Multivariable Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 0-53849-787-4
ISBN 13: 978-0-53849-787-9

Chapter 14 - Partial Derivatives - 14.4 Exercises - Page 946: 13

Answer

The partial derivatives $f_{x}$ and $f_{y}$ exist near ($2,1$), and are continuous functions for $y\neq x$ so $f$ is differentiable at ($2,1$). (by Theorem 8) $L(x, y)==\displaystyle \frac{1}{9}x-\frac{2}{9}y+\frac{2}{3}$

Work Step by Step

$f(x, y)=\displaystyle \frac{x}{x+y}$ $\left.\begin{array}{lllll} f_{x}(x, y) & =\dfrac{1(x+y)-x(1)}{(x+y)^{2}} & ... & f_{y}(x, y) & =x(-1)(x+y)^{-2}\cdot 1\\ & =\dfrac{y}{(x+y)^{2}} & & & =\dfrac{-x}{(x+y)^{2}}\\ & & & & \\ f_{x}(2,1) & =1/9 & & f_{y}(2,1) & =-2/9 \end{array}\right.$ The partial derivatives $f_{x}$ and $f_{y}$ exist near ($2,1$), and are continuous functions for $y\neq x$ so $f$ is differentiable at ($2,1$). (by Theorem 8) The linearization of $f$ at (a,b): $f(x, y)\approx L(x, y)=f(a, b)+ dz$, where $dz=f_{x}(a, b)(x-a)+f_{y}(a, b)(y-b)$ $L(x, y)==f(2,1)+f_{x}(2,1)(x-2)+f_{y}(2,1)(y-1)$ $L(x, y)=\displaystyle \frac{2}{3}+\frac{1}{9}(x-2)-\frac{2}{9}(y-1)$ $L(x, y)=\displaystyle \frac{2}{3}+\frac{1}{9}x-\frac{2}{9}-\frac{2}{9}y+\frac{2}{9}$ $L(x, y)==\displaystyle \frac{1}{9}x-\frac{2}{9}y+\frac{2}{3}$
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