Найдите производную в точке X0: а) y=3x^2 - 7x. x0= -2 б) y= sin^2 x + cos^2x x0=0

f(х)\' = (4х^2 - 9х + 1)’ = (4х^2)’ – (9х)’ + (1)’ = 4 * 2 * х^(2 – 1) – 9 * х^(1 - 1) – 0 = 8х – 9.

f(х)\' = (х^2 - 3х)’ = (х^2)’ - (3х)’ = 2 * х^(2 – 1) - 3* х^(1 – 1) = 2 * х^1 – 3 * х^0 = 2 * х – 3 * 1 = 2х - 3.

f(х)\' = (2х^4 – х^2 + 1)’ = (2х^4)’ – (х^2)’ + (1)’ = 2 * 4 * х^(4 – 1) – 2 * х^(2 - 1) – 0 =

8 * х^3 – 2 * х^1 = 8х^3 – 2х.

f(х)\' = ((1 / 2) * х^2 – (1 / 5) * х^5)’ = ((1 / 2) * х^2)’ – ((1 / 5) * х^5)’ = (1 / 2) * 2 * х^(2 - 1) – (1 / 5) * 5 * х^(5 - 1) = 1 * х^1 – 1 * х^4 = х – х^4.

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

f(х)\' = (3 – 2х + 4х^2 – 7х^3)’ = (3)’ – (2х)’ + (4х^2)’ – (7х^3)’ = 0 – 2 * х^(1 – 1) + 4 * 2 * х^(2 - 1) – 7 * 3 * х^(3 – 1) = -2 * х^0 + 8 * х^1 – 21 * х^2 = -2 + 8х – 21х^2.

f(х)\' = (х^5 + х^3 – х^2 – 1)’ = (х^5)’ + (х^3)’ – (х^2)’ – (1)’ = 5 * х^(5 – 1) + 3 * х^(3 – 1) – 2 * х^(2 – 1) – 0 = 5х^4 + 3х^2 – 2х.