Solve the problem.
-Evaluate $\frac { \mathrm { dw } } { \mathrm { dt } }$ at $\mathrm { t } = 5$ for the function $\mathrm { w } = \mathrm { e } ^ { \mathrm { y } } - \ln \mathrm { x } ; \mathrm { x } = \mathrm { t } ^ { 2 } , \mathrm { y } = \ln \mathrm { t }$.
A) 3
B) $- \frac { 3 } { 5 }$
C) $\frac { 3 } { 5 }$
D) $\frac { 4 } { 5 }$

Question

Quizplus · Accepted Answer

We can use the chain rule and the given values of $x$ and $y$ to find $\frac{dw}{dt}$:
\begin{align*}
\frac{\mathrm{dw}}{\mathrm{dt}} &= \frac{\partial \mathrm{w}}{\partial \mathrm{x}}\cdot \frac{\partial \mathrm{x}}{\partial \mathrm{t}} + \frac{\partial \mathrm{w}}{\partial \mathrm{y}}\cdot \frac{\partial \mathrm{y}}{\partial \mathrm{t}} \
&= \frac{1}{x} \cdot 2t + e^{y} \cdot \frac{1}{t} \
&= \frac{2}{t} + e^{\ln t} \cdot \frac{1}{t} \
&= \frac{2}{t} + \frac{1}{t} \
&= \frac{3}{t}
\end{align*}
Evaluating at $t=5$, we have $\frac{\mathrm{dw}}{\mathrm{dt}} \bigg|_{t=5} = \frac{3}{5}$. Therefore, the answer is $\boxed{	extbf{(C)}}$.

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Solve the Problem $\frac { \mathrm { dw } } { \mathrm { dt } }$