If `y= sin^(-1)"" sqrt""(1-x)+ cos ^(-1) sqrt""x` then ` dy//dx =`

Options

A. `(1)/( sqrt""[x(1-x)])`
B. `(-1)/(sqrt""[x(1-x)])`
C. `(1)/([x(1-x)])`
D. None

Answer

To find the derivative \( \frac{dy}{dx} \) of the function \( y = \sin^{-1}(\sqrt{1-x}) + \cos^{-1}(\sqrt{x}) \), we will differentiate each term separately using the chain rule.

### Step-by-Step Solution:

1. **Identify the function:**
\[ y = \sin^{-1}(\sqrt{1-x}) + \cos^{-1}(\sqrt{x}) \]

2. **Differentiate the first term \( \sin^{-1}(\sqrt{1-x}) \):**
- The derivative of \( \sin^{-1}(u) \) is \( \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx} \).
- Let \( u = \sqrt{1-x} \), then \( \frac{du}{dx} = \frac{1}{2\sqrt{1-x}} \cdot (-1) = -\frac{1}{2\sqrt{1-x}} \).
- Now, we need \( 1-u^2 = 1 - (1-x) = x \).
- Thus, the derivative of the first term is:
\[ \frac{d}{dx} \left( \sin^{-1}(\sqrt{1-x}) \right) = \frac{1}{\sqrt{x}} \cdot \left(-\frac{1}{2\sqrt{1-x}}\right) = -\frac{1}{2\sqrt{x(1-x)}} \]

3. **Differentiate the second term \( \cos^{-1}(\sqrt{x}) \):**
- The derivative of \( \cos^{-1}(v) \) is \( -\frac{1}{\sqrt{1-v^2}} \cdot \frac{dv}{dx} \).
- Let \( v = \sqrt{x} \), then \( \frac{dv}{dx} = \frac{1}{2\sqrt{x}} \).
- Now, we need \( 1-v^2 = 1 - x \).
- Thus, the derivative of the second term is:
\[ \frac{d}{dx} \left( \cos^{-1}(\sqrt{x}) \right) = -\frac{1}{\sqrt{1-x}} \cdot \left(\frac{1}{2\sqrt{x}}\right) = -\frac{1}{2\sqrt{x(1-x)}} \]

4. **Combine the derivatives:**
\[ \frac{dy}{dx} = -\frac{1}{2\sqrt{x(1-x)}} - \frac{1}{2\sqrt{x(1-x)}} \]
\[ \frac{dy}{dx} = -\frac{2}{2\sqrt{x(1-x)}} = -\frac{1}{\sqrt{x(1-x)}} \]

### Final Answer:
\[ \frac{dy}{dx} = -\frac{1}{\sqrt{x(1-x)}} \]