If ` y=tan ^(-1) (sqrt( 1+x^(2)) +x ),then (dy)/(dx) =`

Options

A. ` (-1)/( 2( 1+x^(2)))`
B. ` (1)/( 2( 1+x^(2)))`
C. ` (-2)/( 2( 1+x^(2)))`
D. ` (2)/( 2( 1+x^(2)))`

Answer

To find the derivative of the function \( y = \tan^{-1}(\sqrt{1+x^2} + x) \), we will apply the chain rule and the derivative of the inverse tangent function.

### Step-by-Step Solution:

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

2. **Differentiate using the chain rule**:
The derivative of \( \tan^{-1}(u) \) is given by:
\[ \frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
where \( u = \sqrt{1+x^2} + x \).

3. **Find \( \frac{du}{dx} \)**:
To differentiate \( u = \sqrt{1+x^2} + x \), we need to differentiate each term:
- The derivative of \( \sqrt{1+x^2} \) is:
\[ \frac{d}{dx}(\sqrt{1+x^2}) = \frac{1}{2\sqrt{1+x^2}} \cdot (2x) = \frac{x}{\sqrt{1+x^2}} \]
- The derivative of \( x \) is:
\[ \frac{d}{dx}(x) = 1 \]
Therefore,
\[ \frac{du}{dx} = \frac{x}{\sqrt{1+x^2}} + 1 \]

4. **Substitute \( u \) and \( \frac{du}{dx} \) into the derivative formula**:
Now we substitute \( u \) into the derivative:
\[ \frac{dy}{dx} = \frac{1}{1 + (\sqrt{1+x^2} + x)^2} \cdot \left(\frac{x}{\sqrt{1+x^2}} + 1\right) \]

5. **Simplify \( 1 + (\sqrt{1+x^2} + x)^2 \)**:
Expanding \( (\sqrt{1+x^2} + x)^2 \):
\[ (\sqrt{1+x^2} + x)^2 = 1 + x^2 + 2x\sqrt{1+x^2} \]
Thus,
\[ 1 + (\sqrt{1+x^2} + x)^2 = 1 + 1 + x^2 + 2x\sqrt{1+x^2} = 2 + x^2 + 2x\sqrt{1+x^2} \]

6. **Final expression for \( \frac{dy}{dx} \)**:
Putting everything together:
\[ \frac{dy}{dx} = \frac{\frac{x}{\sqrt{1+x^2}} + 1}{2 + x^2 + 2x\sqrt{1+x^2}} \]

### Final Answer:
\[ \frac{dy}{dx} = \frac{\frac{x + \sqrt{1+x^2}}{\sqrt{1+x^2}}}{2 + x^2 + 2x\sqrt{1+x^2}} \]