calculer la dérivée des fonctions suivantes
f(t) = 9 t - 1 + 2/(2t - 3) la fonction existe pour t ≠ 3/2
f '(t) = 9 + (- 4/(2t - 3)²)
g(x) = (3 x - 2)(x + 4)²
posons u = 3 x - 2 et v = (x+4)²
(u x v)' = u'v + u v'
u = 3 x - 2 ⇒ u' = 3
v = (x+ 4)² ⇒ v' = 2(x + 4)
g '(x) = 3(x+4)² + 2(x +4)(3 x - 2)
= 3(x²+8 x + 16) + 2(3 x² + 10 x - 8)
= 3 x² + 24 x + 48 + 6 x² + 20 x - 16
g'(x) = 9 x² + 44 x + 32
h(t) = (2t-3)/(1 - 4t)²
(u/v)' = (u'v - v'u)/v²
on pose u = 2t-3 ⇒ u' = 2
v = (1 - 4t)² ⇒ v' = 2*(- 4)(1- 4t) = - 8(1-4t)
h '(t) = 2(1-4t)² + 8(1-4t)(2t - 3)]/(1-4t)⁴
= 2(1 - 8t + 16t² + 8(2t - 3 - 8t² + 12t)]/(1-4t)⁴
= (- 32 t² + 96 t - 6)/(1-4t)⁴
