Question

If the value of \[y={{\sin }^{-1}}\left( x.\sqrt{1-x}+\sqrt{x}\sqrt{1-{{x}^{2}}} \right)\], then the value of \[\dfrac{dy}{dx}\] is
(a) \[\dfrac{-2x}{\sqrt{1-{{x}^{2}}}}+\dfrac{1}{2\sqrt{x-{{x}^{2}}}}\]
(b) \[\dfrac{-1}{\sqrt{1-{{x}^{2}}}}-\dfrac{1}{2\sqrt{x-{{x}^{2}}}}\]
(c) \[\dfrac{1}{\sqrt{1-{{x}^{2}}}}+\dfrac{1}{2\sqrt{x-{{x}^{2}}}}\]
(d) None of these

Answer 1

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 –
\[\begin{align}
  & \dfrac{d}{dx}\left( {{\sin }^{-1}}\sqrt{x} \right)=\dfrac{d}{dx}\left( {{\sin }^{-1}}\sqrt{x} \right)=\dfrac{1}{\sqrt{1-{{\left( \sqrt{x} \right)}^{2}}}}\dfrac{d}{dx}\sqrt{x} \\
 & \Rightarrow \dfrac{d}{dx}\left( {{\sin }^{-1}}\sqrt{x} \right)=\dfrac{1}{\sqrt{1-x}}\dfrac{1}{2\sqrt{x}} \\
\end{align}\]