Let `y= sin^(-1)"" (2 x)/(1 +x^2)` where ` 0 lt x lt 1` and `0 lt y lt pi/2` then ` ( dy)/(dx)` is equal to

Options

A. `(2)/(1 + x^2)`
B. `(2x)/(1+x^2)`
C. `(1)/(1+x^2)`
D. `(-x)/(1+x^2)`

Answer

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

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

### Step 2: Use the double angle identity for sine
Using the identity \(\sin(2\theta) = \frac{2\tan(\theta)}{1 + \tan^2(\theta)}\), we can simplify \(y\):
\[ y = \sin^{-1}(\sin(2\theta)) \]
Since \(0 < \theta < \frac{\pi}{4}\), we have:
\[ y = 2\theta \]

### Step 3: Express \(\theta\) in terms of \(x\)
From our substitution \(x = \tan(\theta)\), we can express \(\theta\) as:
\[ \theta = \tan^{-1}(x) \]
Thus, we can rewrite \(y\):
\[ y = 2\tan^{-1}(x) \]

### Step 4: Differentiate \(y\) with respect to \(x\)
Now, we differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\tan^{-1}(x)) \]
Using the derivative of \(\tan^{-1}(x)\), which is \(\frac{1}{1 + x^2}\), we have:
\[ \frac{dy}{dx} = 2 \cdot \frac{1}{1 + x^2} \]

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

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