If `y= tan ^(-1) ((x)/(1+ sqrt(1- x ^(2)))), |x|le 1, `then `(dy)/(dx) at ((1)/(2))` is:

Options

A. `(1)/(sqrt3)`
B. 3
C. `(sqrt3)/(2)`
D. `(2)/(sqrt3)`

Answer

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

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

### Step 2: Simplify the expression
Using the identity \(\sqrt{1 - \sin^2 \theta} = \cos \theta\), we can rewrite the expression:
\[ y = \tan^{-1}\left(\frac{\sin \theta}{1 + \cos \theta}\right) \]

### Step 3: Further simplify the fraction
Using the identity \(1 + \cos \theta = 2 \cos^2\left(\frac{\theta}{2}\right)\) and \(\sin \theta = 2 \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right)\), we can rewrite \(y\) as:
\[ y = \tan^{-1}\left(\frac{2 \sin\left(\frac{\theta}{2}\right) \cos\left(\frac{\theta}{2}\right)}{2 \cos^2\left(\frac{\theta}{2}\right)}\right) = \tan^{-1}\left(\tan\left(\frac{\theta}{2}\right)\right) \]

### Step 4: Simplify using the properties of \(\tan^{-1}\)
Since \(\tan^{-1}(\tan(x)) = x\) for \(x\) in the principal range, we have:
\[ y = \frac{\theta}{2} \]

### Step 5: Substitute back for \(\theta\)
Recall that \(\theta = \sin^{-1}(x)\), so:
\[ y = \frac{1}{2} \sin^{-1}(x) \]

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

### Step 7: Evaluate \(\frac{dy}{dx}\) at \(x = \frac{1}{2}\)
Substituting \(x = \frac{1}{2}\):
\[ \frac{dy}{dx} = \frac{1}{2} \cdot \frac{1}{\sqrt{1 - \left(\frac{1}{2}\right)^2}} = \frac{1}{2} \cdot \frac{1}{\sqrt{1 - \frac{1}{4}}} = \frac{1}{2} \cdot \frac{1}{\sqrt{\frac{3}{4}}} = \frac{1}{2} \cdot \frac{2}{\sqrt{3}} = \frac{1}{\sqrt{3}} \]

### Final Answer
Thus, \(\frac{dy}{dx}\) at \(x = \frac{1}{2}\) is:
\[ \frac{1}{\sqrt{3}} \]