Find $\frac {dy}{dx}$ if $y=\frac{x\sin^{-1}x}{\sqrt{1-x^2}}$ taking JDs advice i used $(fg)'=f'g+fg'$ rule
$$f=\frac x{\sqrt{(1-x^2)}}$$ $$f'=\frac 1{\sqrt{(1-x)}^3}$$
$$g=sin^{-1}x$$ $$g'=\frac{1}{\sqrt{1-x^2}}$$
so anyway adding together
we get
$$\frac 1{(\sqrt{(1-x)}^3}*sin^-x+\frac x{\sqrt{(1-x^2)}}*\frac{1}{\sqrt{1-x^2}}$$
which can be simplified
$$\frac {sin^{-1}x}{(1-x)^3}+\frac x{\sqrt{(1-x^2)^2}}$$
i then multiplied $$\frac x{\sqrt{(1-x^2)^2}}$$ with $\sqrt{(1-x^2)}$ on both numerator and denominator
getting
$$\frac {x(\sqrt{(1-x^2)})}{\sqrt{(1-x^2)^3}}$$
and combining them both we get
$$\frac{sin^{-1}x+x(\sqrt{(1-x^2)}}{\sqrt{(1-x^2)^3}}$$
ps
i never used arcsin before and am compeletely unfamilliar with it

*

*No.

*Since it is false, this is irrelevant.

*I would write the function like this $y=x\arcsin x(1-x^2)^{-1/2}$. By triple product rule,
$$y'=\arcsin x(1-x^2)^{-1/2}+x\frac{1}{\sqrt{1-x^2}}(1-x^2)^{-1/2}+\\x\arcsin x(-1/2)(1-x^2)^{-3/2}(-2x)$$
$$y'=\frac{\arcsin x}{(1-x^2)^{1/2}}+\frac{x}{{1-x^2}}+\frac{x^2\arcsin x}{(1-x^2)^{3/2}}$$
First and third terms can be combined to give:
$$y'=\frac{x}{{1-x^2}}+\frac{\arcsin x}{(1-x^2)^{3/2}}.$$ This is WA.

