If y = sin^{-1}sqrt{1 - x} + cos^{-1}sqrt{x}, | vedclass

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(B) We know that $\sin^{-1}(\sqrt{1 - x}) = \cos^{-1}(\sqrt{x})$.Substituting this into the given equation:$y = \cos^{-1}(\sqrt{x}) + \cos^{-1}(\sqrt{

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$-\frac{1}{\sqrt{x(1 - x)}}$

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(B) We know that $\sin^{-1}(\sqrt{1 - x}) = \cos^{-1}(\sqrt{x})$.Substituting this into the given equation:$y = \cos^{-1}(\sqrt{x}) + \cos^{-1}(\sqrt{x}) = 2\cos^{-1}(\sqrt{x})$.Now,differentiate with respect to $x$:$\frac{dy}{dx} = 2 \cdot \left( -\frac{1}{\sqrt{1 - (\sqrt{x})^2}} \right) \cdot \frac{d}{dx}(\sqrt{x})$$\frac{dy}{dx} = -\frac{2}{\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x}}$$\frac{dy}{dx} = -\frac{1}{\sqrt{x(1 - x)}}$.

If $y = \sin^{-1}\sqrt{1 - x} + \cos^{-1}\sqrt{x}$,then $\frac{dy}{dx} = $

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