Key Concepts and Formulas

• Logarithmic Differentiation: If $y = [u(x)]^{v(x)}$, then $\ln y = v(x) \ln[u(x)]$. Differentiating implicitly gives $\frac{1}{y} \frac{dy}{dx} = \frac{d}{dx} \left( v(x) \ln[u(x)] \right)$.
• Product Rule of Differentiation: $\frac{d}{dx}[u(x)v(x)] = u'(x)v(x) + u(x)v'(x)$.
• Chain Rule of Differentiation: $\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$.

Step-by-Step Solution

Step 1: Define the Function and Apply Logarithm Let the given function be $f(x)$ or $y$: $y = f(x) = \left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^{2} x}, \quad x \in\left(0, \frac{\pi}{2}\right)$ Take the natural logarithm of both sides: $\ln y = \ln \left[ \left(\frac{\sqrt{3 e}}{2 \sin x}\right)^{\sin ^{2} x} \right]$ Using the logarithm property $\ln(a^b) = b \ln a$: $\ln y = \sin^2 x \cdot \ln \left(\frac{\sqrt{3 e}}{2 \sin x}\right)$ Simplify the $\ln$ term further using $\ln(a/b) = \ln a - \ln b$: $\ln y = \sin^2 x \left[ \ln(\sqrt{3 e}) - \ln(2 \sin x) \right]$ $\ln y = \sin^2 x \left[ \frac{1}{2} \ln(3 e) - (\ln 2 + \ln(\sin x)) \right]$ $\ln y = \sin^2 x \left[ \frac{1}{2} (\ln 3 + \ln e) - \ln 2 - \ln(\sin x) \right]$ $\ln y = \sin^2 x \left[ \frac{1}{2} (\ln 3 + 1) - \ln 2 - \ln(\sin x) \right]$

Step 2: Differentiate with Respect to $x$ Differentiate both sides of the equation with respect to $x$. $\frac{d}{dx}(\ln y) = \frac{d}{dx}\left(\sin^2 x \left[ \frac{1}{2} (\ln 3 + 1) - \ln 2 - \ln(\sin x) \right]\right)$ Using the product rule: $\frac{1}{y} \frac{dy}{dx} = (2 \sin x \cos x) \left[ \frac{1}{2} (\ln 3 + 1) - \ln 2 - \ln(\sin x) \right] + \sin^2 x \left[ 0 - 0 - \frac{\cos x}{\sin x} \right]$ $\frac{1}{y} \frac{dy}{dx} = (2 \sin x \cos x) \left[ \frac{1}{2} (\ln 3 + 1) - \ln 2 - \ln(\sin x) \right] - \sin x \cos x$ $\frac{1}{y} \frac{dy}{dx} = \sin x \cos x \left[ 2 \left( \frac{1}{2} (\ln 3 + 1) - \ln 2 - \ln(\sin x) \right) - 1 \right]$ $\frac{1}{y} \frac{dy}{dx} = \sin x \cos x \left[ (\ln 3 + 1) - 2\ln 2 - 2\ln(\sin x) - 1 \right]$ $\frac{1}{y} \frac{dy}{dx} = \sin x \cos x \left[ \ln 3 - \ln 4 - \ln(\sin^2 x) \right]$ $\frac{1}{y} \frac{dy}{dx} = \sin x \cos x \left[ \ln \left(\frac{3}{4 \sin^2 x}\right) \right]$

Step 3: Find Critical Points by Setting $\frac{dy}{dx} = 0$ For local maximum or minimum values, we set the first derivative to zero: $\frac{dy}{dx} = 0$ Since $x \in (0, \frac{\pi}{2})$, $\sin x \neq 0$ and $\cos x \neq 0$. Therefore, we must have $\ln \left(\frac{3}{4 \sin^2 x}\right) = 0$ $\frac{3}{4 \sin^2 x} = e^0 = 1$ $3 = 4 \sin^2 x$ $\sin^2 x = \frac{3}{4}$ $\sin x = \frac{\sqrt{3}}{2}$ (Since $x \in (0, \frac{\pi}{2})$, $\sin x > 0$) $x = \frac{\pi}{3}$

Step 4: Find the Local Maximum Value Now we substitute $x = \frac{\pi}{3}$ into the original function to find the local maximum value: $f\left(\frac{\pi}{3}\right) = \left(\frac{\sqrt{3 e}}{2 \sin \frac{\pi}{3}}\right)^{\sin^2 \frac{\pi}{3}} = \left(\frac{\sqrt{3 e}}{2 \cdot \frac{\sqrt{3}}{2}}\right)^{\left(\frac{\sqrt{3}}{2}\right)^2} = \left(\frac{\sqrt{3 e}}{\sqrt{3}}\right)^{\frac{3}{4}} = (\sqrt{e})^{\frac{3}{4}} = e^{\frac{3}{8}}$ Given that the local maximum value is $\frac{k}{e}$, we have: $\frac{k}{e} = e^{\frac{3}{8}}$ $k = e^{1 + \frac{3}{8}} = e^{\frac{11}{8}}$

Step 5: Evaluate the Expression We need to find the value of $\left(\frac{k}{e}\right)^{8}+\frac{k^{8}}{e^{5}}+k^{8}$: $\left(\frac{k}{e}\right)^{8} = \left(e^{\frac{3}{8}}\right)^{8} = e^3$ $k^8 = \left(e^{\frac{11}{8}}\right)^{8} = e^{11}$ $\frac{k^8}{e^5} = \frac{e^{11}}{e^5} = e^6$ Therefore, $\left(\frac{k}{e}\right)^{8}+\frac{k^{8}}{e^{5}}+k^{8} = e^3 + e^6 + e^{11}$

Common Mistakes & Tips

• Remember to use the chain rule correctly when differentiating composite functions, especially logarithmic functions.
• Be careful with algebraic manipulations and simplifications, especially with exponents and logarithms.
• Double-check your calculations to avoid errors.

Summary We started by taking the natural logarithm of the given function to simplify the differentiation process. Then, we differentiated implicitly with respect to $x$ and set the derivative equal to zero to find the critical points. We found that $x = \frac{\pi}{3}$ is a critical point. We then evaluated the original function at this point to find the local maximum value, which was given as $\frac{k}{e}$. We solved for $k$ and then substituted the value of $k$ into the given expression to obtain the final answer.

Final Answer The final answer is $e^{3}+e^{6}+e^{11}$, which corresponds to option (A).