Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 14 - Section 14.3 - Partial Derivatives - 14.3 Exercise - Page 927: 104

Answer

$1$

Work Step by Step

The formula for finding the derivative of a function at some point $(x,y)$ can be defined as: $\lim\limits_{h \to 0}\dfrac{f(x+h,y)-f(x,y)}{h}=\dfrac{\partial f }{\partial x}$ At the point $(0,0)$, we have $\dfrac{\partial f }{\partial x}(0,0)=\lim\limits_{h \to 0}\dfrac{f(0+h,0)-f(0,0)}{h}$ As per our given statement, we have $\lim\limits_{h \to 0}\dfrac{\sqrt [3] {h^3+0^2}-\sqrt [3] {0^3+0^2}}{h}=\dfrac{\partial f }{\partial x}(0,0)$ Hence, we get $\lim\limits_{h \to 0}\dfrac{h}{h}=\dfrac{\partial f }{\partial x}(0,0)=1$ or, $f_x(0,0)=1$
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