`log (sin ^(-1)x)`

Answer

To find the derivative of the function \( y = \log(\sin^{-1}(x)) \) with respect to \( x \), we can follow these steps:

### Step-by-Step Solution:

1. **Identify the function**: We have \( y = \log(\sin^{-1}(x)) \).

2. **Differentiate using the chain rule**: The derivative of \( \log(u) \) with respect to \( x \) is given by \( \frac{1}{u} \cdot \frac{du}{dx} \). Here, \( u = \sin^{-1}(x) \).

\[ \frac{dy}{dx} = \frac{1}{\sin^{-1}(x)} \cdot \frac{d}{dx}(\sin^{-1}(x)) \]

3. **Differentiate \( \sin^{-1}(x) \)**: The derivative of \( \sin^{-1}(x) \) is \( \frac{1}{\sqrt{1 - x^2}} \).

\[ \frac{d}{dx}(\sin^{-1}(x)) = \frac{1}{\sqrt{1 - x^2}} \]

4. **Substitute back into the derivative**: Now substitute this result back into the derivative we found in step 2.

\[ \frac{dy}{dx} = \frac{1}{\sin^{-1}(x)} \cdot \frac{1}{\sqrt{1 - x^2}} \]

5. **Combine the results**: Thus, we can express the final derivative as:

\[ \frac{dy}{dx} = \frac{1}{\sin^{-1}(x) \cdot \sqrt{1 - x^2}} \]

### Final Answer:
\[ \frac{dy}{dx} = \frac{1}{\sin^{-1}(x) \cdot \sqrt{1 - x^2}} \]