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 807: 28


$\dfrac{1}{|x+yz| \sqrt {(x+yz)^2-1}}$ $\dfrac{z}{|x+yz| \sqrt {(x+yz)^2-1}}$ $\dfrac{y}{|x+yz| \sqrt {(x+yz)^2-1}}$

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: $f_x=\dfrac{1}{|x+yz| \sqrt {(x+yz)^2-1}} \times \dfrac{\partial (x+yz)}{\partial x}=\dfrac{1}{|x+yz| \sqrt {(x+yz)^2-1}}$ $f_y=\dfrac{1}{|x+yz| \sqrt {(x+yz)^2-1}} \times \dfrac{\partial (x+yz)}{\partial y}=\dfrac{z}{|x+yz| \sqrt {(x+yz)^2-1}}$ $f_z=\dfrac{1}{|x+yz| \sqrt {(x+yz)^2-1}} \times \dfrac{\partial (x+yz)}{\partial z}=\dfrac{y}{|x+yz| \sqrt {(x+yz)^2-1}}$
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