This is nothing crazy complicated if you get the pattern; you can do the following (I presented this in regular prime notation and function composition notation):
color(green)("Overall Function")
color(purple)("First Inner Function")
color(darkblue)("Innermost Function")
color(highlight)("Derivative")
sin^2(2x) = (sin(2x))^2
Let:
color(green)(f{g[h(x)]}) = f[(g @ h)(x)] color(green)(= {color(purple)(g[h(x)])}^2)
color(purple)(g[h(x)]) = (g @ h)(x) color(purple)(= sin[color(darkblue)(h(x))])
color(darkblue)(h(x) = 2x)
d/(dx)(color(green)(f{g[h(x)]})) = color(highlight)(f'){color(purple)(g[h(x)])}*color(highlight)(g')[color(darkblue)(h(x))]*color(highlight)(h'color(black)((x)))
= color(highlight)(2)[color(purple)(sin(2x))]*color(highlight)(cos)(color(darkblue)(2x))*color(highlight)(2)
= 2(2sin(2x)cos(2x))
= color(blue)(2sin(4x))
...since sin(2x) = 2sinxcosx and thus sin(4x) = 2sin(2x)cos(2x).
