If where then is equal to

JEE MainMathematicsModertaeDiffercntiation

If $y = 2\sin^{- 1}\sqrt{1 - x} - \sin^{- 1}(2\sqrt{x - x^{2}})$
where $x \in (0,\frac{1}{2})$ then $\frac{dy}{dx}$ is equal to

Show answer and solution

y = 2\cos^{- 1}\sqrt{x} - \cos^{- 1}(1 - 2x)

y = 2\cos^{- 1}\sqrt{x} - \pi + \cos^{- 1}(2x - 1)

Let

\cos^{- 1}\sqrt{x} = \theta \Rightarrow x = \cos^{2}\theta\theta \in (\frac{\pi}{4},\frac{\pi}{2})

y = 2\theta - \pi + \cos^{- 1}(2\cos^{2}\theta - 1)

y = 2\theta - \pi + \cos^{- 1}cos2\theta

y = 2\theta - \pi + 2\theta

y = 4\cos^{- 1}\sqrt{x} - \pi

\frac{dy}{dx} = \frac{- 4(1)}{\sqrt{1 - x}} \cdot \frac{1}{2\sqrt{x}}

\frac{dy}{dx} = - \frac{2}{\sqrt{x - x^{2}}}

Source: Grass-G / M-12 Mathematics Paper 2 Nov.docx

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