Réponse :
1) f : x → (x + 3)√(x + 3)
f '(x) = (u*v)' = u'v + v'u
u(x) = x + 3 ⇒ u '(x) = 1
v(x) = √(x + 3) ⇒ v '(x) = 1/2√(x+3)
f '(x) = √(x+3) + (x+3)/2√(x + 3)
= [2√(x+3) * √(x+3) + (x + 3)]/2√(x+3) or x + 3 > 0
= [2(x+3) + x + 3)]/2√(x+3)
= (2 x + 6 + x + 3)/2√(x+3)
f '(x) = (3 x + 9)/2√(x+3)
2) f : x → 1/(x⁸ + 1)
f '(x) = - 8 x⁷/(x⁸+1)
3) f ; x → √(x² + 2 x + 1)
f '(x) = u'/2√u
u (x) = x² + 2 x + 1 ⇒ u '(x) = 2 x + 2
f '(x) = (2 x + 2)/2√(x²+2 x + 1) = 2(x + 1)/2√(x + 1)² or x + 1 > 0
donc f '(x) = (x + 1)/(x + 1) = 1
4) f '(x) = eˣ/2√(eˣ - 1)
Explications étape par étape :
