Answer
$2x\sqrt{1-x^{2}}$
Work Step by Step
Let $y=\arccos x=\cos^{-1}x.$
Then
$\cos y=x\ $ and $\ y\in[0,\pi]$
Apply the double angle identity
$\sin(2y)=2\sin y\cdot\cos y$
$\cos y=x$, as we defined it to be so.
For $ \sin y=\sin(\cos^{-1}x)$,
we know that y is from either the 1st or 2nd quadrant, where sine is positive, so
we take the + sign when solving $\quad\sin^{2}y+\cos^{2}=1$
$\sin y=+\sqrt{1-\cos^{2}y}$
Because of the way we defined y, this equals $\sqrt{1-x^{2}}.$
So,
$\sin(2y)=2\sqrt{1-x^{2}}\cdot x,\qquad$ or
$\sin(2\arccos x)=2x\sqrt{1-x^{2}}$
