Question

Найти d^2y/dx^2
y = ln(tgx)

Answer 1

Ответ:

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