Réponse :
1) f(x) = 3 x³ - 2√x Df ' = ]0 ; + ∞[
f '(x) = 9 x² - 2/2√x
= 9 x² - 1/√x
= 9 x² - √x/x
= (9 x³ - √x)/x
2) f(x) = (x² - 5)/(x - 4) Df = R\{4}
f '(x) = (u/v)' = (u'v - v'u)/v²
u(x) = x² - 5 ⇒ u'(x) = 2 x
v(x) = x - 4 ⇒ v'(x) = 1
f '(x) = (2 x(x - 4) - (x² - 5))/(x - 4)²
= (2 x² - 8 x - x² + 5)/(x - 4)²
= (x² - 8 x + 5)/(x - 4)²
3) f(x) = (3 + x)/(3 - x) Df = R \{3}
f '(x) = ((3 - x) - (- (3 + x))/(3 - x)²
= (3 - x + 3 + x)/(3 - x)²
= 6/(3 - x)²
Explications étape par étape :