Найти [latex] d^{2} y/d x^{2} [/latex] y=ln(tgx)

[latex]y=ln(tgx)\\\\\frac{dy}{dx}=\frac{1}{tgx}\cdot \frac{1}{cos^2x}=\frac{cosx}{sinx}\cdot \frac{1}{cos^2x}=\frac{1}{sinx\cdot cosx}=\frac{2}{sin2x}\\\\\frac{d^2y}{dx^2}=-\frac{2\cdot (sin2x)'}{sin^22x}=-\frac{2\cdot 2cos2x}{sin^22x}=-\frac{4}{tg2x\cdot sin2x}=-\frac{4ctg2x}{sin2x}[/latex]