If ` Y= tan^(-1) "" [{sqrt""(1+x^2)-1}//x]` then

Options

A. `Y'(0) =1`
B. `y'(0) =1/2`
C. `y'(0) =0`
D. None of these

Answer

To solve the problem, we need to differentiate the function \( Y = \tan^{-1} \left( \frac{\sqrt{1+x^2}-1}{x} \right) \) and find the value of the derivative at \( x = 0 \).

### 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, the expression inside the arctangent becomes:
\[ \frac{\sqrt{1+x^2}-1}{x} = \frac{\sec(\theta)-1}{\tan(\theta)} \]

2. **Simplifying the Expression**:
We can rewrite \( \tan(\theta) \) as \( \frac{\sin(\theta)}{\cos(\theta)} \):
\[ \frac{\sec(\theta)-1}{\tan(\theta)} = \frac{\sec(\theta)-1}{\frac{\sin(\theta)}{\cos(\theta)}} = \frac{(\sec(\theta)-1)\cos(\theta)}{\sin(\theta)} \]
Now, substituting \( \sec(\theta) = \frac{1}{\cos(\theta)} \):
\[ = \frac{\frac{1}{\cos(\theta)} - 1}{\frac{\sin(\theta)}{\cos(\theta)}} = \frac{1 - \cos(\theta)}{\sin(\theta)} \]

3. **Using Trigonometric Identity**:
We can use the identity \( 1 - \cos(\theta) = 2\sin^2\left(\frac{\theta}{2}\right) \):
\[ = \frac{2\sin^2\left(\frac{\theta}{2}\right)}{\sin(\theta)} = \frac{2\sin^2\left(\frac{\theta}{2}\right)}{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)} = \frac{\sin\left(\frac{\theta}{2}\right)}{\cos\left(\frac{\theta}{2}\right)} = \tan\left(\frac{\theta}{2}\right) \]

4. **Final Expression for Y**:
Thus, we have:
\[ Y = \tan^{-1}\left(\tan\left(\frac{\theta}{2}\right)\right) = \frac{\theta}{2} \]

5. **Substituting Back for x**:
Since \( \theta = \tan^{-1}(x) \), we get:
\[ Y = \frac{1}{2} \tan^{-1}(x) \]

6. **Differentiating Y**:
Now, we differentiate \( Y \) with respect to \( x \):
\[ \frac{dY}{dx} = \frac{1}{2} \cdot \frac{1}{1+x^2} \]

7. **Finding the Value at x = 0**:
Now, substituting \( x = 0 \):
\[ \frac{dY}{dx} \bigg|_{x=0} = \frac{1}{2} \cdot \frac{1}{1+0^2} = \frac{1}{2} \]

### Final Answer:
The value of the derivative \( \frac{dY}{dx} \) at \( x = 0 \) is \( \frac{1}{2} \).

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