Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 14 - Partial Derivatives - 14.3 Partial Derivatives - 14.3 Exercises - Page 964: 39


For every $p\in\{1,2,\ldots,n\}$ $$\frac{\partial}{\partial x_p}u=\frac{2x_p}{2\sqrt{x_1^2+\ldots+x_{p-1}^2+x_p^2+x_{p+1}^2+\ldots+x_n^2}}$$

Work Step by Step

We will differentiate with respect to some $x_p$where $p=\overline{1,n}$ and we will hold all other $x_i$, $i\neq p$ constant: $$\frac{\partial}{\partial x_p}u=\frac{\partial}{\partial x_p}\sqrt{x_1^2+\ldots+x_{p-1}^2+x_p^2+x_{p+1}^2+\ldots+x_n^2}=\frac{\frac{\partial}{\partial x_p}(x_1^2+\ldots+x_{p-1}^2+x_p^2+x_{p+1}^2+\ldots+x_n^2)}{2\sqrt{x_1^2+\ldots+x_{p-1}^2+x_p^2+x_{p+1}^2+\ldots+x_n^2}}=\frac{\underbrace{0+0+\ldots+0+\frac{\partial}{\partial x_p}x_p^2+0+\ldots+0}_{n-terms}}{2\sqrt{x_1^2+\ldots+x_{p-1}^2+x_p^2+x_{p+1}^2+\ldots+x_n^2}}=\frac{2x_p}{2\sqrt{x_1^2+\ldots+x_{p-1}^2+x_p^2+x_{p+1}^2+\ldots+x_n^2}}.$$ Note that this is true for every $p\in\{1,2,\ldots,n\}$ so for every individual first partial derivative we want, we just put that particular value for $p$.
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