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: 22

Answer

$$\frac{\partial}{\partial x}f(x,y)=\frac{y-x}{(x+y)^3};$$ $$\frac{\partial}{\partial y}f(x,y)=-\frac{2x}{(x+y)^3}.$$

Work Step by Step

1) Regarding $y$ as constant and differentiating with respect to $x$ we get $$\frac{\partial}{\partial x}f(x,y)=\frac{\partial}{\partial x}\left(\frac{x}{(x+y)^2}\right)=\frac{(x)'_x(x+y)^2-x\left((x+y)^2\right)'_x}{(x+y)^4}=\frac{(x+y)^2-x\cdot 2(x+y)(x+y)'_x}{(x+y)^4}=\frac{(x+y)^2-2x(x+y)}{(x+y)^4}=\frac{(x+y)-2x}{(x+y)^3}=\frac{y-x}{(x+y)^3}$$ 2) Regarding $x$ as constant and differentiating with respect to $y$ we get $$\frac{\partial}{\partial y}f(x,y)=\frac{\partial}{\partial y}\left(\frac{x}{(x+y)^2}\right)=x\frac{\partial}{\partial y}\left(\frac{1}{(x+y)^2}\right)=x\left(-\frac{2}{(x+y)^3}(x+y)'_y\right)=-\frac{2x}{(x+y)^3}.$$
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