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  • Publié le : 18 novembre 2010
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TES Octobre 2010page 1 Ex13p88: "x > 0 ln(x) + 4 a) f (x) = x ! 5 + !# !Df = ]0;1["]1;+#[. x !1 $x ! 1 "x > 0 1 ! x2 b) f (x) = !Df = R+*. + x ln(x) ! # 2x $x ! 0 c) f (x) = ln(3 + x) + x 2 ! x ! 3 +x > 0 !Df = ]$3;+#[.

h) f (x) =

2x +1 " x > 0 • # !Df = ]0;1["]1;+#[. ln x $ln x ! 0 1 u = 2x +1 u'= 2 v = ln x v' = x 1 2 ln x ! (2x +1) x f '(x) = ( ln x )2

x !1 x !1 )! > 0 !x % ]1;2[!Df= ]1;2[. 2!x 2!x "x > 2 % 2 e) f (x) = 3ln(x ! 2) ! 2 ln(x !1) ! #" x < !1! ! Df = ]2;+#[. %# $$ou!x > 1
d) f (x) = ln(

f '(x) =

2x ln x ! 2x !1 et Df ' = ]0;1["]1;+#[. x ( ln x )2

Ex24p89:Df = ]0;+[ f (x) = !x +1 ! ln x Equation de la tangente en A, xA = e ! yA = f (e) = !e +1 !1 = !e

1 • x > 0 ! Df = R+*. ( ln x )2 2 1 1 1 • u = ln x , u ' = , f = u 2 ! f ' = 2u ! u ' x 2 2 ln x !f '(x) = et Df ' = R+*. x b) f (x) = 2x(1 ! ln x) • x > 0 ! Df = R+*.
Ex21p89: a) f (x) =

1 1 ! a = f '(1) = !1 ! x e 1 b = yA ! ax A = !e ! (!1 ! )e = !e + e +1 = 1 e 1 ! (TA) : y = !(1+ )x +1 .e f '(x) = !1 !
Ex38p90: a) ln(x 2 + 3) = ln(4x) !x2 + 3 > 0 • " ! Df = R+* (1). $ 4x > 0 # • ln(x 2 + 3) = ln(4x) ! x 2 + 3 = 4x & x 2 ! 4x + 3 = 0 & (x !1)(x ! 3) = 0 & x ! {1; 3} (2) • (1) et (2)! ln(x 2 + 3) = ln(4x) & x ! {1; 3} . b) ln(!2x + 3) = 0

u = 2x u ' = 2

v = 1 ! ln x

v' = !

1 x

# 1& f = u ! v ! f = u '! v + u ! v ' ! f '(x) = 2(1 ! ln x) + 2x " %! ( $ x'
! f '(x) =!2 ln x et Df '= R+*.

x c) f (x) = ! +1+ 2 ln x • x > 0 ! Df = R+*. 2 1 2 f '(x) = ! + et Df '= R+*. 2 x 2x 2 d) f (x) = • x > 0 ! Df = R+*. ! x ln x 5 4x f '(x) = ! (ln x +1) et Df '= R+*. 5 x !ln x e) f (x) = • x > 0 ! Df = R+*. x 1 v=x u = x ! ln x v' =1 u ' = 1! x 1 (1 ! )x ! (x ! ln x) u u ' v ! uv ' x ! f '(x) = f= !f= x2 v v2 ln x !1 ! f '(x) = et Df ' = R+*. x2 "x > 0 e f) f (x) = • #! Df = ]0;1[U]1;+#[. ln x $ln x ! 0 1 !e " !e " u ' 1 e x u = ln x u ' = ! f = ! f ' = = u2 (ln x)2 x u !e f '(x) = et Df ' = ]0;1[U]1;+#[. x(ln x)2
g) f (x) =

3 • !2x + 3 > 0 ! Df =] ! "; [...
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