<s>[INST] 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
 [/INST] Here a substitution can help out with calculations. Let $ x = \sin t$, i.e. $t = \arcsin x$. First note that $x\in \langle -1,1 \rangle$, so $t\in \langle -\pi/2,\pi/2 \rangle$ and $\cos t>0$. This is important because we can then simplify $\sqrt{\cos^2t} = |\cos t| = \cos t$, which is used couple of times from now on.
Write $$y =\frac{t\sin t}{\sqrt{1-\sin^2t}}=t\tan t. $$ We can now use the chain rule \begin{align}\frac{dy}{dx} &= \frac{dy}{dt} \cdot
\frac{dt}{dx}\\ &= (\tan t+\frac{t}{\cos^2t})\frac{dt}{dx} = \left(\frac{x}{\sqrt{1-x^2}}+\frac{\arcsin x}{1-x^2}\right)\frac{1}{\sqrt{1-x^2}}=\frac{x}{1-x^2}+\frac{\arcsin x}{(1-x^2)^{3/2}}\end{align}
where we used $\cos t = \sqrt{1-x^2}$ which is derived directly from $\sin^2t+\cos^2t = 1$.
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