y=sin^-1(root 1-x + root 1+x  / 2) find dy/dx

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$$\begin{gathered} We have: y = {\sin }^{-1}\left(\frac{\sqrt{1 - x} + \sqrt{1 + x}}{2}\right) \\ Let x = \cos z \Rightarrow {\cos }^{-1}x = z \\ \therefore y = {\sin }^{-1}\left(\frac{\sqrt{1 - x} + \sqrt{1 + x}}{2}\right) \\ = {\sin }^{-1}\left(\frac{\sqrt{1 - \cos z} + \sqrt{1 + \cos z}}{2}\right) \\ = {\sin }^{-1}\left(\frac{\sqrt{2{\sin }^2{\displaystyle \frac{z}{2}}} + \sqrt{2{\cos }^2{\displaystyle \frac{z}{2}}}}{2}\right) \left[\because 1 - \cos x = 2{\sin }^2\frac{x}{2}, 1 + \cos x = 2{\cos }^2\frac{x}{2}\right] \\ = {\sin }^{-1}\left(\frac{\sqrt{2}\left(\sin {\displaystyle \frac{z}{2}} + \cos {\displaystyle \frac{z}{2}}\right)}{2}\right) \\ = {\sin }^{-1}\left(\frac{\sin \frac{z}{2} + \cos \frac{z}{2}}{\sqrt{2}}\right) \\ = {\sin }^{-1}\left(\frac{1}{\sqrt{2}}\sin \frac{z}{2} + \frac{1}{\sqrt{2}}\cos \frac{z}{2}\right) \\ = {\sin }^{-1}\left(\cos \frac{\mathrm{\pi }}{4}\sin \frac{z}{2} + \sin \frac{\mathrm{\pi }}{4}\cos \frac{z}{2}\right) \\ = {\sin }^{-1}\left(\sin \left(\frac{\mathrm{\pi }}{4} + \frac{z}{2}\right)\right) \left[\because \sin \left(x + y\right) = \sin x\cos y + \cos x\sin y\right] \\ = \frac{\mathrm{\pi }}{4} + \frac{z}{2} = \frac{\mathrm{\pi }}{4} + \frac{{\cos }^{-1}x}{2} = y \left[\because {\sin }^{-1}\left(\sin x\right) = x, z = {\cos }^{-1}x\right] \\ \therefore \frac{dy}{dx} = 0 + \frac{1}{2}\left(\frac{-1}{\sqrt{1 - x^2}}\right) = \boxed{\frac{\mathbf{-}\mathbf{1}}{\mathbf{2}\sqrt{\mathbf{1}\mathbf{ }\mathbf{-}\mathbf{ }{\mathbf{x}}^{\mathbf{2}}}}} \end{gathered}$$

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