demo6.py

Created on January 11, 2022

443 Bytes

from maths import*

La fonction ln est dérivable sur ]0; +∞[ et pour tout réel x > 0 : ln0
(x) = 1/x
 
 
On admet que x->ln x est déribale sur R+*

On démontre que lnx'=1/x
R+*-->R
Soit x --> e^(lnx) (=x)
 f(x)=lnx est dérivable sur R+*
 x-> e^x  est derivable sur R
 par composition de fc f(x)=e^(lnx) 
 est dérivable sur R+*
 
 Vxqui appartient à R+*, 
 f'(x)=1
 f'(x)=(lnx)'e^(lnx)
 = lnx'*x
alors lnx'x=1
alors lnx'=1/x

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