Differentiate the following w.r.t. x :
`(sin^(-1)x)^(x)`

Answer

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

### Step 1: Take the natural logarithm of both sides
We start by taking the natural logarithm of both sides to simplify the differentiation process:
\[ \ln y = \ln((\sin^{-1} x)^x) \]

### Step 2: Apply the power rule of logarithms
Using the power rule of logarithms, we can rewrite the right-hand side:
\[ \ln y = x \cdot \ln(\sin^{-1} x) \]

### Step 3: Differentiate both sides with respect to \( x \)
Now we differentiate both sides with respect to \( x \). Remember to use the product rule on the right-hand side:
\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}(x \ln(\sin^{-1} x)) \]
Using the chain rule on the left side:
\[ \frac{1}{y} \frac{dy}{dx} = \ln(\sin^{-1} x) + x \cdot \frac{d}{dx}(\ln(\sin^{-1} x)) \]

### Step 4: Differentiate \( \ln(\sin^{-1} x) \)
To differentiate \( \ln(\sin^{-1} x) \), we use the chain rule:
\[ \frac{d}{dx}(\ln(\sin^{-1} x)) = \frac{1}{\sin^{-1} x} \cdot \frac{d}{dx}(\sin^{-1} x) \]
We know that:
\[ \frac{d}{dx}(\sin^{-1} x) = \frac{1}{\sqrt{1 - x^2}} \]
Thus:
\[ \frac{d}{dx}(\ln(\sin^{-1} x)) = \frac{1}{\sin^{-1} x} \cdot \frac{1}{\sqrt{1 - x^2}} \]

### Step 5: Substitute back into the equation
Now we substitute this back into our differentiation equation:
\[ \frac{1}{y} \frac{dy}{dx} = \ln(\sin^{-1} x) + x \cdot \left(\frac{1}{\sin^{-1} x} \cdot \frac{1}{\sqrt{1 - x^2}}\right) \]

### Step 6: Solve for \( \frac{dy}{dx} \)
Now, we can solve for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = y \left( \ln(\sin^{-1} x) + \frac{x}{\sin^{-1} x \sqrt{1 - x^2}} \right) \]

### Step 7: Substitute \( y \) back in
Recall that \( y = (\sin^{-1} x)^x \), so we substitute back:
\[ \frac{dy}{dx} = (\sin^{-1} x)^x \left( \ln(\sin^{-1} x) + \frac{x}{\sin^{-1} x \sqrt{1 - x^2}} \right) \]

### Final Answer
Thus, the derivative of \( y = (\sin^{-1} x)^x \) with respect to \( x \) is:
\[ \frac{dy}{dx} = (\sin^{-1} x)^x \left( \ln(\sin^{-1} x) + \frac{x}{\sin^{-1} x \sqrt{1 - x^2}} \right) \]

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