Differentiate the following w.r.t. x :
`cot^(-1)((sqrt(1+x^(2))-1)/(x)),xne0`

Answer

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

### Step 1: Simplify the expression inside the inverse cotangent
We start with:
\[ y = \cot^{-1} \left( \frac{\sqrt{1+x^2}-1}{x} \right) \]

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

### Step 2: Rewrite the expression
Using the identities \( \sec(\theta) = \frac{1}{\cos(\theta)} \) and \( \tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)} \), 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 3: Use a trigonometric identity
Using the identity \( 1 - \cos(\theta) = 2\sin^2\left(\frac{\theta}{2}\right) \), we can further simplify:
\[ \frac{1 - \cos(\theta)}{\sin(\theta)} = \frac{2\sin^2\left(\frac{\theta}{2}\right)}{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)} = \tan\left(\frac{\theta}{2}\right) \]
Thus, we have:
\[ y = \cot^{-1} \left( \tan\left(\frac{\theta}{2}\right) \right) \]

### Step 4: Use the cotangent identity
Using the identity \( \cot^{-1}(\tan(x)) = \frac{\pi}{2} - x \):
\[ y = \frac{\pi}{2} - \frac{\theta}{2} \]

### Step 5: Substitute back for \( \theta \)
Since \( \theta = \tan^{-1}(x) \):
\[ y = \frac{\pi}{2} - \frac{1}{2} \tan^{-1}(x) \]

### Step 6: Differentiate with respect to \( x \)
Now, we differentiate \( y \):
\[ \frac{dy}{dx} = 0 - \frac{1}{2} \cdot \frac{1}{1+x^2} = -\frac{1}{2(1+x^2)} \]

### 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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