Найти y'' и d^2yY=lnctg4x

Решите задачу:

$y=ln(ctg4x) \\ y'=(ln(ctg4x))'= \frac{1}{ctg4x}*(- \frac{1}{sin^24x} )*4= \frac{-4sin4x}{cos4x}* \frac{1}{sin^24x}= \\ = \frac{-8}{2cos4xsin4x}= \frac{-8}{sin8x}=-8sin^{-1}8x \\ y''=(y')'=(-8sin^{-1}8x)'=-8*(-1)sin^{-2}8x*8= \frac{64}{sin^28x} \\ \\ dy=y'dx=-8sin^{-1}8xdx \\ d^2y=d(dy)=d(-8sin^{-1}8xdx)=(-8sin^{-1}8xdx)'dx^2=\frac{64}{sin^28x}dx^2 \\ d^2y=\frac{64}{sin^28x}dx^2 \\ .$