Sujet

f I a I f I f a
l a I
lim
x→a
f(x)−f(a)
x−a=l=f′(a)
lim
h→0
f(a+h)−f(a)
h=l=f′(a)
f f′
f(x) = 1
xf′(x) = −1 x2
fn(x) = xnf′
n(x) = n xn−1n
f(x)x0
x= 0 x=π
2
f′I f′′ f′
f′′ = (f′)′
f′′ f
f(x) = 1
xf′′ (x) = 2 x3
f(x) = xnf′′ (x) = n(n−1) xn−2n≥1

f
f(2) =f′′ f(3)
f(3) = (f(2))′
f(4) = (f(3))′
f(n)
f(x)=3x3+ 5x2+ 4
f(x) = 1
xf(n)(x) = (−1)nn!1 xn+1 n! = n×(n−1) ×(n−2) ×... ×2×1
n≤1
Pn(x)
f(x) = Pn(x) + Rn(x) = f(a) + f′(a)
1! (x−a) + f(2)(a) 2! (x−a)2+· · · +f(n)(a) n!(x−a)n+Rn(x)
n! = n×(n−1) ×(n−2) ×... ×2×1Rn(x)x
a
x≈a f(x)≈Pn(x) =
n
∑
k=0
f(k)(a)
k!(x−a)k0! = 1! = 1 f(0) =f f(1) =f′
Pn(x)k= 0 k= 1
f(x) = sin(x)a= 0
P1(x)P3(x)P5(x)
P1(x) = x
P3(x) = x−x3
6
P5(x) = x−x3
6+x5 120