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

Answer

To differentiate the function \( y = (\tan^{-1} x)^x \) with respect to \( x \), we can follow these steps:

### Step 1: Take the natural logarithm of both sides
We start by taking the natural logarithm of both sides to simplify the differentiation process:
\[ \ln y = \ln((\tan^{-1} x)^x) \]
Using the property of logarithms, we can rewrite this as:
\[ \ln y = x \cdot \ln(\tan^{-1} x) \]

### Step 2: Differentiate both sides
Now, we differentiate both sides with respect to \( x \). We will use the product rule on the right side:
\[ \frac{d}{dx}(\ln y) = \frac{d}{dx}(x \cdot \ln(\tan^{-1} x)) \]
Using the chain rule on the left side:
\[ \frac{1}{y} \frac{dy}{dx} = \frac{d}{dx}(x) \cdot \ln(\tan^{-1} x) + x \cdot \frac{d}{dx}(\ln(\tan^{-1} x)) \]
The derivative of \( x \) is \( 1 \), so we have:
\[ \frac{1}{y} \frac{dy}{dx} = \ln(\tan^{-1} x) + x \cdot \frac{1}{\tan^{-1} x} \cdot \frac{d}{dx}(\tan^{-1} x) \]

### Step 3: Differentiate \( \tan^{-1} x \)
The derivative of \( \tan^{-1} x \) is:
\[ \frac{d}{dx}(\tan^{-1} x) = \frac{1}{1 + x^2} \]
Substituting this back into our equation gives:
\[ \frac{1}{y} \frac{dy}{dx} = \ln(\tan^{-1} x) + x \cdot \frac{1}{\tan^{-1} x} \cdot \frac{1}{1 + x^2} \]

### Step 4: Solve for \( \frac{dy}{dx} \)
Now, we multiply both sides by \( y \) to isolate \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = y \left( \ln(\tan^{-1} x) + \frac{x}{\tan^{-1} x(1 + x^2)} \right) \]
Recall that \( y = (\tan^{-1} x)^x \), so we substitute back:
\[ \frac{dy}{dx} = (\tan^{-1} x)^x \left( \ln(\tan^{-1} x) + \frac{x}{\tan^{-1} x(1 + x^2)} \right) \]

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