Answer is: option4

\( (\sin x)^{\frac{1}{x}} \left[ \frac{x \cot x - \ln (\sin x)}{x^2} \right] \)

Solution:

We are given:

\[ y = (\sin x)^{\frac{1}{x}} \]

Taking the natural logarithm on both sides:

\[ \ln y = \frac{1}{x} \ln (\sin x) \] Using the derivative rule: \[ \frac{d}{dx} (\ln y) = \frac{d}{dx} \left( \frac{1}{x} \ln (\sin x) \right) \] Using the product rule and chain rule: \[ \frac{1}{y} \frac{dy}{dx} = \left( \frac{1}{x} \cdot \frac{\cos x}{\sin x} \right) + \left( \ln (\sin x) \cdot \frac{d}{dx} \left(\frac{1}{x} \right) \right) \]

Since \( \frac{d}{dx} \left( \frac{1}{x} \right) = -\frac{1}{x^2} \), we get:

\[ \frac{1}{y} \frac{dy}{dx} = \frac{\cos x}{x \sin x} - \frac{\ln (\sin x)}{x^2} \]

Multiplying both sides by \( y = (\sin x)^{\frac{1}{x}} \), we obtain:

\[ y' = (\sin x)^{\frac{1}{x}} \left[ \frac{\cos x}{x \sin x} - \frac{\ln (\sin x)}{x^2} \right] \]

Since \( \frac{\cos x}{\sin x} = \cot x \), we rewrite:

\[ y' = (\sin x)^{\frac{1}{x}} \left[ \frac{x \cot x - \ln (\sin x)}{x^2} \right] \]