`(d)/(dx)tan^(-1)((x)/(1-sqrt(1+x^(2))))]=`

Options

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

Answer

To differentiate the function \( y = \tan^{-1}\left(\frac{x}{1 - \sqrt{1 + x^2}}\right) \), we will follow a series of steps involving trigonometric substitution and differentiation.

### Step-by-Step Solution:

1. **Substitution**:
Let \( x = \tan(\theta) \). Then, we have:
\[ \sqrt{1 + x^2} = \sqrt{1 + \tan^2(\theta)} = \sec(\theta) \]
Therefore, we can rewrite the function:
\[ y = \tan^{-1}\left(\frac{\tan(\theta)}{1 - \sec(\theta)}\right) \]

2. **Simplifying the Argument**:
The expression inside the arctangent becomes:
\[ \frac{\tan(\theta)}{1 - \sec(\theta)} = \frac{\tan(\theta)}{1 - \frac{1}{\cos(\theta)}} = \frac{\tan(\theta)}{\frac{\cos(\theta) - 1}{\cos(\theta)}} = \frac{\tan(\theta) \cdot \cos(\theta)}{\cos(\theta) - 1} \]
Using the identity \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \), we get:
\[ \frac{\sin(\theta)}{\cos(\theta) - 1} \]

3. **Differentiating**:
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{1 + \left(\frac{\sin(\theta)}{\cos(\theta) - 1}\right)^2} \cdot \frac{d}{dx}\left(\frac{\sin(\theta)}{\cos(\theta) - 1}\right) \]

4. **Finding the Derivative of the Inner Function**:
We need to apply the chain rule and the quotient rule here. The derivative of \( \sin(\theta) \) with respect to \( x \) is:
\[ \frac{d}{dx}(\sin(\theta)) = \cos(\theta) \cdot \frac{d\theta}{dx} \]
and for \( \cos(\theta) - 1 \):
\[ \frac{d}{dx}(\cos(\theta) - 1) = -\sin(\theta) \cdot \frac{d\theta}{dx} \]
Thus, using the quotient rule:
\[ \frac{d}{dx}\left(\frac{\sin(\theta)}{\cos(\theta) - 1}\right) = \frac{(\cos(\theta)(\cos(\theta) - 1) - \sin(\theta)(-\sin(\theta)))}{(\cos(\theta) - 1)^2} \cdot \frac{d\theta}{dx} \]

5. **Finding \( \frac{d\theta}{dx} \)**:
Since \( x = \tan(\theta) \), we have:
\[ \frac{d\theta}{dx} = \frac{1}{1 + x^2} \]

6. **Combining Everything**:
Substitute \( \frac{d\theta}{dx} \) back into the derivative expression and simplify.

7. **Final Result**:
After simplification, we find:
\[ \frac{dy}{dx} = \frac{1}{2(1 + x^2)} \]

### Final Answer:
\[ \frac{d}{dx}\tan^{-1}\left(\frac{x}{1 - \sqrt{1 + x^2}}\right) = \frac{1}{2(1 + x^2)} \]