if $y = \tan u$, $u = v-\frac{1}{v}$, and $v=ln x$, what is the value of $\frac{dy}{dx}$ at $x = e$?\na 0\nb…

Explanation:

Step1: Apply chain - rule

By the chain - rule, $\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dv}\cdot\frac{dv}{dx}$.

Step2: Differentiate $y$ with respect to $u$

If $y = \tan u$, then $\frac{dy}{du}=\sec^{2}u$.

Step3: Differentiate $u$ with respect to $v$

If $u=v - \frac{1}{v}=v - v^{-1}$, then $\frac{du}{dv}=1 + v^{-2}=1+\frac{1}{v^{2}}$.

Step4: Differentiate $v$ with respect to $x$

If $v=\ln x$, then $\frac{dv}{dx}=\frac{1}{x}$.

Step5: Substitute and simplify

$\frac{dy}{dx}=\sec^{2}u\cdot(1 + \frac{1}{v^{2}})\cdot\frac{1}{x}$. Since $u = v-\frac{1}{v}$ and $v=\ln x$, when $x = e$, $v=\ln e = 1$. Then $u=1 - 1=0$. Substitute $u = 0$ and $v = 1$ and $x = e$ into $\frac{dy}{dx}$: $\frac{dy}{dx}=\sec^{2}(0)\cdot(1 + 1)\cdot\frac{1}{e}$. Since $\sec(0)=1$, $\sec^{2}(0)=1$. So $\frac{dy}{dx}=1\times2\times\frac{1}{e}=\frac{2}{e}$.

Answer:

D. $\frac{2}{e}$