Find the derivative of the following function with respect to 'x'
`2 xtan ^(-1)x-log(1+x^(2))`

Answer

To find the derivative of the function \( y = 2x \tan^{-1}(x) - \log(1 + x^2) \) with respect to \( x \), we will apply the product rule and the chain rule of differentiation.

### Step-by-Step Solution:

1. **Identify the function**:
\[ y = 2x \tan^{-1}(x) - \log(1 + x^2) \]

2. **Differentiate the first term \( 2x \tan^{-1}(x) \)**:
- This is a product of two functions: \( u = 2x \) and \( v = \tan^{-1}(x) \).
- Using the product rule \( \frac{d}{dx}(uv) = u \frac{dv}{dx} + v \frac{du}{dx} \):
- \( \frac{du}{dx} = 2 \)
- \( \frac{dv}{dx} = \frac{1}{1 + x^2} \)
- Applying the product rule:
\[ \frac{d}{dx}(2x \tan^{-1}(x)) = 2x \cdot \frac{1}{1 + x^2} + \tan^{-1}(x) \cdot 2 \]
- Simplifying this gives:
\[ \frac{d}{dx}(2x \tan^{-1}(x)) = \frac{2x}{1 + x^2} + 2 \tan^{-1}(x) \]

3. **Differentiate the second term \( -\log(1 + x^2) \)**:
- Using the chain rule:
\[ \frac{d}{dx}(-\log(1 + x^2)) = -\frac{1}{1 + x^2} \cdot \frac{d}{dx}(1 + x^2) = -\frac{1}{1 + x^2} \cdot 2x \]
- This simplifies to:
\[ -\frac{2x}{1 + x^2} \]

4. **Combine the derivatives**:
- Now, we combine the results from steps 2 and 3:
\[ \frac{dy}{dx} = \left( \frac{2x}{1 + x^2} + 2 \tan^{-1}(x) \right) - \frac{2x}{1 + x^2} \]
- The \( \frac{2x}{1 + x^2} \) terms cancel out:
\[ \frac{dy}{dx} = 2 \tan^{-1}(x) \]

5. **Final answer**:
\[ \frac{dy}{dx} = 2 \tan^{-1}(x) \]