## Thomas' Calculus 13th Edition

Published by Pearson

# Chapter 14: Partial Derivatives - Practice Exercises - Page 864: 32

#### Answer

$\frac{\partial w}{\partial u}=\frac{2}{5}$ $\frac{\partial w}{\partial v}=0$

#### Work Step by Step

$x=2e^{u}cos(v)=2$ when $u=v=0$ $\frac{dx}{du}=2e^{u}cos(v)=2$ when $u=v=0$ $\frac{dx}{dv}=-2e^{u}sin(v)=0$ when $u=v=0$ $w=\ln\sqrt{1+x^2}-tan^{-1}x=\frac{1}{2}\ln(1+x^2)-tan^{-1}x$ $\frac{dw}{dx}=\frac{x}{{1+x^2}}-\frac{1}{{1+x^2}}=\frac{1}{5}$ when $u=v=0$ $\frac{\partial w}{\partial u}=\frac{dw}{dx}\frac{dx}{du}=\frac{1}{5}\times2=\frac{2}{5}$ when $u=v=0$ $\frac{\partial w}{\partial v}=\frac{dw}{dx}\frac{dx}{dv}=\frac{1}{5}\times0=0$ when $u=v=0$

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