Find the derivative of the following w.r.t. x :
`xtan^(-1)x`

Answer

To find the derivative of the function \( y = x \tan^{-1}(x) \) with respect to \( x \), we will use the product rule of differentiation. The product rule states that if you have two functions multiplied together, say \( u \) and \( v \), then the derivative is given by:

\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]

### Step-by-Step Solution:

1. **Identify the Functions**:
Let \( u = x \) and \( v = \tan^{-1}(x) \).

2. **Differentiate Each Function**:
- The derivative of \( u \) with respect to \( x \) is:
\[ \frac{du}{dx} = 1 \]
- The derivative of \( v \) with respect to \( x \) is:
\[ \frac{dv}{dx} = \frac{1}{1 + x^2} \]
(This is a standard derivative of the inverse tangent function.)

3. **Apply the Product Rule**:
Now, substituting into the product rule formula:
\[ \frac{dy}{dx} = u \frac{dv}{dx} + v \frac{du}{dx} \]
Substituting \( u \), \( v \), \( \frac{du}{dx} \), and \( \frac{dv}{dx} \):
\[ \frac{dy}{dx} = x \cdot \frac{1}{1 + x^2} + \tan^{-1}(x) \cdot 1 \]

4. **Simplify the Expression**:
This gives us:
\[ \frac{dy}{dx} = \frac{x}{1 + x^2} + \tan^{-1}(x) \]

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