Differentiate `tan^(-1)(( sqrt(1+x^(2))-1)/(x))`

Answer

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

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

### Step 2: Use a trigonometric substitution
We can set \( x = \tan(\theta) \). Then, we have:
\[ \sqrt{1+x^2} = \sqrt{1+\tan^2(\theta)} = \sec(\theta) \]
Thus, we can rewrite \( y \):
\[ y = \tan^{-1}\left(\frac{\sec(\theta) - 1}{\tan(\theta)}\right) \]

### Step 3: Simplify the expression
Now, we can simplify the fraction:
\[ \frac{\sec(\theta) - 1}{\tan(\theta)} = \frac{\sec(\theta) - 1}{\frac{\sin(\theta)}{\cos(\theta)}} = \frac{(\sec(\theta) - 1) \cos(\theta)}{\sin(\theta)} \]
This simplifies to:
\[ \frac{1 - \cos(\theta)}{\sin(\theta)} \]

### Step 4: Use the identity for \( 1 - \cos(\theta) \)
Using the identity \( 1 - \cos(\theta) = 2\sin^2\left(\frac{\theta}{2}\right) \):
\[ y = \tan^{-1}\left(\frac{2\sin^2\left(\frac{\theta}{2}\right)}{\sin(\theta)}\right) \]
And since \( \sin(\theta) = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) \):
\[ y = \tan^{-1}\left(\frac{2\sin^2\left(\frac{\theta}{2}\right)}{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)}\right) = \tan^{-1}\left(\frac{\sin\left(\frac{\theta}{2}\right)}{\cos\left(\frac{\theta}{2}\right)}\right) = \tan^{-1}(\tan\left(\frac{\theta}{2}\right)) \]

### Step 5: Simplify further
This implies:
\[ y = \frac{\theta}{2} \]
Since \( \theta = \tan^{-1}(x) \):
\[ y = \frac{1}{2} \tan^{-1}(x) \]

### Step 6: Differentiate with respect to \( x \)
Now, we differentiate \( y \) with respect to \( x \):
\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{1+x^2} \]

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

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