2
Exercice 2. Dresser le graphe dévaluation et calculer la dérivée des fonctions
usuelles symboliques suivantes. Donner l’ensemble de définition des fonctions
usuelles strictes associées ainsi que de leurs dérivées.
p
a(X) =
sin(X)
cos(X)
sin(X)
+1
b(X) = exp − X 2
c(X) = ln(ln(ln(X)))
Exercices d’analyse, feuille 2
Licence de Mathématiques / Informatique / Économie
1e` re année, semestre 1
1
cos(3X 2 )+1
d(X) =
sin(X+2)
e(X) = q
sin(cos(ln(2X 2 ) + 3))
X−1
g(X) =
πX + X+2
2
2 − 1)))
h(X) = X sin(cos(ln(3X
i(X) = exp sin(X + 1) − cos(X 2 )) ln(2 + sin(X)
Ensembles de définition
Exercice 1. Pour chacune des fonctions usuelles symboliques suivantes, représenter le graphe d’évaluation puis déterminer l’ensemble de définition de la fonction
usuelle stricte associée.
√
1. 3 X
j(X)
4. ln P + ln(2 − Q)
5.
6.
3
sin T cos T
7. ln(1 − ln X)
8.
3
sin X cos Y
q
1
4
+ cos X sin X
X
10. exp − sin
2
X
9.
√3X+cos Y
X 2 +Y 2 −1
13. (ln (1 + X) − 2)X
exp(sin X)
sin(exp X)
k(X)
=
o(X)
p(X)
q(X)
r(X)
s(X)
t(X)
=
=
=
=
=
=
3
eX +2X+2 ln(cos(X) sin(X))
√
2X 3eX + 2
pe −
√
X − X + 6
cosh(X)
exp
X
ln(tanh(X))
2X +3X
pln(X)
u(X) =
tanh(X) + π
v(X) = cos(sin(cos(X)))
3 +π
w(X) = X
, où α est un nombre réel.
X 2 +α 2
y(X) = X exp − X2
X+1
z(X) = sin X+2014
11. ln (ln (ln S))
12.
=
cos(3X) cos(2X) + e3X+1 cos(X) + eX−1 + X 3
P
Xk
l(X) = e−X 2014
Q2014 k=0 k!
m(X) =
k=0
Q (X − k) n(X) = ln 2014
k=0 (X − k)
2. sin2 (X 3 + 1)
√
3. 1 − sin T
ln(X−1)
ln(1−X 2 )
Dérivation
Exercice 3. Calculer les dérivées partielles des fonctions usuelles symboliques suivantes.
√
14. (exp (2T ) − 3 exp T + 2) T
r
q
q
1+X 2
1+X
15. sin 1−X 2 + 2 + X 1 + 1−X
1. a (X, Y ) = 2XY + X 3
2. b (S, T ) = exp (2T ) + exp 2S 2 + T
1
3. f (X, Y ) = X 2Y
Dl =
l 0 (X) =
=
Dm =
m0 (X) =
Dn =
0
n (X) =
Do =
o0 (X) =
Dp =
p0 (X) =
Dq =
2Y 3Z4
4. g (X, Y , Z) = cos X + 2Y 2 − X 1+Z
3
Correction partielle de l’Exercice 2
Da
a0 (X)
= R \ π2 Z
cos2 (X)−sin2 (X)
1−2 sin2 (X)
√
=
= √
2 sin(X) cos(X)+1
Db =
b0 (X) =
Dc =
c0 (X) =
Dd =
d 0 (X)
De
0
e (X)
2 sin(X) cos(X)+1
R \ {0}
X
cos X
exp(− sin
) × 2 sin X−X
X2
X3
]e, +∞[
1
X ln(X) ln(ln(X))
R \ {−2 + kπ | k ∈ Z}
−6X sin(3X 2 ) sin(X+2)+(cos(3X 2 )+1) cos(X+2)
