Question

If $y=si{n}^{-1}(x\sqrt{1-x}+\sqrt{x}\sqrt{1-x^2})$ then $\frac{dy}{dx}=$

• A. $\frac{-2x}{\sqrt{1-x^2}}+\frac{1}{2\sqrt{x-x^2}}$
• B. $\frac{-1}{\sqrt{1-x^2}}-\frac{1}{2\sqrt{x-x^2}}$
• C. $\frac{1}{\sqrt{1-x^2}}+\frac{1}{2\sqrt{x-x^2}}$
• D. $Noneofthese$

Answer 1

The correct option is C

$\frac{1}{\sqrt{1-x^2}}+\frac{1}{2\sqrt{x-x^2}}$

Putting $x=sinA$ and $\sqrt{x}=sinB$
$\Rightarrow y=si{n}^{-1}(sinA\sqrt{1-si{n}^{2}B}+sinB\sqrt{1-si{n}^{2}A})$
$=si{n}^{-1}\left[sin\right(A+B\left)\right]=A+B=si{n}^{-1}x+si{n}^{-1}\sqrt{x}$
$\Rightarrow \frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}+\frac{1}{2\sqrt{x-x^2}}$