✅ Corrigé détaillé — Exercice 2
Partie A
| Fonction f(x) | f'(x) | Domaine de dérivabilité |
|---|---|---|
| a) f(x) = 7 | 0 | ℝ |
| b) f(x) = −4x + 3 | −4 | ℝ |
| c) f(x) = x⁵ | 5x⁴ | ℝ |
| d) f(x) = 1/x³ = x⁻³ | −3x⁻⁴ = −3/x⁴ | ℝ* (ou ]−∞;0[ ∪ ]0;+∞[) |
| e) f(x) = √x | 1/(2√x) | ]0 ; +∞[ |
Partie B
2a) f'(x) = 6x − 5.
2b) g'(x) = 3x² + 8x.
2c) h'(x) = 2×(1/(2√x)) + 3×(−1/x²) = 1/√x − 3/x².
Partie C
3a) En développant : f(x) = x³ − 3x + x² − 3 → f'(x) = 3x² − 3 + 2x = 3x² + 2x − 3.
En utilisant (uv)’ avec u = x+1, v = x²−3 : u’=1, v’=2x → f'(x) = 1×(x²−3) + (x+1)×2x = x²−3+2x²+2x = 3x²+2x−3 ✓.
3b) u = x², v = √x, u’ = 2x, v’ = 1/(2√x).
g'(x) = 2x×√x + x²×(1/(2√x)) = 2x√x + x²/(2√x).
En factorisant par √x : g'(x) = √x(2x + x/(2)) = √x × (5x/2) = (5x/2)√x = (5/2)x^(3/2).
(Ou : g(x) = x^(5/2) → g'(x) = (5/2)x^(3/2) directement.)
Partie D
4a) u = x²+1, v = x, u’ = 2x, v’ = 1.
f'(x) = (2x×x − (x²+1)×1) / x² = (2x²−x²−1)/x² = (x²−1)/x².
4b) u = 3x−1, v = x²+1, u’ = 3, v’ = 2x.
g'(x) = (3(x²+1) − (3x−1)×2x) / (x²+1)²
= (3x²+3 − 6x²+2x) / (x²+1)²
= (−3x²+2x+3) / (x²+1)².
Partie E
5a) g(x) = x², a = 2, b = 3, g'(x) = 2x.
f'(x) = 2 × 2(2x+3) = 4(2x+3).
5b) g(x) = x³, a = 5, b = −1, g'(x) = 3x².
g'(x) = 5 × 3(5x−1)² = 15(5x−1)².
5c) g(x) = √x, a = 4, b = 1, g'(x) = 1/(2√x).
h'(x) = 4 × 1/(2√(4x+1)) = 2/√(4x+1).
5d) g(x) = 1/x, a = 3, b = −2, g'(x) = −1/x².
k'(x) = 3 × (−1/(3x−2)²) = −3/(3x−2)².