`"Find" (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
Let \(y = \tan^{-1}\left(\frac{\sqrt{1+x^2}-1}{x}\right)\).

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

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

### Step 4: Use the half-angle formula
Using the half-angle identity, we can express \(1 - \cos(\theta)\) as:
\[ 1 - \cos(\theta) = 2\sin^2\left(\frac{\theta}{2}\right) \]
Thus, we have:
\[ y = \tan^{-1}\left(\frac{2\sin^2\left(\frac{\theta}{2}\right)}{\sin(\theta)}\right) \]
And since \(\sin(\theta) = 2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)\), we can simplify:
\[ y = \tan^{-1}\left(\frac{2\sin^2\left(\frac{\theta}{2}\right)}{2\sin\left(\frac{\theta}{2}\right)\cos\left(\frac{\theta}{2}\right)}\right) = \tan^{-1}\left(\frac{\sin\left(\frac{\theta}{2}\right)}{\cos\left(\frac{\theta}{2}\right)}\right) = \frac{\theta}{2} \]

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

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

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

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