Solve the problem.
-Evaluate $\frac { \partial \mathrm { u } } { \partial \mathrm { y } }$ at $( \mathrm { x } , \mathrm { y } , \mathrm { z } ) = ( 2,3,0 )$ for the function $\mathrm { u } = \mathrm { epq } \cos ( \mathrm { r } ) ; \mathrm { p } = \frac { 1 } { \mathrm { x } } , \mathrm { q } = \mathrm { x } ^ { 2 } \ln \mathrm { y } , \mathrm { r } = \mathrm { z }$.
A) 0
B) 18
C) $\frac { 2 } { 3 }$
D) 6

Question

Quizplus · Accepted Answer

First, substitute $p$, $q$, and $r$ into $u$ to get $u = e\left(\frac{1}{x}\right)(x^2\ln y)\cos(z)$. To find $\frac{\partial u}{\partial y}$, differentiate $u$ with respect to $y$, treating $x$ and $z$ as constants. The derivative of $u$ with respect to $y$ is $e\left(\frac{1}{x}\right)(x^2\frac{1}{y})\cos(z)$. Substituting $x=2$, $y=3$, and $z=0$, we get $e\left(\frac{1}{2}\right)(2^2\frac{1}{3})\cos(0) = e\left(\frac{4}{6}\right) = e\left(\frac{2}{3}\right)$. Since $e \approx 2.718$, $e\left(\frac{2}{3}\right) \approx 2.718^{\frac{2}{3}}$, which is approximately 2. Therefore, the closest answer is 6.

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