## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 14: Partial Derivatives - Section 14.3 - Partial Derivatives - Exercises 14.3 - Page 808: 70

#### Answer

$s_u=\dfrac{1+2y}{1-2y}$

#### Work Step by Step

Differentiate $x$ and $y$ with respect to $u$. $1=2x \times x_u-2yy_u .....(a)$ and $0=2x \times x_u -y_u ....(b)$ Now, $[0=2x \times x_u -y_u ....(b)] \times 2y$ and subtract from equation $a$. So, $x_u= \dfrac{1}{2x-4xy}$ Subtract equation (b) from equation $a$. We get $y_u= \dfrac{1}{1-2y}$ Now, differentiate $s$ with respect to $u$. $s_u=2x \times x_u-2y \times y_u =2x \times ( \dfrac{1}{2x-4xy}) -2y \times (\dfrac{1}{2y})$ So, $s_u=\dfrac{1+2y}{1-2y}$

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