University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 13 - Section 13.3 - Partial Derivatives - Exercises - Page 702: 58


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Work Step by Step

Since, $f_x(x_0,y_0)=\lim\limits_{h \to 0} \dfrac{f(x_0+h,y_0)-f(x_0,y_0)}{h}$ and $f_y(x_0,y_0)=\lim\limits_{h \to 0} \dfrac{f(x_0,y_0+h)-f(x_0,y_0)}{h}$ $f_x(-2,1)=\lim\limits_{h \to 0} \dfrac{f(-2+h,1)-f(-2,1)}{h}=\lim\limits_{h \to 0} \dfrac{h-1+1}{h}=1$ and $f_y(-2,1)=\lim\limits_{h \to 0} \dfrac{f(-2,1+h)-f(-2,1)}{h}$ or, $\lim\limits_{h \to 0} \dfrac{2h^2+h-1+1}{h}=\lim\limits_{h \to 0} (2h+1)=1$
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