Let `f (x) = x tan ^(-1) (x^(2))` then find the` f'(x)`

Answer

To find the derivative \( f'(x) \) of the function \( f(x) = x \tan^{-1}(x^2) \), we will use the product rule of differentiation. The product rule states that if you have a function that is the product of two functions \( u(x) \) and \( v(x) \), then the derivative is given by:

\[ (uv)' = u'v + uv' \]

### Step-by-Step Solution:

1. **Identify the Functions**:
Let \( u = x \) and \( v = \tan^{-1}(x^2) \).

2. **Differentiate \( u \)**:
\[ u' = \frac{d}{dx}(x) = 1 \]

3. **Differentiate \( v \)**:
To differentiate \( v = \tan^{-1}(x^2) \), we use the chain rule. The derivative of \( \tan^{-1}(x) \) is \( \frac{1}{1+x^2} \), so:
\[ v' = \frac{d}{dx}(\tan^{-1}(x^2)) = \frac{1}{1+(x^2)^2} \cdot \frac{d}{dx}(x^2) = \frac{1}{1+x^4} \cdot 2x = \frac{2x}{1+x^4} \]

4. **Apply the Product Rule**:
Now we can apply the product rule:
\[ f'(x) = u'v + uv' = (1)(\tan^{-1}(x^2)) + (x)\left(\frac{2x}{1+x^4}\right) \]

5. **Simplify the Expression**:
\[ f'(x) = \tan^{-1}(x^2) + \frac{2x^2}{1+x^4} \]

Thus, the derivative \( f'(x) \) is:
\[ f'(x) = \tan^{-1}(x^2) + \frac{2x^2}{1+x^4} \]