If `y = tan^(-1) {(sqrt(1 + x^(2)) -1)/(x)} " then " (dy)/(dx)=` ?

Options

A. `(1)/((1 + x^(2)))`
B. `(2)/((1 + x^(2)))`
C. `(1)/(2 (1 +x^(2)))`
D. none of these

Explanation

To solve the problem \( y = \tan^{-1} \left( \frac{\sqrt{1 + x^2} - 1}{x} \right) \) and find \( \frac{dy}{dx} \), we will use a substitution method to simplify the expression before differentiating.

### Step-by-Step Solution:

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

2. **Simplifying the Expression**:
We can rewrite the expression inside the arctangent:
\[ y = \tan^{-1} \left( \frac{\sec(\theta) - 1}{\tan(\theta)} \right) = \tan^{-1} \left( \frac{\frac{1}{\cos(\theta)} - 1}{\frac{\sin(\theta)}{\cos(\theta)}} \right) \]
Simplifying further:
\[ = \tan^{-1} \left( \frac{1 - \cos(\theta)}{\sin(\theta)} \right) \]

3. **Using Trigonometric Identities**:
We can use the identity \( 1 - \cos(\theta) = 2\sin^2\left(\frac{\theta}{2}\right) \) to rewrite the expression:
\[ y = \tan^{-1} \left( \frac{2\sin^2\left(\frac{\theta}{2}\right)}{\sin(\theta)} \right) \]
Since \( \sin(\theta) = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) \), we have:
\[ y = \tan^{-1} \left( \frac{2\sin^2\left(\frac{\theta}{2}\right)}{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)} \right) = \tan^{-1} \left( \frac{\sin\left(\frac{\theta}{2}\right)}{\cos\left(\frac{\theta}{2}\right)} \right) \]
Thus, we can simplify:
\[ y = \frac{\theta}{2} \]

4. **Finding \( \frac{dy}{dx} \)**:
Now, since \( x = \tan(\theta) \), we have:
\[ \frac{dx}{d\theta} = \sec^2(\theta) \]
Therefore, using the chain rule:
\[ \frac{dy}{dx} = \frac{dy}{d\theta} \cdot \frac{d\theta}{dx} \]
We know \( \frac{dy}{d\theta} = \frac{1}{2} \), and \( \frac{d\theta}{dx} = \frac{1}{\sec^2(\theta)} = \cos^2(\theta) \).

5. **Final Expression**:
Thus, we get:
\[ \frac{dy}{dx} = \frac{1}{2} \cdot \cos^2(\theta) \]
Since \( \cos(\theta) = \frac{1}{\sqrt{1 + x^2}} \), we have:
\[ \frac{dy}{dx} = \frac{1}{2} \cdot \left(\frac{1}{\sqrt{1 + x^2}}\right)^2 = \frac{1}{2(1 + x^2)} \]

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