(sin(X+2))2
= −
= R \ {0}
cos(cos(ln(2X 2 )+3)) sin(ln(2X 2 )+3)
= −2
i h
i h
√ X
√
Dg = 21 2π + 1 − 4π2 + 8π + 1 , 2 ∪ 12 2π + 1 + 4π2 + 8π + 1 , +∞
2
2 +X
1
1
3X
g 0 (X) = 2g(X)
π + X +2X−X
=
π
+
2g(X)
(X+2)2
(X+2)2
i√
h
i
√ h
3, +∞
Dh = −∞, 3 ∪
h0 (X) = 2X sin(cos(ln(3X 2 − 1))) − X 2 cos(cos(ln(3X 2 − 1))) sin(ln(3X 2 − 1)) 3X6X
2 −1
6X 3 cos(cos(ln(3X 2 −1))) sin(ln(3X 2 −1))
= 2X sin(cos(ln(3X 2 − 1))) −
3X 2 −1
Di = R i 0 (X) = exp sin(X + 1) − cos(X 2 ) × (cos(X + 1) + 2X sin(X 2 )) × ln(2 + sin(X))
cos X
+ exp sin(X + 1) − cos(X 2 ) × 2+sin
X
cos X
2
= exp sin(X + 1) − cos(X ) × ln(2 + sin(X))(cos(X + 1) + 2X sin(X 2 )) + 2+sin
X
Dj = R \ {ln(kπ) | k ∈ N>0 }
sin(exp X) cos(X) exp(sin X)−exp(sin X) cos(exp X) exp X
j 0 (X) =
2
=
Dk
k 0 (X)
exp(sin X) ×
sin (exp X)
sin(exp X) cos(X)−cos(exp X) exp X
sin2 (exp X)
R
P
Xk
−X P2014 kX k−1
−e−X 2014
k=1
k=0 k! + e
k!
P
k
X
−X P2013 X k = −e−X X 2014
−e−X 2014
+
e
k=0 k!
k=0 k!
2014!
R
P2014 Qt−1
Q2014
t=0
k=0 (X − k) × k=t+1 (X − k)
] − ∞, 0[ ∪ ]1, 2[ ∪ ]3, 4[ ∪ . . . ∪ ]2014, +∞[
P2014 1
Sk=0 X−k
k∈Z ]2kπ,
π/2 + 2kπ[ ∪ ] − π + 2kπ, −π/2 +2 2kπ[2 3
sin X+cos X
eX +2X+2 (3X 2 + 2) ln(cos(X) sin(X)) + −cos(X)
sin(X)
] − ∞, 0[ ∪ ] ln 2 + ∞[
2X
X
√2e −3e
2 e2X −3eX +2
]3, +∞[
1−
√1
2 X+6
√
2 X+6−1
√ √
√
4 X+6 X− X+6
=
Dr
0
r (X)
Ds
= R \ {0}
cosh(X) X sinh(X)−cosh(X)
=
· exp
X
X2
= ]0, +∞[
s0 (X)
Dt
=
=
t 0 (X)
=
Du
0
u (X)
= R
=
√
√
2 X− X+6
=
]0, +∞[
(ln(2)2X +ln(3)3X ) ln(X)− X1 (2X +3X )
ln2 (X)
1−tanh
√
2
=
(ln(2)2X +ln(3)3X )X ln(X)−(2X +3X )
X ln2 (X)
(X)
2 tanh(X)+π
Dv
v 0 (X)
= R
= − sin(sin(cos(X))) × cos(cos(X)) × sin(X)
= − sin(X) cos(cos(X)) sin(sin(cos(X))
n √ √ o
Si α ≤ 0alors Dw = R \ − |α|, |α| , sinon Dw = R,
w0 (X)
=
Dy
(X 2 +α)×3X 2 −(X 3 +π)×2X
2
X 2 +α
(
)
=
= R
−X 2
(X+2014)
2
4 X( X+6−1)−6
1−tanh2 (X)
tanh(X)
y 0 (X) = (1 − X 2 )e 2
Dz = R \ {−2014} 2013
X+1
z0 (X) =
2 cos X+2014
= R
= −3 sin(3X) cos(2X) + e3X+1 cos(X) + eX−1 + X 3
+ cos(3X) −2 sin(2X) + 3e3X+1 cos(X) + eX−1 + X 3
+ cos(3X) cos(2X) + e3X+1 − sin(X) + eX−1 + 3X 2
√
= √ 2 √X+6−1
q0 (X)
X 4 +3αX 2 −2αX
2
(X 2 +α )
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