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

Answer

To differentiate the function \( y = \log(\sin^{-1}(x)) \) with respect to \( x \), we will use the chain rule. Here’s a step-by-step solution:

### Step 1: Define the function
Let:
\[ y = \log(\sin^{-1}(x)) \]

### Step 2: Apply the chain rule
To differentiate \( y \) with respect to \( x \), we apply the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} \]
where \( u = \sin^{-1}(x) \).

### Step 3: Differentiate \( y \) with respect to \( u \)
The derivative of \( y \) with respect to \( u \) is:
\[ \frac{dy}{du} = \frac{1}{u} = \frac{1}{\sin^{-1}(x)} \]

### Step 4: Differentiate \( u \) with respect to \( x \)
Next, we differentiate \( u = \sin^{-1}(x) \) with respect to \( x \):
\[ \frac{du}{dx} = \frac{1}{\sqrt{1 - x^2}} \]

### Step 5: Combine the derivatives
Now, we combine the derivatives using the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx} = \frac{1}{\sin^{-1}(x)} \cdot \frac{1}{\sqrt{1 - x^2}} \]

### Step 6: Write the final result
Thus, the derivative of \( y = \log(\sin^{-1}(x)) \) with respect to \( x \) is:
\[ \frac{dy}{dx} = \frac{1}{\sin^{-1}(x) \cdot \sqrt{1 - x^2}} \]