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 each term separately.

### Step 1: Differentiate \(\sin^{-1}(\sqrt{1-x})\)

Using the chain rule, the derivative of \(\sin^{-1}(u)\) is given by

\[ \frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}. \]

Let \(u = \sqrt{1-x}\). Then,

\[ u^2 = 1-x \implies 1-u^2 = x. \]

Now, we need to find \(\frac{du}{dx}\):

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

Now substituting back, we have:

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

### Step 2: Differentiate \(\cos^{-1}(\sqrt{x})\)

Using the chain rule again, the derivative of \(\cos^{-1}(u)\) is given by

\[ -\frac{1}{\sqrt{1-u^2}} \cdot \frac{du}{dx}. \]

Let \(v = \sqrt{x}\). Then,

\[ v^2 = x \implies 1-v^2 = 1-x. \]

Now, we need to find \(\frac{dv}{dx}\):

\[ \frac{dv}{dx} = \frac{1}{2\sqrt{x}}. \]

Now substituting back, we have:

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

### Step 3: Combine the derivatives

Now we can combine the derivatives from both parts:

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

### Final Result

Thus, the derivative \(\frac{dy}{dx}\) is

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

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