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: 65



Work Step by Step

We need to take the first partial derivatives of the given function. In order to find the partial derivative, we will differentiate with respect to $x$, by keeping $y$ and $z$ as a constant, and vice versa: $f_x= \dfrac{\partial (xy+z^3 x-2yz)}{\partial x}=0 \\ \implies y+3xz \dfrac{\partial (z)}{\partial x}+z^3-2y\dfrac{\partial (z)}{\partial x}=0$ Now, $f_x(1,1,1)= 1+3(1)(1) \dfrac{\partial (z)}{\partial x}+(1)^3-2 (1)\dfrac{\partial (z)}{\partial x}=0$ This implies that $\dfrac{\partial (z)}{\partial x}=-2$
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