Последние пять заданий

1)
$f'(x)=(tg(x))'+(\pi)'=\frac{1}{cos^2x}\\f'(x_0)=f'(\frac{3\pi}{4})=\frac{1}{(cos\frac{3\pi}{4})^2}=\frac{1}{(-\frac{\sqrt{2}}{2})^2}=2$

2)
$f'(x)=(ctg(x)+3x+8)'=-\frac{1}{sin^2x}+3\\f'(x_0)=f'(-\frac{\pi}{6})=3-\frac{1}{(sin(-\frac{\pi}{6}))^2}=3-\frac{1}{(-\frac{1}{2})^2}=3-4=-1$

3)
$f'(x)=(sin3x)'=cos3x*(3x)'=3cos3x\\f'(x_0)=f'(\frac{\pi}{6})=3cos(\frac{3\pi}{6})=3cos\frac{\pi}{2}=3*0=0$

4)
$f'(x)=(2cos\frac{x}{2}-1)'=2(-sin\frac{x}{2})*(\frac{x}{2})'=-sin\frac{x}{2}\\f'(x_0)=f'(\pi)=-sin\frac{\pi}{2}=-1$

5)
$f'(x)=(\frac{tg2x}{2}+5)'=\frac{\frac{1}{cos^2(2x)}*(2x)'}{2}=\frac{1}{cos^22x}\\f'(x_0)=f'(\frac{\pi}{3})=\frac{1}{(cos\frac{2\pi}{3}^2)}=\frac{1}{(-\frac{1}{2})^2}=4$