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

Options

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

Answer

To find the derivative of the function \( y = \tan^{-1}(\sqrt{1 + x^2} - x) \), we will follow these steps:

### Step 1: Rewrite the function
We start with the given function:
\[ y = \tan^{-1}(\sqrt{1 + x^2} - x) \]

### Step 2: Differentiate using the chain rule
To differentiate \( y \), we apply 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 \( u = \sqrt{1 + x^2} - x \).

### Step 3: Find \( \frac{du}{dx} \)
Now, we need to differentiate \( u \):
\[ u = \sqrt{1 + x^2} - x \]
To differentiate \( u \), we use the derivative of \( \sqrt{1 + x^2} \):
\[ \frac{d}{dx}(\sqrt{1 + x^2}) = \frac{x}{\sqrt{1 + x^2}} \]
Thus,
\[ \frac{du}{dx} = \frac{x}{\sqrt{1 + x^2}} - 1 \]

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

### Step 5: Simplify \( 1 + (\sqrt{1 + x^2} - x)^2 \)
Next, we need to simplify \( 1 + (\sqrt{1 + x^2} - x)^2 \):
\[ (\sqrt{1 + x^2} - x)^2 = 1 + x^2 - 2x\sqrt{1 + x^2} \]
Thus,
\[ 1 + (\sqrt{1 + x^2} - x)^2 = 1 + 1 + x^2 - 2x\sqrt{1 + x^2} = 2 + x^2 - 2x\sqrt{1 + x^2} \]

### Step 6: Final expression for \( \frac{dy}{dx} \)
Now we can write the final expression for \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{\frac{x}{\sqrt{1 + x^2}} - 1}{2 + x^2 - 2x\sqrt{1 + x^2}} \]

### Final Result
Thus, the derivative \( \frac{dy}{dx} \) is:
\[ \frac{dy}{dx} = \frac{x - \sqrt{1 + x^2}}{\sqrt{1 + x^2}(2 + x^2 - 2x\sqrt{1 + x^2})} \]