}x^\frac{y-z}{z}.$$ Regard $x$ and $z$ as constant and differentiate with respect to $y$: $$\frac{\partial}{\partial y}u=\frac{\partial}{\partial y}x^{y/z}=x^{y/z}\ln x\frac{\partial}{\partial y}\frac{y}{z}=\frac{\ln x}{z}x^\frac{y}{z}.$$ Regard $x$ and $y$ and differentiate with respect to $z$: $$\frac{\partial}{\partial z}u=\frac{\partial}{\partial z}x^{y/z}=x^{y/z}\ln x\frac{\partial}{\partial z}\frac{y}{z}=x^\frac{y}{z}\ln x\text{ }y\left(-\frac{1}{z^2}\right)=-\frac{x^\frac{y}{z}y\ln x}{z^2}.$$
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