Question

Let $y={\sin }^{-1}(x\sqrt{1-x}+\sqrt{x}\sqrt{1-x^2})$ and $\frac{dy}{dx}=\frac{1}{\sqrt{1-x^2}}+p$, if $p$ is $\frac{1}{\lambda \sqrt{x(1-x)}}$, then value of $\lambda$.

Answer 1

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