If f(x) = 2 sin^{-1} sqrt{1-x} + sin^{-1} (2 | vedclass

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(B) Let $x = \sin^2 	heta$. Since $x \in (0, 1/2)$,we have $	heta \in (0, \pi/4)$.Then $\sqrt{1-x} = \sqrt{1-\sin^2 	heta} = \cos 	heta$.Also,$\sq

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(B) Let $x = \sin^2 	heta$. Since $x \in (0, 1/2)$,we have $	heta \in (0, \pi/4)$.Then $\sqrt{1-x} = \sqrt{1-\sin^2 	heta} = \cos 	heta$.Also,$\sqrt{x(1-x)} = \sqrt{\sin^2 	heta \cos^2 	heta} = \sin 	heta \cos 	heta = \frac{1}{2} \sin 2	heta$.Substituting these into $f(x)$:$f(x) = 2 \sin^{-1}(\cos 	heta) + \sin^{-1}(2 \cdot \frac{1}{2} \sin 2	heta)$$f(x) = 2 \sin^{-1}(\sin(\pi/2 - 	heta)) + \sin^{-1}(\sin 2	heta)$Since $	heta \in (0, \pi/4)$,then $(\pi/2 - 	heta) \in (\pi/4, \pi/2)$ and $2	heta \in (0, \pi/2)$.Thus,$f(x) = 2(\pi/2 - 	heta) + 2	heta = \pi - 2	heta + 2	heta = \pi$.Since $f(x) = \pi$ is a constant function,its derivative $f'(x) = 0$.

If $f(x) = 2 \sin^{-1} \sqrt{1-x} + \sin^{-1} (2 \sqrt{x(1-x)})$ where $x \in (0, 1/2)$,then $f'(x)$ has the value equal to

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