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

Answer

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

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

### Step 2: Use the substitution
We can use the trigonometric identity to simplify the expression. Set:
\[ x = \tan(\theta) \]
Then, we have:
\[ \sqrt{1+x^2} = \sqrt{1+\tan^2(\theta)} = \sec(\theta) \]
Thus, the expression inside the arctangent becomes:
\[ \frac{\sec(\theta) - 1}{\tan(\theta)} \]

### Step 3: Simplify the expression
Using the identities:
\[ \sec(\theta) = \frac{1}{\cos(\theta)}, \quad \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \]
we can rewrite:
\[ \frac{\sec(\theta) - 1}{\tan(\theta)} = \frac{\frac{1}{\cos(\theta)} - 1}{\frac{\sin(\theta)}{\cos(\theta)}} = \frac{1 - \cos(\theta)}{\sin(\theta)} \]

### Step 4: Use the half-angle identities
Using the half-angle identities:
\[ 1 - \cos(\theta) = 2\sin^2\left(\frac{\theta}{2}\right), \quad \sin(\theta) = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right) \]
we can substitute:
\[ \frac{1 - \cos(\theta)}{\sin(\theta)} = \frac{2\sin^2\left(\frac{\theta}{2}\right)}{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)} = \frac{\sin\left(\frac{\theta}{2}\right)}{\cos\left(\frac{\theta}{2}\right)} = \tan\left(\frac{\theta}{2}\right) \]

### Step 5: Substitute back
Thus, we have:
\[ y = \tan^{-1}\left(\tan\left(\frac{\theta}{2}\right)\right) = \frac{\theta}{2} \]
Since \( \theta = \tan^{-1}(x) \), we can write:
\[ y = \frac{1}{2} \tan^{-1}(x) \]

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

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

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