Derivada
Função
Derivada f ´(u ) = 0
Integral Primitiva
f (u ) = k ,
k ∈ℜ
∫ du = ∫ k du = ∫ u du = n u+c k .u + c u n +1 + c n +1 ln u + c − cos u + c
f (u ) = u n , n ∈ℜ f (u) = k . f (u) f (u ) = u f ( x) = u.v f ( x) = u v
f ´(u ) = n.u n −1 f ´(u ) = k . f ' (u) f '(u ) = 1 f '( x ) = u .v '+ u '.v
∫
du = u
f ' ( x) = f '( x ) =
u '.v − u .v ' v² 1 u ln a u ∫ sen u du = ∫ cos u du = ∫ sec
2
f (u ) = log a u f (u) = a ,0 < a ≠ 1 u sen u + c
f '(u ) = a .ln a f '( u ) = 1 u
u du
2
tg u + c − cotg u + c sec u+c eu + c au + c (a > 0, a ≠ 1) ln a
f (u ) = ln u f (u ) = sen u f (u ) = cos u f (u ) = tg u f (u ) = sec u
∫ cosec u du = ∫ sec u.tg u du = ∫ e du = u f '(u ) = cos u f '(u ) = − sen u f '(u ) = sec 2 u f '( u ) = sec u .tg u
∫a
u
du =
f (u ) = cossec u f (u ) = cotg u f (u ) = e u f (u ) = arc cosu
f '(u ) = -cotg u.cossec u f '( u ) = − cossec 2 u f '(u ) = eu
∫ ∫
1 1− u 1 1+ x
1
2 2
=
arc sen u + c
=
arc cos u + c
f '(u ) = −
1 u2 −1 1 u2 −1
∫ 1+ x
2
=
arc tg u + c
f (u ) = arc sen u
f '(u ) =
f (u ) = arc tg u
f '(u ) =
1 u +1
2