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 - 14.3 Exercises - Page 939: 100

Answer

$1$

Work Step by Step

The formula for finding the derivative of a function at the some point is given by: $\lim\limits_{h \to 0}\dfrac{f(x+h,y)-f(x,y)}{h}=\dfrac{\partial f }{\partial x}$ $\lim\limits_{h \to 0}\dfrac{f(0+h,0)-f(0,0)}{h}=\dfrac{\partial f }{\partial x}(0,0)$ and $\lim\limits_{h \to 0}\dfrac{\sqrt [3] {h^3+0^2}-\sqrt [3] {0^3+0^2}}{h}=\dfrac{\partial f }{\partial x}(0,0)$ This gives $\lim\limits_{h \to 0}\dfrac{h}{h}=\dfrac{\partial f }{\partial x}(0,0)=1$
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