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

Options

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

Answer

To solve the problem, we need to differentiate the function given by:

\[ y = \sin^{-1} \left( x \sqrt{1-x} + \sqrt{x} \sqrt{1-x^2} \right) \]

### Step 1: Simplify the expression inside the inverse sine function

Let’s denote:
- \( x = \sin a \)
- \( \sqrt{x} = \sin b \)

From these, we have:
- \( \sqrt{1-x^2} = \cos a \)
- \( \sqrt{1-x} = \cos b \)

Substituting these into the expression for \( y \):

\[ y = \sin^{-1} \left( \sin a \cos b + \cos a \sin b \right) \]

### Step 2: Use the sine addition formula

The expression \( \sin a \cos b + \cos a \sin b \) can be recognized as \( \sin(a + b) \). Thus, we can rewrite \( y \):

\[ y = \sin^{-1}(\sin(a + b)) \]

### Step 3: Simplify further

Since \( \sin^{-1} \) and \( \sin \) are inverse functions, we have:

\[ y = a + b \]

### Step 4: Substitute back for \( a \) and \( b \)

Recall that:
- \( a = \sin^{-1}(x) \)
- \( b = \sin^{-1}(\sqrt{x}) \)

Thus, we can express \( y \) as:

\[ y = \sin^{-1}(x) + \sin^{-1}(\sqrt{x}) \]

### Step 5: Differentiate \( y \)

Now we differentiate \( y \):

\[ \frac{dy}{dx} = \frac{d}{dx} \left( \sin^{-1}(x) \right) + \frac{d}{dx} \left( \sin^{-1}(\sqrt{x}) \right) \]

Using the derivative of \( \sin^{-1}(u) \), which is \( \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \):

1. For \( \sin^{-1}(x) \):
\[ \frac{d}{dx} \left( \sin^{-1}(x) \right) = \frac{1}{\sqrt{1-x^2}} \]

2. For \( \sin^{-1}(\sqrt{x}) \):
Let \( u = \sqrt{x} \), then \( \frac{du}{dx} = \frac{1}{2\sqrt{x}} \):
\[ \frac{d}{dx} \left( \sin^{-1}(\sqrt{x}) \right) = \frac{1}{\sqrt{1-x}} \cdot \frac{1}{2\sqrt{x}} = \frac{1}{2\sqrt{x}\sqrt{1-x}} \]

### Step 6: Combine the derivatives

Thus, we have:

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

### Final Result

The final expression for \( \frac{dy}{dx} \) is:

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