Find `(dy)/(dx)` for the function: `y=sin^(-1)sqrt((1-x))+cos^(-1)sqrt(x)`

Explanation

To find \(\frac{dy}{dx}\) for the function

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

we will differentiate it step by step.

### Step 1: Rewrite the Function
We start with the given function:

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

### Step 2: Use the Identity for Inverse Functions
We know that \(\sin^{-1}(a) + \cos^{-1}(a) = \frac{\pi}{2}\) for any \(a\) in the range. However, in this case, we have two different arguments. We can simplify the expression using the fact that:

\[ \sin^{-1}(\sqrt{1-x}) + \cos^{-1}(\sqrt{x}) = \frac{\pi}{2} \text{ when } \sqrt{1-x} = \sqrt{x}. \]

But we will differentiate directly without using this identity for now.

### Step 3: Differentiate Each Term
Now we differentiate each term separately.

1. **Differentiate \(\sin^{-1}(\sqrt{1-x})\)**:
- Let \(u = \sqrt{1-x}\).
- Then, \(\frac{du}{dx} = \frac{-1}{2\sqrt{1-x}}\).
- The derivative of \(\sin^{-1}(u)\) is \(\frac{1}{\sqrt{1-u^2}}\).
- Thus, applying the chain rule:

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

2. **Differentiate \(\cos^{-1}(\sqrt{x})\)**:
- Let \(v = \sqrt{x}\).
- Then, \(\frac{dv}{dx} = \frac{1}{2\sqrt{x}}\).
- The derivative of \(\cos^{-1}(v)\) is \(-\frac{1}{\sqrt{1-v^2}}\).
- Thus, applying the chain rule:

\[ \frac{d}{dx}(\cos^{-1}(\sqrt{x})) = -\frac{1}{\sqrt{1-(\sqrt{x})^2}} \cdot \frac{dv}{dx} = -\frac{1}{\sqrt{1-x}} \cdot \left(\frac{1}{2\sqrt{x}}\right) = -\frac{1}{2\sqrt{x(1-x)}}. \]

### Step 4: Combine the Derivatives
Now we combine the derivatives from both terms:

\[ \frac{dy}{dx} = -\frac{1}{2\sqrt{x(1-x)}} - \frac{1}{2\sqrt{x(1-x)}} = -\frac{1}{\sqrt{x(1-x)}}. \]

### Final Answer
Thus, the derivative \(\frac{dy}{dx}\) is:

\[ \frac{dy}{dx} = -\frac{1}{\sqrt{x(1-x)}}. \]

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