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

Options

A. `(1)/(4)
B. 0
C. 1
D. -1

Answer

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

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

### Step 2: Substitute \( x = \tan \theta \)
To simplify the expression, we can use the substitution \( x = \tan \theta \). Then we have:
\[ \sqrt{1+x^2} = \sqrt{1+\tan^2 \theta} = \sec \theta \]
Thus, the expression becomes:
\[ y = \tan^{-1} \left( \frac{\sec \theta - 1}{\tan \theta} \right) \]

### Step 3: Simplify the expression
Using the identity \( \sec \theta = \frac{1}{\cos \theta} \) and \( \tan \theta = \frac{\sin \theta}{\cos \theta} \), we can rewrite \( y \):
\[ y = \tan^{-1} \left( \frac{\frac{1}{\cos \theta} - 1}{\frac{\sin \theta}{\cos \theta}} \right) = \tan^{-1} \left( \frac{1 - \cos \theta}{\sin \theta} \right) \]

### Step 4: Use the identity for \( 1 - \cos \theta \)
We know that:
\[ 1 - \cos \theta = 2 \sin^2 \left( \frac{\theta}{2} \right) \]
Thus, we can write:
\[ y = \tan^{-1} \left( \frac{2 \sin^2 \left( \frac{\theta}{2} \right)}{\sin \theta} \right) \]

### Step 5: Rewrite \( \sin \theta \)
Using the double angle identity for sine:
\[ \sin \theta = 2 \sin \left( \frac{\theta}{2} \right) \cos \left( \frac{\theta}{2} \right) \]
We substitute this back into our expression for \( y \):
\[ y = \tan^{-1} \left( \frac{2 \sin^2 \left( \frac{\theta}{2} \right)}{2 \sin \left( \frac{\theta}{2} \right) \cos \left( \frac{\theta}{2} \right)} \right) = \tan^{-1} \left( \frac{\sin \left( \frac{\theta}{2} \right)}{\cos \left( \frac{\theta}{2} \right)} \right) = \frac{\theta}{2} \]

### Step 6: Substitute back for \( \theta \)
Since \( \theta = \tan^{-1} x \), we have:
\[ y = \frac{1}{2} \tan^{-1} x \]

### Step 7: Differentiate \( y \)
Now we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{1+x^2} \]

### Step 8: Evaluate at \( x = 1 \)
Now we substitute \( x = 1 \):
\[ y'(1) = \frac{1}{2} \cdot \frac{1}{1+1^2} = \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{4} \]

Thus, the final answer is:
\[ \boxed{\frac{1}{4}} \]