Calculus 8th Edition

Published by Cengage

Chapter 14 - Partial Derivatives - 14.3 Partial Derivatives - 14.3 Exercises: 24

Answer

$$\frac{\partial w}{\partial u}=-\frac{e^v}{(u+v^2)^2};$$ $$\frac{\partial w}{\partial v}=\frac{u-2v+v^2}{(u+v^2)^2}e^v.$$

Work Step by Step

Regard $v$ as a constant and differentiate with respect to $u$: $$\frac{\partial w}{\partial u}=\frac{\partial}{\partial u}\left(\frac{e^v}{u+v^2}\right)=e^v\frac{\partial}{\partial u}\left(\frac{1}{u+v^2}\right)=e^v\left(-\frac{1}{(u+v^2)^2}(u+v^2)'_u\right)=-\frac{e^v}{(u+v^2)^2}.$$ Regard $u$ as a constant and differentiate with respect to $v$: $$\frac{\partial w}{\partial v}=\frac{\partial}{\partial v}\left(\frac{e^v}{u+v^2}\right)=\frac{(e^v)'_v(u+v^2)-e^v(u+v^2)'_v}{(u+v^2)^2}=\frac{e^v(u+v^2)-2ve^v}{(u+v^2)^2}=\frac{u-2v+v^2}{(u+v^2)^2}e^v$$

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