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

Options

A. `(2-4x^(2))/sqrt(1-x^(2))`
B. `(2+4x^(2))/sqrt(1-x^(2))`
C. `(2-4x^(2))/sqrt(1+x^(2))`
D. `(2+4x^(2))/sqrt(1+x^(2))`

Answer

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

### Step 1: Apply the double angle formula
We know the double angle formula for sine:
\[ \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \]
Let \(\theta = \sin^{-1}(x)\). Then, we can rewrite \(y\) as:
\[ y = \sin(2\theta) = 2 \sin(\theta) \cos(\theta) \]
Substituting \(\theta\):
\[ y = 2 \sin(\sin^{-1}(x)) \cos(\sin^{-1}(x)) \]
Since \(\sin(\sin^{-1}(x)) = x\), we have:
\[ y = 2x \cos(\sin^{-1}(x)) \]

### Step 2: Find \(\cos(\sin^{-1}(x))\)
Using the identity \(\cos(\theta) = \sqrt{1 - \sin^2(\theta)}\), we find:
\[ \cos(\sin^{-1}(x)) = \sqrt{1 - x^2} \]
Thus, we can rewrite \(y\) as:
\[ y = 2x \sqrt{1 - x^2} \]

### Step 3: Differentiate \(y\) with respect to \(x\)
Now we differentiate \(y\):
\[ \frac{dy}{dx} = \frac{d}{dx}(2x \sqrt{1 - x^2}) \]
Using the product rule:
\[ \frac{dy}{dx} = 2 \left(\sqrt{1 - x^2} + x \frac{d}{dx}(\sqrt{1 - x^2})\right) \]

### Step 4: Differentiate \(\sqrt{1 - x^2}\)
To differentiate \(\sqrt{1 - x^2}\), we use the chain rule:
\[ \frac{d}{dx}(\sqrt{1 - x^2}) = \frac{1}{2\sqrt{1 - x^2}} \cdot (-2x) = \frac{-x}{\sqrt{1 - x^2}} \]
Substituting this back into our derivative:
\[ \frac{dy}{dx} = 2 \left(\sqrt{1 - x^2} + x \left(\frac{-x}{\sqrt{1 - x^2}}\right)\right) \]
\[ = 2 \left(\sqrt{1 - x^2} - \frac{x^2}{\sqrt{1 - x^2}}\right) \]

### Step 5: Simplify the expression
Combining the terms:
\[ \frac{dy}{dx} = 2 \left(\frac{(1 - x^2) - x^2}{\sqrt{1 - x^2}}\right) = 2 \left(\frac{1 - 2x^2}{\sqrt{1 - x^2}}\right) \]
Thus, the final result is:
\[ \frac{dy}{dx} = \frac{2(1 - 2x^2)}{\sqrt{1 - x^2}} \]