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: 8


$f_x=2x^2 (x^3+\dfrac{y}{2})^{-1/3}$ and $f_y=\dfrac{1}{3} (x^3+\dfrac{y}{2})^{-1/3}$

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 $y$ as a constant, and vice versa: $f_x=\dfrac{2}{3} \times (x^3+y/2)^{-1/3}\times 3x^2$ and $f_y=\dfrac{2}{3} \times (x^3+y/2)^{-1/3}\times 1/2$ Our results are: $f_x=2x^2 (x^3+\dfrac{y}{2})^{-1/3}$ and $f_y=\dfrac{1}{3} (x^3+\dfrac{y}{2})^{-1/3}$
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