Question

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

Answer 1

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