Помогите наити производный функции в точке х0

Решите задачу:

$f(x)=tgx-\sqrt3ctgx\\\\f'(x)=\frac{1}{cos^2x}+\frac{\sqrt3}{sin^2x}\\\\f'(\frac{\pi}{6})=\frac{1}{\frac{3}{4}}+\frac{\sqrt3}{\frac{1}{4}}=\frac{4}{3}+4\sqrt3\\\\\\f(x)=ctgx-\frac{\sqrt3}{3}tgx\\\\f'(x)=-\frac{1}{sin^2x}-\frac{\sqrt3}{3}\cdot \frac{1}{cos^2x}\\\\f'(\frac{\pi}{3})=-\frac{1}{\frac{3}{4}}-\frac{\sqrt3}{3}\cdot \frac{1}{\frac{1}{4}}=-\frac{4}{3}-\frac{4\sqrt3}{3}=-\frac{4}{3}(1+\sqrt3)$