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**Hint**: To solve this question we will use various trigonometric identities. Some of them are as follows- \[\sin A.\cos B+\cos A.\sin B=\sin \left( A+B \right)\].

\[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\] and \[\dfrac{d}{dx}{{\sin }^{-1}}x=\dfrac{1}{\sqrt{1-{{x}^{2}}}}\]. First we will make proper substitution then we will use above identities to get the result.

**:**

__Complete step-by-step answer__Given that, \[y={{\sin }^{-1}}\left( x.\sqrt{1-x}+\sqrt{x}\sqrt{1-{{x}^{2}}} \right)\].

First of all we will simplify the given terms for that we will make certain assumptions.

Let \[x=\sin A\] and \[\sqrt{x}=\sin B\].

Now as, \[x=\sin A\].

And we have a trigonometric formula which is given as, \[{{\sin }^{2}}\theta +{{\cos }^{2}}\theta =1\].

\[\Rightarrow {{\sin }^{2}}\theta =1-{{\cos }^{2}}\theta \]

Taking square root both sides we get,

\[\sin \theta =\sqrt{1-\cos \theta }\] - (1)

Now we have, \[x=\sin A\]

Using equation (1), we get

\[\sqrt{1-{{x}^{2}}}=\cos A\]

This is so as, \[x=\sin A\]

Squaring both sides \[\Rightarrow {{x}^{2}}={{\sin }^{2}}A\].

\[\Rightarrow 1-{{x}^{2}}=1-{{\sin }^{2}}A\]

And taking under root we have,

\[\Rightarrow \sqrt{1-{{x}^{2}}}=\sqrt{1-{{\sin }^{2}}A}\]

\[\Rightarrow \sqrt{1-{{x}^{2}}}=\cos A\] - (2)

Similarly we have, \[\sqrt{x}=\sin B\].

Again using equation (i) we have,

\[\sqrt{{{\left( 1-\sqrt{x} \right)}^{2}}}=\cos B\]

\[\Rightarrow \sqrt{1-x}=\cos B\] - (3)

Finally we will use obtained values in the value of y.

Substituting the value of equation (2) and (3) in value of y we get;

\[y={{\sin }^{-1}}\left( \sin A\cos B+\cos A\sin B \right)\]

Now we will use trigonometric identity given as,

\[\sin \left( A+B \right)=\sin A\cos B+\cos A\sin B\]

Using this identity in value of y we get;

\[\begin{align}

& y={{\sin }^{-1}}\left( \sin \left( A+B \right) \right) \\

& \Rightarrow y=\left( A+B \right) \\

\end{align}\]

Now we had, \[x=\sin A\].

Taking \[{{\sin }^{-1}}\] both sides we have, \[{{\sin }^{-1}}x=A\].

And we had \[\sqrt{x}=\sin B\].

Again taking \[{{\sin }^{-1}}\] both sides we have, \[{{\sin }^{-1}}\sqrt{x}=B\].

Substituting the value of A & B in y we get,

\[y={{\sin }^{-1}}x+{{\sin }^{-1}}\sqrt{x}\]

Now finally we will differentiate the obtained values. For that we will the formula of differentiation of \[{{\sin }^{-1}}x\] which is \[\dfrac{d}{dx}{{\sin }^{-1}}x=\dfrac{1}{\sqrt{1-{{x}^{2}}}}\].

Using this formula and differentiating y with respect to x we get,

\[\dfrac{d}{dx}=\dfrac{d}{dx}\left( {{\sin }^{-1}}x+{{\sin }^{-1}}\sqrt{x} \right)\]

Using, \[\dfrac{d}{dx}{{\sin }^{-1}}x=\dfrac{1}{\sqrt{1-{{x}^{2}}}}\] we get,

Applying chain rule of differentiation we get,

\[\dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-{{x}^{2}}}}+\dfrac{1}{\sqrt{1-{{\left( \sqrt{x} \right)}^{2}}}}.\dfrac{1}{2\sqrt{x}}\]

\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-{{x}^{2}}}}+\dfrac{1}{2\sqrt{x}\sqrt{1-\left( x \right)}}\]

Multiplying \[\sqrt{x}\] inside we get,

\[\Rightarrow \dfrac{dy}{dx}=\dfrac{1}{\sqrt{1-{{x}^{2}}}}+\dfrac{1}{2\sqrt{x-{{x}^{2}}}}\]

Hence the value of \[\dfrac{dy}{dx}\] is \[\dfrac{1}{\sqrt{1-{{x}^{2}}}}+\dfrac{1}{2\sqrt{x-{{x}^{2}}}}\],

**So, the correct answer is “Option C”.**

**Note**: Students can get confused at the point where we have to differentiate \[{{\sin }^{-1}}\sqrt{x}\]. Always remember that we apply chain rule of differentiate in such cases,

The derivative is done as follows –

\[\begin{align}

& \dfrac{d}{dx}\left( {{\sin }^{-1}}\sqrt{x} \right)=\dfrac{d}{dx}\left( {{\sin }^{-1}}\sqrt{x} \right)=\dfrac{1}{\sqrt{1-{{\left( \sqrt{x} \right)}^{2}}}}\dfrac{d}{dx}\sqrt{x} \\

& \Rightarrow \dfrac{d}{dx}\left( {{\sin }^{-1}}\sqrt{x} \right)=\dfrac{1}{\sqrt{1-x}}\dfrac{1}{2\sqrt{x}} \\

\end{align}\]

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