Answer is: option4

\( \frac{2}{e} \)

Solution:

Differentiate \( y = \sin u \) with respect to \( x \) \[ \frac{dy}{dx} = \cos u \cdot \frac{du}{dx} \] Differentiate \( u = v - \frac{1}{v} \) with respect to \( x \)

Again, using the chain rule:

\[ \frac{du}{dx} = \left( \frac{dv}{dx} \right) - \left( \frac{d}{dx} \frac{1}{v} \right) \]

Since \( \frac{d}{dx} \frac{1}{v} = -\frac{1}{v^2} \cdot \frac{dv}{dx} \), we get:

\[ \frac{du}{dx} = \frac{dv}{dx} + \frac{1}{v^2} \cdot \frac{dv}{dx} \]

Factor \( \frac{dv}{dx} \):

\[ \frac{du}{dx} = \frac{dv}{dx} \left(1 + \frac{1}{v^2} \right) \] Differentiate \( v = \ln x \) \[ \frac{dv}{dx} = \frac{1}{x} \]

Substituting this into the previous equation:

\[ \frac{du}{dx} = \frac{1}{x} \left( 1 + \frac{1}{v^2} \right) \] First, compute \( v \) at \( x = e \): \[ v = \ln e = 1 \]

Substituting \( v = 1 \) into \( \frac{du}{dx} \):

\[ \frac{du}{dx} = \frac{1}{e} \left( 1 + \frac{1}{1^2} \right) = \frac{1}{e} \times 2 = \frac{2}{e} \] Now, compute \( u \) at \( x = e \): \[ u = v - \frac{1}{v} = 1 - \frac{1}{1} = 0 \]

Thus,

\[ \cos u = \cos 0 = 1 \] Finally, \[ \frac{dy}{dx} = 1 \times \frac{2}{e} = \frac{2}{e} \] Final Answer: \[ \boxed{\frac{2}{e}} \]