Question

Вычислите f'(x0), если:
а) f(x)=3cos2x, x0=-2π/3
б) f(x)=4tg3x, x0=-π/3

Answer 1

$a) f(x)=3cos2x\\f'(x)=-3sin2x*2=-6sin2x\\f'(-\frac{\pi }{3})=-6sin(-\frac{2\pi }{3})=-6*(-\frac{\sqrt{3} }{2})=3\sqrt{3}\\b) f(x)=4tg3x\\f'(x)=4*\frac{1}{cos^23x}*3=\frac{12}{cos^23x}\\ f'(-\frac{\pi }{3})=\frac{12}{cos^2(-\pi)}=12$