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

A faster way is logarithmic differentiation
$$y=\frac{x \sin ^{-1}(x)}{\sqrt{1-x^2}}\implies \log(y)=\log(x)+\log(\sin ^{-1}(x))-\frac 12 \log(1-x^2)$$
$$\frac{y'}y=\frac 1 x+\frac{1 } {\sin ^{-1}(x)\sqrt{1-x^2}}+\frac x{1-x^2}=\frac {x \sqrt{1-x^2} +\sin ^{-1}(x)}{x \left(1-x^2\right) \sin ^{-1}(x) }$$
$$y'=\frac{y'}y \times y=\frac {x \sqrt{1-x^2} +\sin ^{-1}(x)}{x \left(1-x^2\right) \sin ^{-1}(x) } \times \frac{x \sin ^{-1}(x)}{\sqrt{1-x^2}}$$ Just simplify
