Exercice 1

DERIVEES/EXERCICES
Exercices
Dérivées - Fonctions logarithmiques et exponentielles
Chercher les fonctions dérivées des fonctions numériques f définies dans R par :
f (x) = xlnx
f (x) = lnx2
f (x) = alnx
f (x) = ae
x
f (x) = xx
√
f (x) = ln 1 − x2
√
f (x) = ln(x + 1 + x2 )
a+x
)
f (x) = ln(
a−x
r
1+x
f (x) = ln
1−x
f (x) = ln(lnx)
f (x) = ln2 x
f (x) = xx
√
x2 + 1 − x
f (x) = ln √
x2 + 1 + x
f (x) = ex (x − 1)
f (x) = ex (x2 − 2x + 2)
ex − 1
f (x) = x
e +1
f (x) = ex lnx
x
☞ ici les réponses
Référence: derivees-e0003.pdf
√
2
2
f (x) = eln a +x
ex
f (x) =
1+x
x
f (x) = ln √
2
x +1+x
√
f (x) = ln(x + a + 2ax + x2 )
ax
f (x) = x
x
√
f (x) = x(a2 + x2 ) a2 − x2
f (x) = (ax + 1)2
ax − 1
f (x) = x
a +1
f (x) = lnsinx
f (x) = ln(sin2 x)
f (x) = lncosx
f (x) = lntanx
f (x) = lncotx
1
)
f (x) = ln(
cosx
1
f (x) = ln(
)
sinx
f (x) = ex cosx
☞ ici les réponses
DERIVEES/EXERCICES
Exercices
Réponses :
f ′ (x) = (xlnx)′ = lnx + 1
2
f ′ (x) = (lnx2 )′ =
x
alnx lna
f ′ (x) = (alnx )′ =
x
x
x
f ′ (x) = (ae )′ = ae ex lna
1
x
x
f ′ (x) = (xx )′ = xx [lnx(1 + lnx) + ]xx
x
√
x
′
′
f (x) = (ln 1 − x2 ) = −
1 − x2
√
1
f ′ (x) = (ln(x + 1 + x2 ))′ = √
1 + x2
2a
a+x ′
)) = 2
f ′ (x) = (ln(
a−x
a − x2
r
1
1+x ′
f ′ (x) = (ln
) =
1−x
1 − x2
2lnx
f ′ (x) = (ln2 x)′ =
x
′
x ′
x
f (x) = (x ) = x (lnx + 1)
√
x2 + 1 − x ′
−2
f ′ (x) = (ln √
) =√
2
x +1+x
x2 + 1
f ′ (x) = (ex (x − 1))′ = xex
f ′ (x) = (ex (x2 − 2x + 2))′ = x2 ex
2ex
ex − 1 ′
) = x
f ′ (x) = ( x
e +1
(e + 1)2
1
f ′ (x) = (ex lnx)′ ex (lnx + )
x
☞ Retour
Référence: derivees-e0003.pdf
f ′ (x) = (ln(lnx))′ =
1
xlnx
Exercices
DERIVEES/EXERCICES
Réponses :
√
a2 +x2 ′
f ′ (x) = (eln
) =√
x
+ x2
a2
xex
ex ′
) =
1+x
1 + x2
1
x
1
)′ = − √
f ′ (x) = (ln √
2
2
x
x +1+x
x +1
√
1
f ′ (x) = (ln(x + a + 2ax + x2 ))′ = √
2ax + x2
x
a
a
a
f ′ (x) = ( x )′ = ( )x (ln − 1)
x
x
x
√
a4 + a2 x2 − 4x4
′
′
2
2
2
2
√
f (x) = (x(a + x ) a − x ) =
a2 − x 2
f ′ (x) = ((ax + 1)2 )′ = 2ax (ax + 1)lna
2ax lna
ax − 1 ′
) = x
f ′ (x) = ( x
a +1
(a + 1)2
f ′ (x) = (lnsinx)′ = cotx
f ′ (x) = (ln(sin2 x))′ = 2cotx
f ′ (x) = (lncosx)′ = −tanx
2
f ′ (x) = (lntanx)′ =
sin2x
2
f ′ (x) = (lncotx)′ = −
sin2x
1
))′ = tanx
f ′ (x) = (ln(
cosx
1
f ′ (x) = (ln(
))′ = −cotx
sinx
f ′ (x) = (ex cosx)′ = ex (cosx − sinx)
f ′ (x) = (
☞ Retour
Référence: derivees-e0003.pdf