maths.py

Created on December 13, 2023

504 Bytes

u.v=AB*AC*Cos(BAC)

u.v=1/2([u+v]**2-u**2-v**2)
u.v=1/4([u+v]**2-[u-v]**2)
u.v=1/2(u**2+v**2-[u-v]**2)

orthogonalité : AB.AC=0

ab=sqrt(x**2+y**2+z**2)



y=f'(a)(x-a)+f(a)

f'(1/x)=-1/x**2
f'(sqrt(x))=1/2sqrt(x)

u.v=u'.v+u.v'
u/v=(u'.v-u.v')/v**2

(e**u)'=u'.e**u
(u**n)'=n*u'*u**n-1
(sqrt(u))'=u'/2sqrt(u)
(1/u(x))'=-u'(x)/(u(x))**2


->f est continue sur l'intervalle et est strictement 
decroi./croissante
->donner image
->dire que la solution appartient 
a l'intervalle des images

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