Answer:
Given express can be written as \[y={{\sin }^{-1}}[\sqrt{x}\sqrt{1-{{x}^{2}}}-x\sqrt{1-(\sqrt{{{x}^{2}}})}]\] Now, put \[x=\sin \theta \] and \[x=\sin \phi \] Then, the given expression reduces to \[y={{\sin }^{-1}}[sin\phi \sqrt{1-{{\sin }^{2}}\theta }-\sin \theta \sqrt{1-{{\sin }^{2}}\phi ]}\] \[={{\sin }^{-1}}[sin\phi cos\theta -cos\phi sin\theta ]\] \[[\because 1-{{\sin }^{2}}A={{\cos }^{2}}A]\] \[={{\sin }^{-1}}[sin(\phi -\theta )]\] \[[\because \sin A\cos B-\cos A\sin B=\sin (A-B)]\] \[=\phi -\theta ={{\sin }^{-1}}\sqrt{x}-{{\sin }^{-1}}x\] \[[\because \,\,x=\sin \theta \,\,\,\Rightarrow \,\,\,\theta ={{\sin }^{-1}}x\] and \[\sqrt{x}=sin\phi \,\,\,\Rightarrow \,\,\,\phi =si{{n}^{-1}}\sqrt{x}]\] \[\therefore \] \[\frac{dy}{dx}=\frac{1}{\sqrt{1-{{(\sqrt{x})}^{2}}}}\cdot \frac{1}{2\sqrt{x}}-\frac{1}{\sqrt{1-{{x}^{2}}}}\] \[=\frac{1}{2\sqrt{x}\sqrt{1-x}}-\frac{1}{\sqrt{1-{{x}^{2}}}}\]
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