Functie Primitieve F(x) = Afgeleide f ' (x) = ! ! = ! ! ! ! ! = 3! f (x) = (x ! 4) 3 f (x) = 3(2 ! x)4 f ' (x) = !12(2 ! x)3 1 1 3 F(x) = 3• (2 ! x)5 • + c = ! (2 ! x)5 + c 5 !1 5 1 1 1 5 F(x) = (2x ! 3)5 • + c = ( 2x ! 3) + c 5 2 10 ! ! = (2! − 3)! ! ! = (3 − 2!)! f (x) = 4(3! 2x) •!2 = !8(3! 2x) ' 3 3 ! ! = 5(2! − 3)! f (x) = e 1 1 1 5 F(x) = (3! 2x)5 • + c = ! (3! 2x ) + c 5 !2 10 1 1 1 5 F(x) = 5• (2x ! 3)5 • + c = ( 2x ! 3) + c 5 2 2 x f (x) = e 2x f (x) = 4e 3x f (x) = e 2 x+3 f (x) = 4e 2 x+3 f (x) = 2e 3!4 x f (x) = ln x f (x) = ln 2x Niet nodig 1 2 1 f (x) = •2 = = 2x 2x x 1 2 4 f ' (x) = 4 • • 2 = 4 • = 2x 2x x 1 2 8 f ' (x) = 4 • •2 = 4• = 2x !1 2x !1 2x !1 1 1 1 f ' (x) = •!1 = ! = e! x e! x x!e Niet nodig ' f (x) = 4 ln 2x f (x) = 4 ln(2x !1) f (x) = ln(e ! x) f (x) = ln x ! e Niet nodig Niet Nodig Niet nodig Niet nodig f (x) = ln x ! e x Niet nodig 1 f (x) = x F(x) = ln | x | +c 1 f (x) = x+2 3 f (x) = x+2 F(x) = ln | x + 2 | +c F(x) = 3ln | x + 2 | +c f (x) = 3 2 ! 4x f !(x) = "1• f (x) = !3 4x ! 2 f !(x) = "1• 3 (2 " 4x ) 2 "3 ( 4x " 2) 2 •"4 = •4 = 12 (2 " 4x)2 F(x) = 3ln | 2 ! 4x |• 1 3 + c = ! ln | 2 ! 4x | +c !4 4 1 3 F(x) = !3ln | 4x ! 2 |• + c = ! ln | 4x ! 2 | +c 4 4 12 (4x " 2)2 f (x) = sin x f (x) = sin 2x f (x) = sin(2x + 3) f (x) = sin 3x + 2 1 f (x) = sin(2x + ! ) 3 1 f (x) = sin( ! ! 2x) 3 f (x) = cos x f (x) = cos2x f (x) = cos(2x + 3) !" ! = −2cos ( − 2!) !" 3 sin ! ! ! ! − 2! !" = ! cos(! ! − 2!) + ! 1 cos 2! !" = sin 2! + ! 2 f (x) = cos3x + 2 ! (cos 3! + 2) !" = ! sin 3x +2x+c ! 1 f (x) = cos(2x + ! ) 3 1 f (x) = cos( ! ! 2x) 3 !" !" f (x) = 3 x f (x) = 3x + 4 f (x) = 452 x = 2025 x !" !" f (x) = 2 log x f (x) = 3 log x ! ! = −2 ∗ − sin ! ! − 2! = 2sin (! ! − 2!) !" = ln 3 ∗ 3! !" !" = ln 3 ∗ 3! !" = ln 2025 ∗ 2025! !" 1 1 1 = ∗ = !" ln 2 ! ! ∗ ln 2 1 1 1 cos ! − 2! !" = − sin − 2! + ! 3 2 3! 3! !" = f (x) = log(x + 3) f (x) = log(3x + 2) f (x) = log(x 2 + 3) 2x + 5 x!4 4x ! 2 f (x) = 2x ! 2 !! (3! + 4 )!" = !" ! + 4! + ! 2025! 2025! !" = + ! ln 2025 f (x) = 3 log(x + 2) f (x) = 3! + ! ln 3 !5x +1 f (x) = x+2 f (x) = (2x +1)e x f (x) = (2x +1)e3x f (x) = e x (1! x 2 ) f (x) = e x ln x f (x) = (x ! e)ln(e ! x) x2 ex ln x f (x) = x e f (x) = (x ! 3)(x + 4) 2 f (x) =