Quiz : Maîtrisez la dérivabilité et les fonctions dérivées !

Question 1 sur 10 10:00

[{"id":30528,"question":"Soit f(x) = x² + 3x - 5. Quelle est la dérivée f'(x) ?","option_a":"A. f'(x) = 2x + 3","option_b":"B. f'(x) = x² + 3","option_c":"C. f'(x) = 2x - 5","option_d":"D. f'(x) = 2x + 3x","option_e":"","option_f":"","bonne_reponse":"A","explication":"La dérivée d'une fonction polynôme se calcule terme à terme : (x²)' = 2x, (3x)' = 3 et (-5)' = 0. Donc f'(x) = 2x + 3.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":30529,"question":"La fonction f(x) = |x| est-elle dérivable en x = 0 ?","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"B","explication":"La fonction valeur absolue n'est pas dérivable en 0 car elle n'admet pas de nombre dérivé en ce point (la limite à gauche et à droite du taux d'accroissement n'est pas la même).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":30530,"question":"Soit f(x) = (2x + 1)³. Quelle est la dérivée f'(x) ?","option_a":"A. f'(x) = 3(2x + 1)²","option_b":"B. f'(x) = 6(2x + 1)²","option_c":"C. f'(x) = 6x(2x + 1)²","option_d":"D. f'(x) = 3(2x + 1)","option_e":"","option_f":"","bonne_reponse":"B","explication":"On applique la règle de la chaîne : f'(x) = 3(2x + 1)² × 2 = 6(2x + 1)².","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":30531,"question":"Soit f(x) = e^(3x). Quelle est la dérivée f'(x) ?","option_a":"A. f'(x) = e^(3x)","option_b":"B. f'(x) = 3e^(3x)","option_c":"C. f'(x) = e^(x)","option_d":"D. f'(x) = 3e^x","option_e":"","option_f":"","bonne_reponse":"B","explication":"La dérivée de e^(u(x)) est u'(x)e^(u(x)). Ici, u(x) = 3x, donc u'(x) = 3. D'où f'(x) = 3e^(3x).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":30532,"question":"La dérivée d'une fonction constante est toujours nulle.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"C'est vrai : si f(x) = k (constante), alors f'(x) = 0 pour tout x.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":30533,"question":"Soit f(x) = ln(5x² + 1). Quelle est la dérivée f'(x) ?","option_a":"A. f'(x) = 10x \/ (5x² + 1)","option_b":"B. f'(x) = 10x","option_c":"C. f'(x) = 5x \/ (5x² + 1)","option_d":"D. f'(x) = 10x \/ (x² + 1)","option_e":"","option_f":"","bonne_reponse":"A","explication":"On utilise la dérivée de ln(u(x)) = u'(x)\/u(x). Ici, u(x) = 5x² + 1, donc u'(x) = 10x. Ainsi, f'(x) = 10x \/ (5x² + 1).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":30534,"question":"Soit f(x) = x * e^x. Quelle est la dérivée f'(x) ?","option_a":"A. f'(x) = e^x","option_b":"B. f'(x) = (x + 1)e^x","option_c":"C. f'(x) = x e^x","option_d":"D. f'(x) = e^x + x","option_e":"","option_f":"","bonne_reponse":"B","explication":"On applique la règle du produit : (u*v)' = u'v + uv'. Ici, u = x, v = e^x, donc u' = 1 et v' = e^x. D'où f'(x) = e^x + x e^x = (x + 1)e^x.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":30535,"question":"Si f'(a) = 0, alors la fonction f admet un extremum en x = a.","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"B","explication":"Faux. f'(a) = 0 est une condition nécessaire mais pas suffisante. Il faut aussi étudier le signe de f' autour de a pour confirmer un extremum.","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":30536,"question":"Soit f(x) = sin(2x). Quelle est la dérivée f'(x) ?","option_a":"A. f'(x) = 2cos(2x)","option_b":"B. f'(x) = cos(2x)","option_c":"C. f'(x) = 2sin(x)","option_d":"D. f'(x) = -2cos(2x)","option_e":"","option_f":"","bonne_reponse":"A","explication":"La dérivée de sin(u(x)) est u'(x)cos(u(x)). Ici, u(x) = 2x, donc u'(x) = 2. Ainsi, f'(x) = 2cos(2x).","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0},{"id":30537,"question":"La dérivée de la fonction f(x) = 1\/x est f'(x) = -1\/x².","option_a":"Vrai","option_b":"Faux","option_c":"","option_d":"","option_e":"","option_f":"","bonne_reponse":"A","explication":"Vrai. En utilisant la formule de dérivation des fonctions puissances : (x^n)' = n x^(n-1), avec n = -1, on obtient f'(x) = -1 * x^(-2) = -1\/x².","points":1,"type":"qcm","actif":1,"section_id":null,"ordre":0}]