Calcul différentiel

fC1
(∂f
∂x (x, y) = xy2
∂f
∂y (x, y) = x2y
∂f
∂x (x, y) = x
√x2+y2
∂f
∂y (x, y) = y
√x2+y2(∂f
∂x (x, y) = x x2+y2
∂f
∂y (x, y) = −y x2+y2
Rnf:Rn→Rn
C1f(0) = 0
df(x)x∈Rn
f
ϕ∈ C1(Rn,Rn) dϕ(0)
V ϕ V
URnf, g, h:U→R
∀x∈U, f(x)≤g(x)≤h(x)
f h a ∈U f(a) = h(a)g
a
(r, θ)7→ (rcos θ, r sin θ)
(r, θ, ϕ)7→ (rsin θcos ϕ, r sin θsin ϕ, r cos θ)
(u, v)7→ (u+v, uv)C1
U=(u, v)∈R2u<v
V
ϕ: (x, y)7→ (x+1
2cos y, y +1 2cos x)
C1R2
k∈]0 ; 1[ f∈ C1(R,R)
∀x∈R,f0(x)≤k
ϕ:R2→R2
ϕ(x, y)=(y+f(x), x +f(y))
ϕC1R2
Rnf:R+→R
C1f(0) = 1 f0(0) = 0
F(x) = fkxkx
N:x7→ kxk C1Rn\ {0}
FC1Rn
∀(x, h)∈Rn×Rn,(dF(x)(h)|h)≥f(kxk)khk2
FRnRn

S(a, 0)
x2
a2+y2 b2= 1
M, N (SMN)
f(x, y) = yexp(x) + xexp(y)
Rn
C2
RnnN∗
fR
R3
(x, y, z)7→ (2x+y−z)(x+y+ 2z)
fR3
E f g
E fg
E f g
f(x, y) = y(x2+ (ln y)2)R×]0 ; +∞[
f(x, y) = y(x2+ (ln y)2)
f(x, y) = y(x2+ (ln y)2)
ω= (x+y) dx+ (x−y) dy
ω
x+y+ (x−y)y0= 0
y x
ω(x, y) = (xy −y2+ 1) dx+ (x2−xy −1) dy
f:R→R
ω(x, y)f(xy)
ω(x, y) = x2dy+y2dx
ω
ω
ω

ω=xdy+ydx ω =xdy−ydx
(x−y)2ω=xdx+ydy x2+y2−ydy
ω(x, y) = xdy−ydx
x2+y2
R2\(0,0)ω
ω O
ω
n∈N∗
ω(x, y) = e−y
x2+y2(xsin x−ycos x) dx+ (xcos x+ysin x) dy
ω
n→+∞
Z+∞
0
sin x
xdx
O, A, B 0, r, r exp(iπ/4) r > 0
ΓrC
[O;A]O A
CrO r A B
[B;O]B O
Ir=IΓr e−(x+iy)2(dx+ i dy)
r→+∞
Jr=ZCr e−(x+iy)2(dx+ i dy)
I=IΓ xdy+ydx
Γy=x2O A(2,4)
I=IΓ x2dy+y2dx
Γ Ω(a, b)R > 0
I=IΓ x2dy+y2dx
Γ (OIJ)I(1,0) J(0,1)

f(x, y) = 1
2x2y2+Cte R2
f(x, y) = px2+y2+Cte R2\(0,0)
∂f
∂x (x, y) = x x2+y2
f(x, y) = 1
2ln(x2+y2) + C(y)
y
x2+y2+C0(y) = −y x2+y2C
y
a, b ∈Rnϕ: [0 ; 1] →Rn
ϕ(t) = fa+t(b−a)
ϕC1
ϕ0(t)=dfa+t(b−a).(b−a)
f(b)−f(a) = ϕ(1) −ϕ(0)
f(b)−f(a) = Z1
0
dfa+t(b−a).(b−a) dt
f(b)−f(a) ≤Z1
0
dfa+t(b−a).(b−a) dt≤ kb−ak
f
fC1
±1
fC1RnRn
∀y∈Rn,d(f−1)(y) = df(x)−1x=f−1(y)
f
f−1
f−1
∀c, d ∈Rn,
f−1(d)−f−1(c) ≤ kd−ck
∀a, b ∈Rn,
f(b)−f(a) =kb−ak
f(0) = 0
∀a∈Rn,
f(a) =kak
f(b)−f(a)
2=
f(b)
2−2(f(a)|f(b)) + 
f(a)
2
∀a, b ∈Rn,(f(a)|f(b)) = (a|b)
a, b, h ∈Rnt6= 0
1
t(f(a+t.h)−f(a)|f(b))= (h|b)
t→0
(df(a).h|f(b)) = (h|b)
f
∀a, a0, c, h ∈Rn,(df(a).h|c) = (df(a0).h|c)
∀a, a0∈Rn,df(a)=df(a0)
f `
∀a∈Rn, f(a) = f(0) + Z1
0
df(0 + t.a).a dt=Z1
0
`(a) dt=`(a)
f
fC1URn
V
V
B(0, R)V
U f C1

U B(0, R)
∀a, b ∈U, 
f(b)−f(a) ≤ kb−ak
∀c, d ∈B(0, R),
f−1(d)−f−1(c) ≤ kd−ck
∀a, b ∈U, 
f(b)−f(a) =kb−ak
∀a, b ∈U, (f(a)|f(b)) = (a|b)
f(b)Rn
f U
x0∈Rng:x7→ f(x0+x)−f(x0)
f
dϕ(0) = IdRn
ψ:x7→ ϕ(x)−x
ψC1dψ(0) = ˜ 0B
∀x∈B, 
dψ(x) ≤1
2
∀x, y ∈B, 
ψ(y)−ψ(x) ≤1
2ky−xk
x, y ∈B ϕ(x) = ϕ(y)ψ(y)−ψ(x) = y−x
ky−xk ≤ 1
2ky−xk
y=x
θ= (dϕ)−1(0) ◦ϕC1
dθ(0) = (dϕ−1)(0) ◦dϕ(0) = IdRn
V θ
V
ϕ V
h−f a
h−f a
h−f a f
h a `
u→0
f(a+u) = f(a) + `(u)+o1(u)h(a+u) = h(a) + `(u)+o2(u)
g(a) + `(u)+o1(u)≤g(a+u)≤g(a) + `(u)+o2(u)
g(a+u) = g(a) + `(u) + o(u)
g a a `
C1
r
r2sin θ
ϕ: (u, v)7→ (u+v, uv)C1UR2
(s, p)∈R2
(s, p) = ϕ(u, v)u v x2−sx +p= 0
∆ = s2−4p > 0
ϕ
V=(s, p)∈R2s2−4p > 0
(s, p)∈V(u, v)u<v
ϕ(u, v)=(s, p)x2−sx +p= 0
u=s−ps2−4p
2v=s+ps2−4p 2
ϕ U V
U V
ϕ ϕ−1
C1