`"F i n d"(dy)/(dx)"if"y=tan^(-1)((sqrt(1+x^2)-1)/x),` where `x!=0`

Explanation

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

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

### Step 2: Substitute \(x = \tan(\theta)\)
Let \(x = \tan(\theta)\). Then, we have:
\[ \sqrt{1+x^2} = \sqrt{1+\tan^2(\theta)} = \sec(\theta) \]
Thus, we can rewrite \(y\) as:
\[ y = \tan^{-1}\left(\frac{\sec(\theta)-1}{\tan(\theta)}\right) \]

### Step 3: Rewrite the Expression
Using the identity \(\sec(\theta) = \frac{1}{\cos(\theta)}\) and \(\tan(\theta) = \frac{\sin(\theta)}{\cos(\theta)}\), we can simplify:
\[ y = \tan^{-1}\left(\frac{\frac{1}{\cos(\theta)} - 1}{\frac{\sin(\theta)}{\cos(\theta)}}\right) \]
This simplifies to:
\[ y = \tan^{-1}\left(\frac{1 - \cos(\theta)}{\sin(\theta)}\right) \]

### Step 4: Use Trigonometric Identities
Using the identity \(1 - \cos(\theta) = 2\sin^2\left(\frac{\theta}{2}\right)\) and \(\sin(\theta) = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)\):
\[ y = \tan^{-1}\left(\frac{2\sin^2\left(\frac{\theta}{2}\right)}{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)}\right) \]
This simplifies to:
\[ y = \tan^{-1}\left(\tan\left(\frac{\theta}{2}\right)\right) \]

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

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

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

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