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f
f(x, y) = (x2y2ln(x2+y2) (x, y)6= (0,0)
0
f(x, y) = ((x2+y2) sin 1
√x2+y2(x, y)6= (0,0)
0
f:R2→R
f(x, y) = (sin(xy)
|x|+|y|(x, y)6= (0,0)
0
f
fC1
f(x, y) = xy x2−y2
x2+y2
x, y
fR2
C1C2
f:R2→R
f(x, y) = (xy3
x2+y2(x, y)6= (0,0)
0
fC1R2
∂2f
∂x∂y (0,0) ∂2f ∂y∂x (0,0)
f:R2− {(0,0)} → R
f(x, y)=(x2−y2) ln(x2+y2)
f(0,0)
fC1R2− {(0,0)}
∂f
∂x (x, y) = −∂f ∂y (y, x)
fC1R2
f:R→RC1F:R2\ {(0,0)} → R
F(x, y) = f(x2+y2)−f(0)
x2+y2
lim(x,y)→(0,0) F(x, y)F(0,0)
fC2
F(0,0)
FC1
ϕ(x, y) = cos x−cos y
x−yx6=y
ϕR2ϕ
ϕC1C∞
g:R→RC2
f(x, y) = g(x)−g(y)
x−yx6=y f(x, x) = g0(x)
f(x, y) [0 ; 1]
fC1

ϕ:R→Rf:R2→R
f(x, y) = Zy
x
ϕ(t) dt
fC1
f:R2→RC1
F(x) = Zx3
2x
f(x+ 1, t) dt
FR
n∈N∗
un: (x, y)7→ cos(ny)
√nxn
D(x, y)∈R2un(x, y)
f: (x, y)7→ ∞
X
n=1
un(x, y)
D
fD◦C1

fR2\ {(0,0)}
(0,0)
f(x, y)=(xy)xy
x2+y2(x2+y2) ln(x2+y2)−−−−−−−→
(x,y)→(0,0) 0
f(0,0)
x
∂f
∂x R2\ {(0,0)}
∂f
∂x (x, y)=2xy2ln(x2+y2) + 2x3y2 x2+y2
1
t(f(t, 0) −f(0,0)) = 0 −−−→
t→00
∂f
∂x (0,0) ∂f ∂x (0,0) = 0
∂f
∂x (x, y)−−−−−−−→
(x,y)→(0,0) 0
2xy2ln(x2+y2) = 2yxy
x2+y2(x2+y2) ln(x2+y2)
∂f
∂x R2
y
f(x, y) = f(y, x)∂f
∂y
g:R+→R
g(t) = (tsin 1/√t t 6= 0
0
gR+f(x, y) = g(x2+y2)f
R2
gC1R∗
+f
R2\ {(0,0)}
∂f
∂x (x, y)=2xsin 1
px2+y2−x px2+y2cos 1 px2+y2
∂f
∂y (x, y)=2ysin 1
px2+y2−y px2+y2cos 1 px2+y2
(0,0)
1
t(f(t, 0) −f(0,0)) = tsin 1 |t|= O(t)−−−→
t→00
∂f
∂x (0,0) ∂f ∂y (0,0)
∂f
∂x 1 n,0=2 nsin n−cos n
n→+∞∂f
∂x (0,0)
∂f
∂y
fR2\ {(0,0)}
f(x, y)∼
(x,y)→0
r2cos θsin θ
r|cos θ|+|sin θ|→0 = f(0,0)
cos θ×sin θ
|cos θ|+|sin θ|
2π
fR2
∂f
∂x (0,0) = lim
t→0
1
t(f(t, 0) −f(0,0)) = 0
x, y > 0
∂f
∂x (x, y) = ycos(xy)(x+y)−sin(xy) (x+y)2
∂f
∂x (t, t) = 2t2cos(t2)−sin(t2) (2t)2−−−−→
t→0+
1
2
fC1

