## Calculus 8th Edition

Published by Cengage

# Chapter 14 - Partial Derivatives - 14.3 Partial Derivatives - 14.3 Exercises: 69

#### Answer

$\frac{∂^{3}w}{∂z∂y∂x}=\frac{4}{(y+2z)^{3}}$ and $\frac{∂^{3}w}{∂x^{2}∂y}=0$

#### Work Step by Step

Consider the function $w=\frac{x}{y+2z}$ Let us start differentiating the function with respect to $x$ keeping $y$ and $z$ constant. $\frac{∂w}{∂x}=\frac{∂}{∂x}[\frac{x}{y+2z}]$ $=\frac{1}{{y+2z}}$ Differentiate with respect to $y$ keeping $x$ and $z$ constant. $\frac{∂^{2}w}{∂y∂x}=\frac{∂}{∂y}[\frac{1}{{y+2z}}]$ $=-\frac{1}{(y+2z)^{2}}$ Differentiate with respect to $z$ keeping $x$ and $y$ constant. $\frac{∂^{3}w}{∂z∂y∂x}=\frac{∂}{∂z}[-\frac{1}{(y+2z)^{2}}]$ $=\frac{2}{(y+2z)^{3}}(2)$ $=\frac{4}{(y+2z)^{3}}$ Thus, $\frac{∂^{3}w}{∂z∂y∂x}=\frac{4}{(y+2z)^{3}}$ Let us start differentiating the function with respect to $y$ keeping $x$ and $z$ constant. $\frac{∂w}{∂y}=\frac{∂}{∂y}[\frac{x}{y+2z}]$ $=-\frac{x}{{(y+2z})^{2}}$ and $\frac{∂^{2}w}{∂x∂y}=\frac{∂}{∂x}[-\frac{x}{{(y+2z})^{2}}]$ $=-\frac{1}{{(y+2z})^{2}}$ Again differentiating $\frac{∂^{2}w}{∂x∂y}$ with respect to $x$, we have Hence, $\frac{∂^{3}w}{∂x^{2}∂y}=0$

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