Fonction arcsin

arcsin
f(x) = arcsin 1 + x
1−xf
f
f
x= 1 1 −x= 0 f
−161 + x
1−x61
1−x1−x
1−x > 0 =⇒x < 1
−161 + x
1−x61
⇐⇒ −1(1 −x)61 + x61−x
⇐⇒ x−161 + x61−x
x−161 + x
1 + x61−x
⇐⇒ −161
x60
⇐⇒ x60
1−x < 0 =⇒x > 1
−161 + x
1−x61
⇐⇒ −1(1 −x)>1 + x>1−x
⇐⇒ x−1>1 + x>1−x
x−1>1 + x
1 + x>1−x
⇐⇒ −1>1
x>0
−1>1

f]−∞,0]
u
t
arcsin 1 −1x= 0
f]−∞,0[
g(x) = 1+x
1−xf(x) = arcsin g(x)f0(x) = g0(x) [(arcsin)0(g(x))]
g0(x) = 1 + x
1−x0
=1−x+1+x
(1 −x)2=2 (1 −x)2
(arcsin)0(x)
∀x∈]−1,1[,sin (arcsin(x)) = x
(arcsin)0(x) [cos (arcsin(x))] = 1
=⇒(arcsin)0(x) = 1
cos (arcsin(x))
∀x∈R,cos2(x) + sin2(x) = 1 ∀x∈[−1,1]
[cos (arcsin(x))]2+ [sin (arcsin(x))]2= 1
=⇒[cos (arcsin(x))]2= 1 −x2
=⇒ |cos (arcsin(x))|=√1−x21−x2>0x∈[−1,1]
cos (arcsin(x)) x∈[−1,1] arcsin(x)∈−π
2,π 2
cos (arcsin(x)) >0
cos (arcsin(x)) = p1−x2(arcsin)0(x) = 1
√1−x2
f0(x) = 2
(1 −x)2
1
s1−1 + x
1−x2
=2
(1 −x)2
1
s1−x
1−x2
−1 + x
1−x2
=2
(1 −x)21
|1−x|
1
p1−2x+x2−(1 + 2x+x2)
=2
|1−x|
1
√−4x
√−4x
f]−∞,0] √−4x

f0(x) = 1
|1−x|√−x=1 (1 −x)√−xx < 0
f f0
f(x) = arcsin 1+x
1−x
x−→ −∞ f