If ` y= sin(2 sin^(-1)"" x). (dy)/(dx)=`

Options

A. `sqrt(((1-y^2)/(1-x^2))`
B. `2 ((1-2x^2))/sqrt((1-x^2))`
C. `sqrt((1-x^2))/(1-y^2))`
D. None

Answer

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

### Step 1: Rewrite the Function
We start with the function:
\[ y = \sin(2 \sin^{-1}(x)) \]
Using the double angle formula for sine, we can rewrite this as:
\[ y = 2x \sqrt{1 - x^2} \]
This comes from the identity:
\[ \sin(2\theta) = 2\sin(\theta)\cos(\theta) \]
where \(\theta = \sin^{-1}(x)\), thus \(\sin(\theta) = x\) and \(\cos(\theta) = \sqrt{1 - x^2}\).

### Step 2: Differentiate the Function
Now we differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = \frac{d}{dx}(2x\sqrt{1 - x^2}) \]
Using the product rule, where \(u = 2x\) and \(v = \sqrt{1 - x^2}\):
\[ \frac{dy}{dx} = u'v + uv' \]
Calculating \(u'\) and \(v'\):
- \(u' = 2\)
- For \(v = (1 - x^2)^{1/2}\), using the chain rule:
\[ v' = \frac{1}{2}(1 - x^2)^{-1/2} \cdot (-2x) = -\frac{x}{\sqrt{1 - x^2}} \]

### Step 3: Substitute and Simplify
Now substituting \(u\), \(u'\), \(v\), and \(v'\) into the product rule:
\[ \frac{dy}{dx} = 2 \cdot \sqrt{1 - x^2} + 2x \left(-\frac{x}{\sqrt{1 - x^2}}\right) \]
This simplifies to:
\[ \frac{dy}{dx} = 2\sqrt{1 - x^2} - \frac{2x^2}{\sqrt{1 - x^2}} \]
Combining the terms over a common denominator:
\[ \frac{dy}{dx} = \frac{2(1 - x^2) - 2x^2}{\sqrt{1 - x^2}} = \frac{2(1 - 2x^2)}{\sqrt{1 - x^2}} \]

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