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Fonction logarithme népérien
Les dérivées avec la fonction
x
↦
ln
(
u
(
x
)
)
x\mapsto \ln \left(u\left(x\right)\right)
x
↦
ln
(
u
(
x
)
)
- Exercice 1
20 min
40
On suppose que toutes les fonctions ci-dessous sont dérivables sur un intervalle
I
I
I
que l'on ne cherchera pas à déterminer.
Question 1
f
(
x
)
=
ln
(
2
x
−
7
)
f\left(x\right)=\ln \left(2x-7\right)
f
(
x
)
=
ln
(
2
x
−
7
)
Correction
(
ln
(
u
)
)
′
=
u
′
u
\left(\ln \left(u\right)\right)^{'} =\frac{u'}{u}
(
ln
(
u
)
)
′
=
u
u
′
On a :
u
(
x
)
=
2
x
−
7
u\left(x\right)=2x-7
u
(
x
)
=
2
x
−
7
et
u
′
(
x
)
=
2
u'\left(x\right)=2
u
′
(
x
)
=
2
Ainsi :
f
′
(
x
)
=
2
2
x
−
7
f'\left(x\right)=\frac{2}{2x-7}
f
′
(
x
)
=
2
x
−
7
2
Question 2
f
(
x
)
=
ln
(
−
5
x
+
1
)
f\left(x\right)=\ln \left(-5x+1\right)
f
(
x
)
=
ln
(
−
5
x
+
1
)
Correction
(
ln
(
u
)
)
′
=
u
′
u
\left(\ln \left(u\right)\right)^{'} =\frac{u'}{u}
(
ln
(
u
)
)
′
=
u
u
′
On a :
u
(
x
)
=
−
5
x
+
1
u\left(x\right)=-5x+1
u
(
x
)
=
−
5
x
+
1
et
u
′
(
x
)
=
−
5
u'\left(x\right)=-5
u
′
(
x
)
=
−
5
Ainsi :
f
′
(
x
)
=
−
5
−
5
x
+
1
f'\left(x\right)=\frac{-5}{-5x+1}
f
′
(
x
)
=
−
5
x
+
1
−
5
Question 3
f
(
x
)
=
ln
(
3
x
)
f\left(x\right)=\ln \left(3x\right)
f
(
x
)
=
ln
(
3
x
)
Correction
(
ln
(
u
)
)
′
=
u
′
u
\left(\ln \left(u\right)\right)^{'} =\frac{u'}{u}
(
ln
(
u
)
)
′
=
u
u
′
On a :
u
(
x
)
=
3
x
u\left(x\right)=3x
u
(
x
)
=
3
x
et
u
′
(
x
)
=
3
u'\left(x\right)=3
u
′
(
x
)
=
3
Ainsi :
f
′
(
x
)
=
3
3
x
f'(x)=\frac{3}{3x}
f
′
(
x
)
=
3
x
3
f
′
(
x
)
=
1
x
f'\left(x\right)=\frac{1}{x}
f
′
(
x
)
=
x
1
Question 4
f
(
x
)
=
ln
(
x
2
+
7
)
f\left(x\right)=\ln \left(x^{2}+7\right)
f
(
x
)
=
ln
(
x
2
+
7
)
Correction
(
ln
(
u
)
)
′
=
u
′
u
\left(\ln \left(u\right)\right)^{'} =\frac{u'}{u}
(
ln
(
u
)
)
′
=
u
u
′
On a :
u
(
x
)
=
x
2
+
7
u\left(x\right)=x^2+7
u
(
x
)
=
x
2
+
7
et
u
′
(
x
)
=
2
x
u'\left(x\right)=2x
u
′
(
x
)
=
2
x
Ainsi :
f
′
(
x
)
=
2
x
x
2
+
7
f'\left(x\right)=\frac{2x}{x^2+7}
f
′
(
x
)
=
x
2
+
7
2
x
Question 5
f
(
x
)
=
ln
(
−
x
+
9
)
f\left(x\right)=\ln \left(-x+9\right)
f
(
x
)
=
ln
(
−
x
+
9
)
Correction
(
ln
(
u
)
)
′
=
u
′
u
\left(\ln \left(u\right)\right)^{'} =\frac{u'}{u}
(
ln
(
u
)
)
′
=
u
u
′
On a :
u
(
x
)
=
−
x
+
9
u\left(x\right)=-x+9
u
(
x
)
=
−
x
+
9
et
u
′
(
x
)
=
−
1
u'\left(x\right)=-1
u
′
(
x
)
=
−
1
Ainsi :
f
′
(
x
)
=
−
1
−
x
+
9
f'(x)=\frac{-1}{-x+9}
f
′
(
x
)
=
−
x
+
9
−
1
f
′
(
x
)
=
−
1
−
x
+
9
f'\left(x\right)=-\frac{1}{-x+9}
f
′
(
x
)
=
−
−
x
+
9
1
Question 6
f
(
x
)
=
ln
(
9
x
−
8
)
f\left(x\right)=\ln \left(9x-8\right)
f
(
x
)
=
ln
(
9
x
−
8
)
Correction
(
ln
(
u
)
)
′
=
u
′
u
\left(\ln \left(u\right)\right)^{'} =\frac{u'}{u}
(
ln
(
u
)
)
′
=
u
u
′
On a :
u
(
x
)
=
9
x
−
8
u\left(x\right)=9x-8
u
(
x
)
=
9
x
−
8
et
u
′
(
x
)
=
9
u'\left(x\right)=9
u
′
(
x
)
=
9
Ainsi :
f
′
(
x
)
=
9
9
x
−
8
f'\left(x\right)=\frac{9}{9x-8}
f
′
(
x
)
=
9
x
−
8
9
Question 7
f
(
x
)
=
ln
(
x
+
1
)
f\left(x\right)=\ln \left(x+1\right)
f
(
x
)
=
ln
(
x
+
1
)
Correction
(
ln
(
u
)
)
′
=
u
′
u
\left(\ln \left(u\right)\right)^{'} =\frac{u'}{u}
(
ln
(
u
)
)
′
=
u
u
′
On a :
u
(
x
)
=
x
+
1
u\left(x\right)=x+1
u
(
x
)
=
x
+
1
et
u
′
(
x
)
=
1
u'\left(x\right)=1
u
′
(
x
)
=
1
Ainsi :
f
′
(
x
)
=
1
x
+
1
f'\left(x\right)=\frac{1}{x+1}
f
′
(
x
)
=
x
+
1
1
Question 8
f
(
x
)
=
2
ln
(
4
x
+
6
)
f\left(x\right)=2\ln \left(4x+6\right)
f
(
x
)
=
2
ln
(
4
x
+
6
)
Correction
(
ln
(
u
)
)
′
=
u
′
u
\left(\ln \left(u\right)\right)^{'} =\frac{u'}{u}
(
ln
(
u
)
)
′
=
u
u
′
On a :
u
(
x
)
=
4
x
+
6
u\left(x\right)=4x+6
u
(
x
)
=
4
x
+
6
et
u
′
(
x
)
=
4
u'\left(x\right)=4
u
′
(
x
)
=
4
Ainsi :
f
′
(
x
)
=
2
×
4
4
x
+
6
f'\left(x\right)=2\times \frac{4}{4x+6}
f
′
(
x
)
=
2
×
4
x
+
6
4
f
′
(
x
)
=
8
4
x
+
6
f'\left(x\right)=\frac{8}{4x+6}
f
′
(
x
)
=
4
x
+
6
8