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: 57


$-13$ and $-2$

Work Step by Step

We know that $f_x(x_0,y_0)=\lim\limits_{h \to 0} \dfrac{f(x_0+h,y_0)-f(x_0,y_0)}{h}$ $f_x(1,2)=\lim\limits_{h \to 0} \dfrac{f(1+h,2)-f(1,2)}{h}=\lim\limits_{h \to 0} \dfrac{-6h^2-13h}{h}=-13$ Also, $f_y(x_0,y_0)=\lim\limits_{h \to 0} \dfrac{f(x_0,y_0+h)-f(x_0,y_0)}{h}$ Now, $f_y(1,2)=\lim\limits_{h \to 0} \dfrac{f(1,2+h)-f(1,2)}{h}=\lim\limits_{h \to 0} \dfrac{(-4-2h)-(-4)}{h}=-2$
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