Bonjour,
f'(x)= 2*2x-3+0
f'(x)= 4x-3
f'(x)= 1/3(3*x²)-6
f'(x)= (3x²/3) -6
f'(x)= x²-6
- f(x)= (2x+3)/x ⇔ 2x/x+3/x = 2+3/x ⇔ 3/x+2
f'(x)= - 3/x²
f(x)= (-2x+1)/(3x+2)
(u/v)'= (u'v-uv')/ v²
u= -2x+1; u'= -2
v= 3x+2; v'= 3
on remplace dans (u'v-uv')/ v²:
f'(x)= [ (-2)(3x+2) -3 (-2x+1) ] / (3x+2)²
f'(x)= (-6x-4+6x-3) / (3x+2)²
f'(x)= - 7/ (3x+2)²
