## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 14: Partial Derivatives - Section 14.3 - Partial Derivatives - Exercises 14.3 - Page 807: 45

#### Answer

$r_{xx}=\frac{-1}{(x+y)^2}$ $r_{yy}=\frac{-1}{(x+y)^2}$ $r_{xy}=r_{yx}=\frac{-1}{(x+y)^2}$

#### Work Step by Step

Take the first partial derivatives of the given function. When taking partial derivative with respect to x, treat y as a constant, and vice versa: $r_x=\frac{1}{x+y}$ $r_y=\frac{1}{x+y}$ Then take the derivative of the first order partial derivatives to find second partial derivatives: $r_{xx}=\frac{(x+y)(0)-1}{(x+y)^2}=\frac{-1}{(x+y)^2}$ $r_{yy}=\frac{(x+y)(0)-1}{(x+y)^2}=\frac{-1}{(x+y)^2}$ Second partial derivatives of first order partial derivative of x with respect to y and y with respect to x are the same: $r_{xy}=r_{yx}=\frac{(x+y)(0)-1}{(x+y)^2}=\frac{-1}{(x+y)^2}$

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