Question

If y = sin ${}^{-1}(x.\sqrt{1-x}+\sqrt{x}\sqrt{1-x^2})$, 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. None of these

Answer 1

The correct option is A $\frac{-2x}{\sqrt{1+x^2}}+\frac{1}{2\sqrt{x-x^2}}$

$y={\sin }^{-1}(x\sqrt{1-x}+\sqrt{x}\sqrt{1-x^2})={\sin }^{-1}x+{\sin }^{-1}\sqrt{x}\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}+\frac{1}{\sqrt{1-x}}\times \frac{1}{2\sqrt{x}}=\frac{1}{\sqrt{1-x^2}}+\frac{1}{2\sqrt{x}\sqrt{1-x}}$