Question

Найти $d^{2} y/d x^{2}$
y=ln(tgx)

Answer 1

$y=ln(tgx)\\frac{dy}{dx}=frac{1}{tgx}cdot frac{1}{cos^2x}=frac{cosx}{sinx}cdot frac{1}{cos^2x}=frac{1}{sinxcdot cosx}=frac{2}{sin2x}\\frac{d^2y}{dx^2}=-frac{2cdot (sin2x)'}{sin^22x}=-frac{2cdot 2cos2x}{sin^22x}=-frac{4}{tg2xcdot sin2x}=-frac{4ctg2x}{sin2x}$