TD2 : Dérivation : taux d`accroissement, élasticités, calcul marginal

f1(x) = x2+ 1 f2(x) = ln(x)f3(x) = 1+x5
1−x
f4(x) = √4−x2f5(x) = ln x+1
x−1.ln x−1 x+1 f6(x) = −x
√x2−1
g1(x) = x3g2(x) = ln(x)g3(x) = 2x2+ 1
g4(x) = eαx g5(x) = ln(2x2+ 1) g6(x) = x+1
x−1
v1(t) = eαt v2(t) = Aγt v3(t) = 4+t
5−t2
v4(t) = ln(βt)v5(t) = √t+ 4 v6(t) = t(1 −t)
α∈[0; 1]
h(x)R
h(x) = [αKx+ (1 −α)Wx]1
x
u(x) = ln[h(x)] g(x) = ln[αKx+ (1 −α)Wx]
g0(x)
u(x) = g(x)−g(0)
x−0
limx→0u(x)limx→0h(x) = KαW1−α
f(x) = ln x−√x
f Df

f0f00 f
f
ln 4 ≈1,39 f(x)Df
∀x∈[1,+∞[,0≤ln(x)
x≤1
√x
lim
x→+∞
ln x
x= 0
CT (x) = x3−2x2+ 4x+ 8
a b c
2x3−2x2−8 = 2(x−2)(ax2+bx +c)
t Y
Y(t) = 20e0,04t
y(t) = Y(t)
P(t)L(t)
P(t)L(t)
P(t) = 100e0,02t
0,01

B0(t) = rB(t) + D(t)
Y0(t) = gY (t)
b(t)
d(t)
b(t) = B(t)
Y(t);d(t) = D(t) Y(t)
b(t)
d(t) = 0,
b0(t) = 0,
f(x) = ex−e−x
2g(x) = 42x−3w(x) = ln(x2+x+ 1)
CT (x) = ln x+ 4x+ 2
x= 5 −1
2p
x p

CT
x
CT (x) = x+ 8
p
x= 12 −2p
p(x)
RT (x)RT (x) = p(x)×x
RM Rm CM
Cm
(p(x) = CM )
v