` Fi nd dy/dx , y=sin^(-1)""(2x)/(1+x^(2)),-1lexle1`

Explanation

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 identity for \(\sin(2\theta)\)
We know from trigonometric identities that:

\[ \frac{2\tan(\theta)}{1+\tan^2(\theta)} = \sin(2\theta) \]

Thus, we can rewrite \(y\):

\[ y = \sin^{-1}(\sin(2\theta)) \]

### Step 3: Simplify \(y\)
Since \(\sin^{-1}(\sin(2\theta)) = 2\theta\) (for \(-\frac{\pi}{2} \leq 2\theta \leq \frac{\pi}{2}\)), we have:

\[ y = 2\theta \]

### Step 4: Differentiate \(y\) with respect to \(x\)
Now, we differentiate both sides with respect to \(x\):

\[ \frac{dy}{dx} = \frac{d}{dx}(2\theta) \]

Using the chain rule, we get:

\[ \frac{dy}{dx} = 2 \cdot \frac{d\theta}{dx} \]

### Step 5: Find \(\frac{d\theta}{dx}\)
Since \(x = \tan(\theta)\), we differentiate:

\[ \frac{dx}{d\theta} = \sec^2(\theta) \]

Thus,

\[ \frac{d\theta}{dx} = \frac{1}{\sec^2(\theta)} = \cos^2(\theta) \]

### Step 6: Substitute back to find \(\frac{dy}{dx}\)
Substituting back into our expression for \(\frac{dy}{dx}\):

\[ \frac{dy}{dx} = 2 \cdot \cos^2(\theta) \]

### Step 7: Express \(\cos^2(\theta)\) in terms of \(x\)
Using the identity \(\sec^2(\theta) = 1 + \tan^2(\theta)\), we have:

\[ \sec^2(\theta) = 1 + x^2 \implies \cos^2(\theta) = \frac{1}{1+x^2} \]

Thus,

\[ \frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} = \frac{2}{1+x^2} \]

### Final Result
Therefore, the derivative \(\frac{dy}{dx}\) is:

\[ \frac{dy}{dx} = \frac{2}{1+x^2} \]

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