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;

& y={{\sin }^{-1}}\left( \sin \left( A+B \right) \right) \\\ & \Rightarrow y=\left( A+B \right) \\\ \end{aligned}$$ Now we had, $$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{aligned} & \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{aligned}$$