corrigé

x√x2=x
x√x2=|x|p(−3)2=√9=3
f:R→Rg:R→R
x y x ≤y f(x)≤f(y)
g(f(x)) ≤g(f(y)) g◦f
−∞
f(x) = sin(x2)R
f0(x) = cos(x2)
fR R
x
f0(x) = 2x.cos(x2)
a a
f(x)−f(a)=(x−a)f(x)−f(a)
x−a
lim
x→a
f(x)−f(a)
x−a=f0(a)
lim
x→a(f(x)−f(a)) = 0
lim
x→af(x) = f(a)f a

exp(−x) = 1
exp(x)
1
exp(x)6= exp(x)x6= 0
1
e6=e
R
x f(−x) = f(x)f(−x) = −f(x)
f(x) = −f(x)f(x)=0
f1(x) = ln(sin(x))
f1xsin(x)
k
2kπ < x < (2k+ 1)π
Df1=[
k∈Z
]2kπ, (2k+ 1)π[ = ... ∪]−2π, −π[∪]0, π[∪]2π, 3π[∪...
f2(x) = √x2−3x−4
f2x x2−3x−4
x2−3x−4 ∆ = 25 = 52
−1 4 x2−3x−4 = (x+ 1)(x−4)
x≤ −1x≥4
Df2=] − ∞,−1] ∪[4,+∞[
f3(x) = 1
ex(x4+x2)
f3ex(x4+x2)
x4+x2= 0
x4+x2=x2(x2+ 1)
x2+ 1 x
x2= 0 x= 0
Df3=R∗=] − ∞,0[ ∪]0,+∞[
f4(x) = p|tan(x)|
f4
xπ
2+kπ
Df4=R−nπ
2+kπ , k ∈Zo

f5(x)=2x
f5x
2x= exp(x.ln(2))
Df5=R
f6(x) = 1
ex−e1−x
f6x ex−e1−x
ex=e1−x
ex=e1−x⇔x= 1 −x⇔2x= 1 ⇔x=1
2
Df6=R−1
2= ] − ∞,1 2[∪]1 2,+∞[
g1(x) = (x2+ 1)ln(x)
g1]0,+∞[
g0
1(x)=2x.ln(x) + x2+ 1
x
g2(x) = exp(x+ sin(x))
g2R
g0
2(x) = (1 + cos(x)).exp(x+ sin(x))
g3(x) = sin(x)
x4+ 1
x4+1 x
g3R R
g0
3(x) = cos(x)(x4+ 1) −sin(x).4x3
(x4+ 1)2
g4(x) = tan(x)
R−π
2+kπ , k ∈Z
g0
4(x) = 1 + tan2(x) = 1
cos2(x)

g5(x) = cos2(x)
g5(x) = (cos(x))2g5
R R
g0
5(x) = −2.sin(x).cos(x) = −sin(2x)
g6(x) = (2x+ 1)3
g6R R
g0
6(x)=3.2.(2x+ 1)2= 6(2x+ 1)2