For $0<x<\frac\pi4$, let $f(x)=\sqrt{1+\sin2x}-\sqrt{1-\sin2x}$. What is $f\left(\frac\pi{12}\right)$?

1. Using $\sin^2x+\cos^2x=1$ and $\sin2x=2\sin x\cos x$: $$1+\sin2x=(\sin x+\cos x)^2,\qquad 1-\sin2x=(\cos x-\sin x)^2.$$
2. For $00$, $\cos x>\sin x$, so both $\sin x+\cos x$ and $\cos x-\sin x$ are positive. Hence $\sqrt{1+\sin2x}=\sin x+\cos x$ and $\sqrt{1-\sin2x}=\cos x-\sin x$.
3. Therefore $f(x)=(\sin x+\cos x)-(\cos x-\sin x)=2\sin x$.
4. $f\left(\dfrac\pi{12}\right)=2\sin15^\circ=2\cdot\dfrac{\sqrt6-\sqrt2}4=\boxed{\dfrac{\sqrt6-\sqrt2}2}$.