Найти значение производной в точке x0 f(x)=5x^3-6x^4+3x^2+1 x0=1 f(x)=(x^2+1)(x^3-2) x0=2 f(x)=2x*sin5 . x0=Π/2

f(х)\' = (2х^3 + 7х^2 +х – 1)’ = (2х^3)’ + (7х^2)’ + (х)’ – (1)’ = 2 * 3 * х^2 + 7 * 2 * х + 1 – 0 = 6х^2 + 14х + 1.

f(х)\' = (2 * х^(-3) – х)’ = (2 * х^(-3))’ – (х)’ = 2 * (-3) * х^(-4) – 1 = -6 * х^(-4) – 1 = (-6 / х^4) – 1.

f(х)\' = (2 * х^(-3))’ = 2 * (-3) * х^(-4) = -6 * х^(-4) = (-6 / х^4).

f(х)\' = ((х^3 – 4х)^4)’ = (х^3 – 4х)’ * ((х^3 – 4х)^4)’ = ((х^3)’ – (4х)’) * ((х^3 – 4х)^4)’ =

(3 * х^2 – 4) * 4 * (х^3 – 4х)^3 = 4 * (3х^2 – 4) * (х^3 – 4х)^3.

f(x)\' = (x^3 * sin (2x))’ = (x^3)’ * sin (2x) + x^3 * (sin (2x))’ = (x^3)’ * sin (2x) + x^3 * (2x)’ * (sin (2x))’ = 3x^2 * sin (2x) + x^3 * 2 * cos (2x) = 3x^2sin (2x) + 2x^3cos (2x).