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 - Page 965: 45

Answer

$f_x=y^2-3x^2y$ $f_y=2xy-x^3$

Work Step by Step

The function f(x,y)=xy^2-x^3y is provided. We use the definition of partial derivatives as limits in order to find the partial derivatives. $f_x(x,y)=\lim\limits_{h \to 0} {\frac{(x+h)y^2-(x+h)^3y-xy^2+x^3y}{h}}=\lim\limits_{h \to 0} {\frac{xy^2+hy^2-x^3y-3x^2hy-3xh^2y+h^3y-xy^2+x^3y}{h}}$ Some of the terms cancel leaving the limits: $\lim\limits_{h \to 0} {\frac{hy^2-3x^2hy-3xhy^2-h^3y}{h}}=\lim\limits_{h \to 0} {y^2-3x^2y-3xhy+h^2y}=[y^2-3x^2y+0+0]=y^2-3x^2y$ Applying a similar procedure for the partial derivative of $y$: $f_y(x,y)=\lim\limits_{h \to 0}{\frac{x(y+h)^2-x^3(y+h)-xy^2+x^3y}{h}}=\lim\limits_{h \to 0} {2xy+xh-x^3}=[2xy-x^3+0]=2xy-x^3$
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