$\displaystyle \frac {d^2y} {dx^2} for y = cos^2 4x$
Use the chain rule.
$\displaystyle y = cos^2 4x$
$\displaystyle \frac{dy}{dx} = 2cos(4x) . (-sin(4x)) . 4$
$\displaystyle = -8cos(4x)(sin(4x))$
You must derive twice. But use the product rule now.
$\displaystyle \frac{dy}{dx} (-8cos(4x)(sin(4x))) = -8cos(4x)' . (sin(4x)) + -8cos(4x) . (sin(4x))' $
$\displaystyle = (-32(-sin(4x))) . (sin(4x)) + (-8(cos(4x)) . (4(cos(4x))) $
I hope that's right...
I don't have much experience with this. Jhevon's still helping me get the hang of these things...