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

Options

A. `1//4`
B. `1//2`
C. `-1//4`
D. `-1//2`

Answer

To find \( y' \) when \( y = \tan^{-1} \left( \frac{\sqrt{1 + x^2} - 1}{x} \right) \), we will follow these steps:

### Step 1: Simplify the Expression
We start with the expression for \( y \):
\[ y = \tan^{-1} \left( \frac{\sqrt{1 + x^2} - 1}{x} \right) \]
Let \( x = \tan(\theta) \). Then, we have:
\[ \sqrt{1 + x^2} = \sqrt{1 + \tan^2(\theta)} = \sec(\theta) \]
Thus, we can rewrite \( y \) as:
\[ y = \tan^{-1} \left( \frac{\sec(\theta) - 1}{\tan(\theta)} \right) \]

### Step 2: Rewrite in Terms of Sine and Cosine
Using the identities for secant and tangent:
\[ y = \tan^{-1} \left( \frac{\sec(\theta) - 1}{\tan(\theta)} \right) = \tan^{-1} \left( \frac{\frac{1}{\cos(\theta)} - 1}{\frac{\sin(\theta)}{\cos(\theta)}} \right) \]
This simplifies to:
\[ y = \tan^{-1} \left( \frac{1 - \cos(\theta)}{\sin(\theta)} \right) \]

### Step 3: Use Half-Angle Identities
Using the half-angle identities:
\[ 1 - \cos(\theta) = 2 \sin^2\left(\frac{\theta}{2}\right) \quad \text{and} \quad \sin(\theta) = 2 \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right) \]
We can rewrite \( y \):
\[ y = \tan^{-1} \left( \frac{2 \sin^2\left(\frac{\theta}{2}\right)}{2 \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right)} \right) = \tan^{-1} \left( \frac{\sin\left(\frac{\theta}{2}\right)}{\cos\left(\frac{\theta}{2}\right)} \right) = \frac{\theta}{2} \]

### Step 4: Substitute Back for \( \theta \)
Since \( \theta = \tan^{-1}(x) \):
\[ y = \frac{1}{2} \tan^{-1}(x) \]

### Step 5: Differentiate \( y \)
Now we differentiate \( y \) with respect to \( x \):
\[ y' = \frac{1}{2} \cdot \frac{1}{1 + x^2} \]

### Step 6: Evaluate at \( x = 1 \)
Now we need to find \( y'(1) \):
\[ y'(1) = \frac{1}{2} \cdot \frac{1}{1 + 1^2} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \]

### Final Answer
Thus, the value of \( y'(1) \) is:
\[ \boxed{\frac{1}{4}} \]