Détails du calcul de 3 dérivées composées

f(x) = x3−3x
3(x2−2x)
x
x
f(x) = x(x2−3)
3x(x−2)
x= 0 x= 2 Df=R− {0,2}
x= 0 x
f(x) = x2−3
3(x−2)
f(x) = 1
3×u v
u=x2−3 ; u0= 2x
v=x−2 ; v0= 1
f0(x) = 1
3×u0v−uv0 v2
f0(x) = 1
3×2x(x−2) −(x2−3) (x−2)2
f0(x) = 2x2−4x−x2+ 3
3(x−2)2
f0(x) = x2−4x+ 3
3(x−2)2
x1= 1 1 −4 + 3 = 0
x2−Sx +P= 0
S=x1+x2= 4 P=x1x2= 3
x2=P/x1= 3.
f0(x) = (x−1)(x−3)
3(x−2)2
x= 0
0/0x= 0
x x 0
f(0) = 1/2

f(x) = 8√x−x
x+√x
√x
√x
f(x) = 8√x−(√x)2
(√x)2+√x
f(x) = √x(8 −√x)
√x(√x+ 1)
x < 0√x x = 0
√x Df=R+∗
x= 0 √x
f(x) = 8−√x
√x+ 1 =u
v
u= 8 −√x v =√x+ 1 u v √x
(√x)0= (x1/2)0= (1/2)x−1/2=1
2√x
u0=−1
2√x
v0=1
2√x
f0(x) = u0v−uv0
v2
f0(x) = −1
2√x(√x+ 1) −(8 −√x)1 2√x
(√x+ 1)2
1
2√x
f0(x) =
1
2√x(−(√x+ 1) −(8 −√x))
(√x+ 1)2
f0(x) = −(√x+ 1) −(8 −√x)
2√x(√x+ 1)2
f0(x) = −√x−1−8 + √x
2√x(√x+ 1)2
f0(x) = −9
2√x(√x+ 1)2

f(x) = 2x
x2−1
x2−1 = (x−1)(x+ 1)
x=±1
f(x) = 2x
x2−1=u v
u= 2x
v=x2−1
u0= 2
v0= 2x
f0(x) = u0v−uv0
v2
f0(x) = 2(x2−1) −2x(2x)
(x2−1)2
f0(x) = 2x2−2−4x2
(x2−1)2
f0(x) = −2x2−2
(x2−1)2
f0(x) = −2(x2+ 1)
(x2−1)2