Réponse :
1. f(x) = 1/x(x³ - 1) ; I = R*
f '(x) = (u*v)' = u'v + v'u
u = 1/x ⇒ u' = - 1/x²
v = x³ - 1 ⇒ v' = 3 x²
f '(x) = - 1/x²(x³ - 1) + 3 x²(1/x)
= - x + 1/x² + 3 x
f '(x) = 2 x + 1/x²
2) f(x) = x²(√x + 1) ; I = [0 ; + ∞[ plutôt I = ]0 ; + ∞[
f '(x) = 2 x(√x + 1) + 1/2√x ( x²)
= 2 x√x + 2 x + x²/2√x
= 4 x² + 4 x√x + x²)/2 x
= 5 x² + 4 x√x)/2 x
= x(5 x + 4√x)/2 x
f '(x) = (5 x + 4√x)/2
3) f (x) = 1/(x² + 1) ; I = R
f '(x) = (1/u)' = - u'/u²
u = x² + 1 ⇒ u' = 2 x donc f '(x) = - 2 x/(x² + 1)²
4) f(x) = 1/√x ; I = ]0 ; + ∞[
f '(x) = (1/u)' = - u'/u²
u = √x ⇒ u' = 1/2√x donc f '(x) = - 1/2√x/(√x)² = - 1/(2 x√x)
Explications étape par étape
