## Calculus, 10th Edition (Anton)

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