Question

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

Answer 1

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

$\frac{d{\sin }^{-1}x}{dx}=\frac{1}{\sqrt{1-x^2}}$

$\frac{d\left(x\sqrt{1-x^2}\right)}{dx}=\sqrt{1-x^2}\frac{dx}{dx}+x\frac{d\sqrt{1-x^2}}{dx}$

$\frac{d\sqrt{1-x^2}}{dx}=\frac{d\sqrt{1-x^2}}{d(1-x^2)}\times \frac{d(1-x^2)}{dx}=\frac{1}{2\sqrt{1-x^2}}\times (-2x)=\frac{-x}{\sqrt{1-x^2}}$

$\frac{d\left(x\sqrt{1-x^2}\right)}{dx}=\sqrt{1-x^2}+x\frac{-x}{\sqrt{1-x^2}}=\frac{1-2x^2}{\sqrt{1-x^2}}$

$\frac{dy}{dx}=\frac{d{\sin }^{-1}x}{dx}+\frac{d\left(x\sqrt{1-x^2}\right)}{dx}=\frac{1}{\sqrt{1-x^2}}+\frac{1-2x^2}{\sqrt{1-x^2}}=2\sqrt{1-x^2}$