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: 26


$f_x=\dfrac{-x}{(x^2+y^2+z^2)^{3/2}}$ $f_y=\dfrac{-y}{(x^2+y^2+z^2)^{3/2}}$ $f_z=\dfrac{-z}{(x^2+y^2+z^2)^{3/2}}$

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: $f_x=-(1/2)(x^2+y^2+z^2)^{-3/2}\times 2x=\dfrac{-x}{(x^2+y^2+z^2)^{3/2}}$ $f_y=-(1/2) \times (x^2+y^2+z^2)^{-3/2}\times 2y=\dfrac{-y}{(x^2+y^2+z^2)^{3/2}}$ $f_z=-(1/2) (x^2+y^2+z^2)^{-3/2}\times 2z=\dfrac{-z}{(x^2+y^2+z^2)^{3/2}}$
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