Examen 2 (solutions) 201-NYA Calcul Diff´erentiel
21 octobre 2019
Professeur : Dimitri Zuchowski
Question 1. (5%)
C’est le taux de variation instantan´e ;f0(a) = lim
∆x→0
f(a+ ∆x)−f(a)
∆x . On peut interpr´eter g´eom´etrique- ment cette valeur comme ´etant la pente de la droite tangente `a la fonction au point (a, f(a)).
Question 2. (10%)
f0(2) = lim
h→0
f(2 +h)−f(2)
h = lim
h→0
p2(2 +h) + 3−p
2(2) + 3
h = lim
h→0
√7 + 2h−√ 7 h
= lim
h→0
(√
7 + 2h−√ 7)(√
7 + 2h+√ 7) h(√
7 + 2h+√
7) = lim
h→0
7 + 2h−7 h(√
7 + 2h+√
7) = lim
h→0
2h h(√
7 + 2h+√ 7)
= lim
h→0
√ 2
7 + 2h+√
7 = 2 2√
7 =
√ 7 7
Question 3. (10%)
f0(x) = 1
x2−1 0
= (x2−1)−10
=−(x2−1)−2(2x) = −2x (x2−1)2
et donc f0(2) = −4
(3)2 = −4
9 . L’´equation de la droite a donc la forme y = −4
9x+b et passe par le point
2,1 3
. D’o`u 1 3 =−4
9(2) +b=⇒b= 1 3 +8
9 = 11
9 . L’´equation de la droite est
y=−4 9x+11
9
Question 4. (35%) a) f0(x) = 21x6−28
3 x13 + 21x−4+ 0
b) f0(x) = 10x4p
3x5+x+(2x5−2)(15x4+ 1) 2√
3x5+x
c) f0(x) = (2x−3)(2x−5)−(x2−3x)(2) (2x−5)2
d) f0(x) = 50(8x3−6x)49(24x2−6)
1
page 2 Examen 2 (solutions)
e) f0(x) =
(12x2+ 5)(x2−5)√3
x7+ 2−(4x3+ 5x)
2x√3
x7+ 2 + (x2−5) 7x6
3√3
(x7+2)2
(x2−5)2p3
(x7+ 2)2
Question 5. (20%) a)
f0(x) = 7(x2−2)6(2x)(5x−7)9+ (x2−2)79(5x−7)8(5) = 14x(x2−2)6(5x−7)9+ 45(x2−2)7(5x−7)8
= (x2−2)6(5x−7)8 14x(5x−7) + 45(x2−2)
= (x2−2)6(5x−7)8 70x2−98x+ 45x2−90
= (x2−2)6(5x−7)8 115x2−98x−90
b)
f0(x) =
1 2√
x(√
x+ 1)−2√1x(√ x−1) (√
x+ 1)2 = (√
x+ 1)−(√ x−1) 2√
x(√
x+ 1)2 = 2 2√
x(√
x+ 1)2 = 1
√x(√ x+ 1)2
Question 6. (10%)
2(x3y2)0− 3x
y 0
= (5)0−(y3)0
2(3x2y2+x32yy0)−3y−3xy0
y2 = 0−3y2y0 2y2(3x2y2+x32yy0)−3y−3xy0 =−3y4y0
6x2y4+ 4x3y3y0−3y−3xy0=−3y4y0 (4x3y3−3x+ 3y4)y0 = 3y−6x2y4
y0 = 3y−6x2y4 4x3y3−3x+ 3y4
Question 7. (10%)
f0(x) = 1
2(x2−3)−12(2x) =x(x2−3)−12 f00(x) = (x2−3)−12 +x
−1
2(x2−3)−32(2x)
= (x2−3)−12 −x2(x2−3)−32
f000(x) =
−1
2(x2−3)−32(2x)
−2x(x2−3)−32 −x2
−3
2(x2−3)−52(2x)
=−x(x2−3)−32 −2x(x2−3)−32 + 3x3(x2−3)−52
=−3x(x2−3)−32 + 3x3(x2−3)−52
Calcul Diff´erentiel – 201-NYA – Automne 2019