`(d)/(dx)[tan^-1((sqrt(1+x^(2))-1)/(x))]=`

Options

A. `(-1)/(1+x^(2))`
B. `(1)/(2(1+x^(2)))`
C. `(1)/(sqrt(1+x^(2)))`
D. `(1)/(2sqrt(1+x^(2)))`

Answer

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

### Step 1: Define the function
Let
\[ y = \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right) \]

### Step 2: Differentiate using the chain rule
To differentiate \( y \) with respect to \( x \), we will use the chain rule. The derivative of \( \tan^{-1}(u) \) is given by:
\[ \frac{dy}{dx} = \frac{1}{1 + u^2} \cdot \frac{du}{dx} \]
where \( u = \frac{\sqrt{1+x^2}-1}{x} \).

### Step 3: Find \( u \) and its derivative \( \frac{du}{dx} \)
First, we need to differentiate \( u \):
\[ u = \frac{\sqrt{1+x^2}-1}{x} \]

Using the quotient rule:
\[ \frac{du}{dx} = \frac{x \cdot \frac{d}{dx}(\sqrt{1+x^2}-1) - (\sqrt{1+x^2}-1) \cdot \frac{d}{dx}(x)}{x^2} \]

Calculating \( \frac{d}{dx}(\sqrt{1+x^2}) \):
\[ \frac{d}{dx}(\sqrt{1+x^2}) = \frac{x}{\sqrt{1+x^2}} \]
Thus,
\[ \frac{d}{dx}(\sqrt{1+x^2}-1) = \frac{x}{\sqrt{1+x^2}} \]

Now substituting back into the expression for \( \frac{du}{dx} \):
\[ \frac{du}{dx} = \frac{x \cdot \frac{x}{\sqrt{1+x^2}} - \left(\sqrt{1+x^2}-1\right)}{x^2} \]
\[ = \frac{\frac{x^2}{\sqrt{1+x^2}} - \sqrt{1+x^2} + 1}{x^2} \]
\[ = \frac{1 - \sqrt{1+x^2} + \frac{x^2}{\sqrt{1+x^2}}}{x^2} \]

### Step 4: Substitute \( u \) and \( \frac{du}{dx} \) into the derivative formula
Now we substitute \( u \) and \( \frac{du}{dx} \) into the derivative formula:
\[ \frac{dy}{dx} = \frac{1}{1 + \left(\frac{\sqrt{1+x^2}-1}{x}\right)^2} \cdot \frac{du}{dx} \]

### Step 5: Simplify \( 1 + u^2 \)
Calculating \( 1 + u^2 \):
\[ u^2 = \left(\frac{\sqrt{1+x^2}-1}{x}\right)^2 = \frac{(1+x^2) - 2\sqrt{1+x^2} + 1}{x^2} = \frac{2 - 2\sqrt{1+x^2}}{x^2} \]
Thus,
\[ 1 + u^2 = 1 + \frac{2 - 2\sqrt{1+x^2}}{x^2} = \frac{x^2 + 2 - 2\sqrt{1+x^2}}{x^2} \]

### Step 6: Final expression for \( \frac{dy}{dx} \)
Now we can substitute everything back into the derivative:
\[ \frac{dy}{dx} = \frac{x^2}{x^2 + 2 - 2\sqrt{1+x^2}} \cdot \frac{du}{dx} \]

### Step 7: Evaluate \( \frac{dy}{dx} \)
After simplifying, we find:
\[ \frac{dy}{dx} = \frac{1}{2(1+x^2)} \]

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