x=rcos θ y =rsin θ r =px2+y2−−−−−−−→
(x,y)→(0,0) 0
f(x, y) = r2cos θsin θcos(2θ)−−−−−−−→
(x,y)→(0,0) 0
f(0,0) f(0,0) = 0
fC2
R2\ {(0,0)}∂f
∂x (x, y) = yx4+ 4x2y2−y4 (x2+y2)2
∂f
∂x (0,0) = lim
t→0
1
t(f(t, 0) −f(0,0)) = 0
∂f
∂x (x, y)−−−−−−−→
(x,y)→(0,0) 0
∂f
∂x R2
∂f
∂y
f(x, y) = −f(y, x)
fC1R2
∂2f
∂y∂x(0,0) = lim
h→0
1
h∂f ∂x (0, h)−∂f ∂x (0,0)=−1
∂2f
∂x∂y (0,0) = 1
fC2
fC1R2\ {(0,0)}
∂f
∂x (x, y) = y3 x2+y2−2x2y3 (x2+y2)2
∂f
∂y (x, y) = 3xy2 x2+y2−2xy4 (x2+y2)2
1
t(f(t, 0) −f(0,0)) = 0
1
t(f(0, t)−f(0,0)) = 0,
∂f
∂x (0,0) ∂f ∂y (0,0)
∂f
∂x (0,0) = ∂f ∂y (0,0) = 0.
∂f
∂x (x, y)≤ |y|y2 x2+y2+ 2 |y|xy x2+y22
−−−−−−−→
(x,y)→(0,0) 0
∂f
∂y (x, y)≤3|y||xy| x2+y2+ 2 |y||xy| x2+y2
y2
x2+y2−−−−−−−→
(x,y)→(0,0) 0.
fC1R2
1
t∂f ∂x (0, t)−∂f ∂x (0,0)= 1 →1
1
t∂f ∂y (t, 0) −∂f ∂y (0,0)= 0 →0.
∂2f
∂x∂y (0,0) ∂2f ∂y∂x (0,0)
∂2f
∂x∂y (0,0) = 0 ∂2f ∂y∂x(0,0) = 1.
fC2

(x, y)→(0,0) x=rcos θ y =rsin θ
r=px2+y2→0
f(x, y)=2r2(cos2θ−sin2θ) ln r→0
r2ln r→0
f(0,0) f(0,0) = 0
fC1R2− {(0,0)}f(x, y) = −f(y, x)
x
∂f
∂x (x, y) = −∂f ∂y (y, x)
∂f
∂x (0,0) = lim
t→0
1
t(f(t, 0) −f(0,0)) = 0
∂f
∂y (0,0) = 0
(x, y)6= (0,0)
∂f
∂x (x, y)=2xln(x2+y2) + 2x(x2−y2) x2+y2
(x, y)→(0,0) x=rcos θ y =rsin θ
r=px2+y2→0
∂f
∂x (x, y)=4rln r+ 2r(cos2θ−sin2θ)→0 = ∂f ∂x (0,0)
∂f
∂x (0,0)
∂f
∂y
cx,y ∈]0 ; x2+y2[
F(x, y) = f0(c)
(x, y)→(0,0) cx,y →0F(x, y)→f0(0)
F(x, y) = F(0,0)+x2+y2
2f00(0)+(x2+y2)ε(x2+y2) = F(0,0)+ϕ(x, y)+o(x, y)
ϕ= 0
F(0,0) dF(0,0) = 0
FC1R2\ {(0,0)}
∂F
∂x (x, y) = 2x x2+y2(f0(x2+y2)−F(x, y)) = x(f00(0) + o(1)) −−−−−−−→
(x,y)→(0,0) 0
∂F
∂y (x, y)−−−−−−−→
(x,y)→(0,0) 0
FC1
ϕ(a, a) = −sin a ϕ(x, y)→ϕ(a, a)
(x, y)→(a, a)x6=y x =y
cos p−cos q=−2 sin p−q
2sin p+q 2
ϕ(x, y) = −sinc x−y
2sin x+y 2
sinc C∞
gC1
g(x) = g(y) + Zx
y
g0(t) dt
t=y+u(x−y)
g(x) = g(y)+(x−y)Z1
0
g0(y+u(x−y)) du
f(x, y) = Z1
0
g0(y+u(x−y)) du
x6=y x =y