Find `(dy)/(dx)` in the following:`y=sin^(-1)((2x)/(1+x^2))`

Options

A. `2/(1+x^2)`
B. `5/(4+x^2)`
C. `1/(3+x^2)`
D. `5/(6+x^2)`

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 \(\tan(\theta)\)
We can use the trigonometric identity to simplify the expression. We know that:
\[ \frac{2\tan(\theta)}{1+\tan^2(\theta)} = \sin(2\theta) \]
Thus, we can set:
\[ y = \sin^{-1}\left(\frac{2\tan(\theta)}{1+\tan^2(\theta)}\right) \]
This implies:
\[ y = \sin^{-1}(\sin(2\theta)) = 2\theta \]

### Step 2: Express \(\theta\) in terms of \(x\)
Since we have set \(x = \tan(\theta)\), we can express \(\theta\) as:
\[ \theta = \tan^{-1}(x) \]
Therefore:
\[ y = 2\tan^{-1}(x) \]

### Step 3: Differentiate both sides 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 get:
\[ \frac{dy}{dx} = 2 \cdot \frac{1}{1+x^2} \]

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

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