Question

Найти y'' и d^2y
Y=lnctg4x

Answer 1

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