Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 14: Partial Derivatives - Section 14.3 - Partial Derivatives - Exercises 14.3 - Page 808: 64



Work Step by Step

We know that $f_x(x_0,y_0,z_0)=\lim\limits_{h \to 0} \dfrac{f(x_0,y_0,z_0+h,y_0)-f(x_0,y_0,z_0)}{h}$ $f_x(-1,0,3)=\lim\limits_{h \to 0} \dfrac{f(-1,h, 3)-f(-1,0,3)}{h}=\lim\limits_{h \to 0} \dfrac{(2h^2+9h)-0}{h}=\lim\limits_{h \to 0} (h+9)=9$
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