Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Section 14.3 - Partial Derivatives - Exercises 14.3 - Page 809: 91


$f$ is not differentiable in $(x,y)=(0,0)$

Work Step by Step

In order to find the partial derivative, we will differentiate with respect to $x$, by keeping $y$ as a constant, and vice versa: $f_x=\dfrac{\partial }{\partial x}(\dfrac{xy^2)}{x^2+y^4})=\dfrac{y^2(x^2+y^4)-xy^2(2x+y^4)}{(x^2+y^4)^2}=\dfrac{y^6-x^2y^2-xy^6}{(x^2+y^4)^2}$ and $f_y=\dfrac{\partial }{\partial y}(\dfrac{xy^2)}{x^2+y^4})=\dfrac{2xy(x^2+y^4)-xy^2(x^2+4y^3)}{(x^2+y^4)^2}=\dfrac{2x^3y-x^3y^2-2xy^5}{(x^2+y^4)^2}$ Now, $\lim\limits_{(y^2,y) \to (0,0)} f_x=\lim\limits_{(y^2,y) \to (0,0)}\dfrac{y^6-y^6-y^8}{(y^4+y^4)^2}=\dfrac{-1}{4}$ and $\lim\limits_{(-y^2,y) \to (0,0)} f_y=\lim\limits_{(-y^2,y) \to (0,0)}\dfrac{y^6-y^6-y^8}{(y^4+y^4)^2}=\dfrac{1}{4}$ Thus, we conclude that $f_x$ and $f_y$ are not continuous in $(0,0)$, so $f$ is not differentiable in $(x,y)=(0,0)$.
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