$y=\dfrac{\sin x}{x}$
$y'=\dfrac{(\sin x)'x-\sin x.(x)'}{x^2}=\dfrac{x\cos x-\sin x}{x^2}$
$y''=\dfrac{(x\cos x-\sin x)'x^2-(x\cos x-\sin x).(x^2)'}{x^4}$
$=\dfrac{[x^2(x\cos x)'-x^2(\sin x)']-2x(x\cos x-\sin x)}{x^4}$
$=\dfrac{x^2.(\cos x+x.\sin x)-x^2.\cos x-2x^2\cos x+2x\sin x}{x^4}$
$=\dfrac{x^3\sin x-2x^2\cos x+2x\sin x}{x^4}$
$=\dfrac{x^2\sin x-2x\cos x+2\sin x}{x^3}$
