If `f(x) = tan^-1 {1/x (sqrt(1+x^2))}` then `f'(0)` is

Options

A. -1
B. 0
C. 1
D. does not exist

Answer

To find \( f'(0) \) for the function \( f(x) = \tan^{-1} \left( \frac{\sqrt{1+x^2}}{x} \right) \), we will follow these steps:

### Step 1: Differentiate \( f(x) \)

We start by using the chain rule for differentiation. The derivative of \( \tan^{-1}(u) \) is given by:

\[ \frac{d}{dx} \tan^{-1}(u) = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]

where \( u = \frac{\sqrt{1+x^2}}{x} \).

### Step 2: Find \( u \) and its derivative \( \frac{du}{dx} \)

First, we need to differentiate \( u \):

\[ u = \frac{\sqrt{1+x^2}}{x} \]

Using the quotient rule, where \( p = \sqrt{1+x^2} \) and \( q = x \):

\[ \frac{du}{dx} = \frac{q \frac{dp}{dx} - p \frac{dq}{dx}}{q^2} \]

Calculating \( \frac{dp}{dx} \):

\[ \frac{dp}{dx} = \frac{1}{2\sqrt{1+x^2}} \cdot 2x = \frac{x}{\sqrt{1+x^2}} \]

And \( \frac{dq}{dx} = 1 \). Now substituting these into the quotient rule:

\[ \frac{du}{dx} = \frac{x \cdot \frac{x}{\sqrt{1+x^2}} - \sqrt{1+x^2} \cdot 1}{x^2} \]

This simplifies to:

\[ \frac{du}{dx} = \frac{x^2 - (1+x^2)}{x^2 \sqrt{1+x^2}} = \frac{-1}{x^2 \sqrt{1+x^2}} \]

### Step 3: Substitute \( u \) and \( \frac{du}{dx} \) into the derivative of \( f(x) \)

Now we can substitute \( u \) and \( \frac{du}{dx} \) back into the derivative of \( f(x) \):

\[ f'(x) = \frac{1}{1 + \left( \frac{\sqrt{1+x^2}}{x} \right)^2} \cdot \frac{-1}{x^2 \sqrt{1+x^2}} \]

Calculating \( 1 + \left( \frac{\sqrt{1+x^2}}{x} \right)^2 \):

\[ 1 + \frac{1+x^2}{x^2} = \frac{x^2 + 1 + x^2}{x^2} = \frac{2+x^2}{x^2} \]

Thus, we have:

\[ f'(x) = \frac{1}{\frac{2+x^2}{x^2}} \cdot \frac{-1}{x^2 \sqrt{1+x^2}} = \frac{-x^2}{2+x^2} \cdot \frac{1}{x^2 \sqrt{1+x^2}} = \frac{-1}{(2+x^2) \sqrt{1+x^2}} \]

### Step 4: Evaluate \( f'(0) \)

Now we need to find \( f'(0) \):

\[ f'(0) = \frac{-1}{(2+0) \sqrt{1+0}} = \frac{-1}{2 \cdot 1} = -\frac{1}{2} \]

### Final Answer

Thus, the value of \( f'(0) \) is:

\[ \boxed{-\frac{1}{2}} \]