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: 66



Work Step by Step

We need to take the first partial derivatives of the given function. The partial derivative of $xz$ is equal to $x+z \dfrac{\partial (x)}{\partial z}+z^3-2y\dfrac{\partial (z)}{\partial x}$ and the partial derivative of $y \ln x$ is equal to $y(1/x) \dfrac{\partial (z)}{\partial x}$ The partial derivative of $-x^2$ is equal to $-2x \dfrac{\partial (z)}{\partial x}$ Now, $ \dfrac{\partial (z)}{\partial x}(z+\dfrac{y}{x}-2x)=-x$ and $ \dfrac{\partial (z)}{\partial x}=\dfrac{-x}{(z+\dfrac{y}{x}-2x)}$ $ \dfrac{\partial (z)}{\partial x}(1,-1,-3)=\dfrac{1}{6}$
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