\(y'=3cosx-4sinx-\frac{1}{cos^2x}\)
\(\Rightarrow y'\left(\frac{\pi}{6}\right)=3cos\left(\frac{\pi}{6}\right)-4sin\left(\frac{\pi}{6}\right)-\frac{1}{cos^2\left(\frac{\pi}{6}\right)}=\frac{-20+9\sqrt{3}}{6}\)
b/ \(y'=-8x^3+\frac{3}{x^4}-\frac{1}{x^2}\)