Question

Пусть f(x)=2+3x/2-3x. Найдите f'(x), f'(0), f'(7), f'(3).

Answer 1

y'=(3(2-3x)-(2+3x)(-3))/(4-6x+9x^2)=12/(4-6x+9x^2)
y'(0)=12/4=3
y'(7)= 12/(4-42+343)=12/305
y'(3)=12/(4-18+81)=12/67