University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 13 - Section 13.3 - Partial Derivatives - Exercises - Page 702: 25


$f_x=1$ $f_y=-\dfrac{y}{\sqrt {y^2+z^2}}$ $f_z=-\dfrac{z}{\sqrt {y^2+z^2}}$

Work Step by Step

Our aim is to take the first partial derivatives of the given function $f(x,y)$ with respect to $x$, by keeping $y$ and $z$ as a constant, and vice versa: $f_x=1$ $f_y=-\dfrac{1}{2} (y^2+z^2)^{-1/2}\times 2y=-\dfrac{y}{\sqrt {y^2+z^2}}$ $f_z=-\dfrac{1}{2}(y^2+z^2)^{-1/2}\times 2z=-\dfrac{z}{\sqrt {y^2+z^2}}$
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