If ` y=sin ^(-1)(2x sqrt (1-x^(2))),then (dy)/(dx)=`

Options

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

Answer

To solve the problem \( y = \sin^{-1}(2x \sqrt{1 - x^2}) \) and find \( \frac{dy}{dx} \), we can follow these steps:

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

### Step 2: Simplify the expression
Using the identity \( 1 - \sin^2 \theta = \cos^2 \theta \), we can simplify:
\[ y = \sin^{-1}(2 \sin \theta \cos \theta) \]
Using the double angle identity \( \sin(2\theta) = 2 \sin \theta \cos \theta \), we have:
\[ y = \sin^{-1}(\sin(2\theta)) \]

### Step 3: Simplify further
Since \( \sin^{-1}(\sin(2\theta)) = 2\theta \) (for \( \theta \) in the appropriate range), we can write:
\[ y = 2\theta \]

### Step 4: Substitute back for \( \theta \)
Since \( \theta = \sin^{-1}(x) \), we can substitute back:
\[ y = 2 \sin^{-1}(x) \]

### Step 5: Differentiate \( y \) with respect to \( x \)
Now we differentiate \( y \):
\[ \frac{dy}{dx} = 2 \cdot \frac{d}{dx}(\sin^{-1}(x)) \]
The derivative of \( \sin^{-1}(x) \) is \( \frac{1}{\sqrt{1 - x^2}} \), so:
\[ \frac{dy}{dx} = 2 \cdot \frac{1}{\sqrt{1 - x^2}} = \frac{2}{\sqrt{1 - x^2}} \]

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

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