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 703: 59

Answer

$\dfrac{1}{2}, \dfrac{3}{4}$

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 $f_y(x_0,y_0)=\lim\limits_{h \to 0} \dfrac{f(x_0,y_0+h)-f(x_0,y_0)}{h}$ $f_x(-2,3)=\lim\limits_{h \to 0} \dfrac{f(-2+h,3)-f(-2,3)}{h}=$ or, $\implies \lim\limits_{h \to 0} \dfrac{2}{\sqrt {2h+4}-2}=\dfrac{1}{2}$ and $f_y(x_0,y_0)=\lim\limits_{h \to 0} \dfrac{f(x_0,y_0+h)-f(x_0,y_0)}{h}$ Next, $f_y(-2,3)=\lim\limits_{h \to 0} \dfrac{f(-2,3+h)-f(-2,3)}{h}$ $\implies \lim\limits_{h \to 0} \dfrac{\sqrt {3h+4}-2}{h}=\lim\limits_{h \to 0} \dfrac{3}{\sqrt {3h+4}-2}=\dfrac{3}{4}$
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