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: 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
A: *

*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.

A: You might see that the function can be rewritten as follows:
$$
y=-\arcsin(x)\Big((1-x^2)^{\tfrac{1}{2}}\Big)'
$$
A: 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 $\sin^2t+\cos^2t = 1$.
