Answer
$\frac{d^2y}{dx^2}=-4\sin x\cos x$
Work Step by Step
$y=sinxcosx$
$\frac{dy}{dx}=(cosx)\times{(cosx)}+(sinx)\times{(-sinx)}$
$\frac{dy}{dx}=cos^2x-sin^2x$
First write it as $\frac{dy}{dx}=(cosx)^2-(sinx)^2$ to spot the chain rule easier.
$\frac{d^2y}{dx^2}=2(cosx)^{2-1}\times\frac{d}{dx}(cosx)-(2(sinx)^{2-1}\times\frac{d}{dx}(sinx))$
$\frac{d^2y}{dx^2}=2(cosx)^1\times{(-sinx)}-(2(sinx)^{1}\times{(cosx)})$
$\frac{d^2y}{dx^2}=-2sinxcosx-2sinxcosx$
$\frac{d^2y}{dx^2}=-4\sin x\cos x$