IV. Dérivées d`ordre supérieur

I a ∈I f :I−→ R
f a f a
f I f I
f(x) = f(a)+(x−a)g(x)g(x) = (f(x)−f(a)
x−ax6=a
f0(a)x=aφ
f a g f a
f I f a ∈I f
a∈I f I
R0
R+0
f I f0I
(f0)0f f00 f(2)
k k ∈N∗kef f(k)
f f(x) = x2−3x+ 4 fR
f0(x) = 2x−3f0RfR
f00(x)=2 f00 RfR
f(3)(x) = 0 ∀k≥3, f(k)(x)=0 f
k∈N∗f∈Ck(I)f CkI f k I ∀j∈[[1; k]] f(j)
I
Ckk ke
f f
f f0f0
f f00 f00
f k f(k−1) f(k−1)
f(k)
f(x) = x2−3x+ 4 C1R
C2C3Ckk

f∈C∞(I)f C∞I f I
I
f C∞I
I
f Ckk∈N∗f Cj1≤j≤k−1
f C∞f Ckk∈N∗
C∞Df0
C∞R
C∞R
C∞]0; +∞[
C∞]− ∞; 0[ ]0; +∞[
R0
]− ∞; 0[ ]0; +∞[C∞
[0; +∞[ 0
]0; +∞[C∞]0; +∞[
CkC∞
Ck
C∞CkC∞
f f(x) = ln(x+ 1) x≥0
x x < 0
f C∞]− ∞; 0[
]− ∞; 0[ f(x) = x f C∞
f C∞]0; +∞[
]0; +∞[f(x) = ln(x+ 1)
x7−→ x+ 1 C∞]0; +∞[ ]1; +∞[X7−→ ln(X)C∞]1; +∞[
f0
lim
x→0−
f(x) = lim
x→0−
x= 0 lim
x→0+f(x) = lim x→0+ln(x+ 1) = ln(0 + 1) = 0
0f f 0fR
f0
f0(x) = 1
x+1 x > 0
1x < 0
lim
x→0+
1
x+ 1 = 1 = lim
x→0−1f0f C1R
f0
f00(x) = −1
(x+1)2x > 0
0x < 0
lim
x→0+
−1
(x+ 1)2=−16= 0 = lim
x→0−0f0f C2R