Answer is: option4

Solution:

Step 1: Recall the inverse function derivative rule

If \( g \) is the inverse of \( f \), then the derivative of the inverse function at any point is given by the formula:

\[ (f^{-1})'(y) = \frac{1}{f'(f^{-1}(y))} \]

Step 2: Solve for \( f^{-1}\left(\frac{\sqrt{3}}{2}\right) \)

We need to find the value of \( x \) such that \( \sin x = \frac{\sqrt{3}}{2} \). This is a standard value from trigonometry. The solution for \( \sin x = \frac{\sqrt{3}}{2} \) is:

\[ x = \frac{\pi}{3} \]

Step 3: Find \( f'(x) \)

The derivative of \( f(x) = \sin x \) is:

\[ f'(x) = \cos x \]

Step 4: Evaluate \( f'\left(\frac{\pi}{3}\right) \)

Substitute \( x = \frac{\pi}{3} \) into \( f'(x) \):

\[ f'\left(\frac{\pi}{3}\right) = \cos\left(\frac{\pi}{3}\right) = \frac{1}{2} \]

Step 5: Apply the inverse function derivative formula

Now, apply the inverse function derivative formula:

\[ (f^{-1})'\left(\frac{\sqrt{3}}{2}\right) = \frac{1}{f'\left(\frac{\pi}{3}\right)} = \frac{1}{\frac{1}{2}} = 2 \]

Final Answer:

The value of \( (f^{-1})'\left(\frac{\sqrt{3}}{2}\right) \) is \( 2 \), so the correct answer is (D).