Lista 2 de Cálculo 2

Prévia do material em texto

Matema´tica para a Economia I - 2a lista de exerc´ıcios
Prof. - Juliana Coelho
1 - Calcule os limites pedidos:
(a) lim
x→−2−
3x− 1
x+ 2
;
(b) lim
x→3+
√
x+ 6
2x− 6 ;
(c) lim
x→−1+
x
x2 + 2x+ 1
;
(d) lim
x→−1−
x
x2 + 2x+ 1
; Existe o limite lim
x→−1
x
x2 + 2x+ 1
?;
(e) lim
x→5−
3x− x2
5− x ;
(f) lim
x→0+
5− x
3x− x2 ;
(g) lim
x→5−
1
x2 − 4x− 5;
(h) lim
x→+∞
2x2 − x;
(i) lim
x→−∞
2x2 − 3x3;
(j) lim
x→−∞
x3 − 2x+ 1
3− x4 ;
(k) lim
x→−∞
3− x4
x3 − 2x+ 1;
(l) lim
x→+∞
x2 − 1
x2 + 1
;
(m) lim
x→+∞
12x+ 3x2 − 4x3
19− 3x+ x3 .
1
2 - Calcule as derivadas pedidas, simplificando o resultado:
(a) f(x) = 12x3 − 3x2 + x− 12, ache f ′(x);
(b) f(x) = x6 − 2x5 + 3x+ 1, ache f ′(x), f ′′(x) e f ′′′(x);
(c) f(x) = 5x−5 + 25x−1, ache f ′(x) e f ′′(x);
(d) f(x) =
x3
2x− 1, ache f
′(x);
(e) f(x) =
2x2 − x+ 1
x+ 1
, ache f ′(x);
(f) f(x) =
3
√
x5, ache f ′(x);
(g) f(x) =
3
√
x5
x− 1, ache f
′(x);
(h) f(x) = (2x2 − x+ 1)7, ache f ′(x);
(i) f(x) =
5
√
12x− x2, ache f ′(x);
(j) f(x) = x2 − 1
x− 1 +
√
x2 − 1, ache f ′(x);
(k) f(x) =
(
2x+ 1
x− 2
)5
, ache f ′(x);
(l) f(x) = (3x+ 1)4(4x− 1)3, ache f ′(x);
(m) f(x) = cos(x2 − 1) + sen2(x), ache f ′(x);
(n) f(x) = cos(x) sen(x), ache f ′(x);
(o) f(x) =
sen(x)
x
, ache f ′(x);
(p) f(x) =
1 + sen(x)
1− sen(x) , ache f
′(x);
(q) f(x) = e5x
3−x, ache f ′(x);
(r) f(x) =
ln(x)
x
, ache f ′(x);
(s) f(x) = e−1/x + ln(4 + 5x), ache f ′(x);
(t) f(x) =
3
√
e4x + 5− 1
ln(x)
, ache f ′(x);
2
Gabarito:
1a QUESTA˜O:
(a) Como lim
x→−2−
3x− 1 = −7 e lim
x→−2−
x+ 2 = 0, precisamos analisar o sinal de g(x) = x+2.
Fazendo a ana´lise de sinal, vemos que no limite a` esquerda (isto e´, x < −2), temos
sinal negativo. Logo
lim
x→−2−
3x− 1
x+ 2
=
−7
−0 = +∞.
(b) lim
x→3+
√
x+ 6
2x− 6 = +∞;
(c) lim
x→−1+
x
x2 + 2x+ 1
= −∞;
(d) lim
x→−1−
x
x2 + 2x+ 1
= −∞; Como os limites laterais coincidem (veja o item (c)), o
limite existe e lim
x→−1
x
x2 + 2x+ 1
= −∞.
(e) lim
x→5−
3x− x2
5− x = −∞;
(f) lim
x→0+
5− x
3x− x2 = +∞;
(g) lim
x→5−
1
x2 − 4x− 5 = −∞.
(h) lim
x→+∞
2x2 − x = lim
x→+∞
2x2 = +∞;
(i) lim
x→−∞
2x2 − 3x3 = lim
x→−∞
−3x3 = (−3) · (−∞) = +∞;
(j) lim
x→−∞
x3 − 2x+ 1
3− x4 = limx→−∞
x3
−x4 = limx→−∞−
1
x
= 0;
(k) lim
x→−∞
3− x4
x3 − 2x+ 1 = limx→−∞
−x4
x3
= lim
x→−∞
−x = +∞;
(l) lim
x→+∞
x2 − 1
x2 + 1
= lim
x→+∞
x2
x2
= lim
x→+∞
1 = 1;
(m) lim
x→+∞
12x+ 3x2 − 4x3
19− 3x+ x3 = limx→+∞
−4x3
x3
= lim
x→+∞
−4 = −4.
3
2a QUESTA˜O:
(a) f ′(x) = 12 · 3x2 − 3 · 2x+ 1 + 0 = 36x2 − 6x+ 1;
(b) f ′(x) = 6x5 − 2 · 5x4 + 3 + 0 = 6x5 − 10x4 + 3,
f ′′(x) = (6x5 − 10x4 + 3)′ = 6 · 5x4 − 10 · 4x3 + 0 = 30x4 − 40x3
f ′′′(x) = (30x4 − 40x3)′ = 30 · 4x3 − 40 · 3x2 = 120x3 − 120x2;
(c) f ′(x) = 5 · (−5)x−6 + 25 · (−1)x−2 = −25x−6 − 25x−2
f ′′(x) = (−25x−6 − 25x−2)′ = −25 · (−6)x−7 − 25 · (−2)x−3 = 150x−7 + 50x−3;
(d) f ′(x) =
(x3)′(2x− 1)− x3(2x− 1)′
(2x− 1)2 =
3x2(2x− 1)− 2x3
4x2 − 4x+ 1 =
4x3 − 3x2
4x2 − 4x+ 1
(e) f ′(x) =
(2x2 − x+ 1)′(x+ 1)− (2x2 − x+ 1)(x+ 1)′
(x+ 1)2
=
(4x− 1)(x+ 1)− (2x2 − x+ 1)
(x+ 1)2
=
=
2x2 + 4x− 2
(x+ 1)2
;
(f) como f(x) =
3
√
x5 = x5/3, temos f ′(x) =
5
3
x2/3 =
5
3
3
√
x2;
(g) usando o item anterior, temos f ′(x) =
(
3
√
x5)′(x− 1)− ( 3√x5)(x− 1)′
(x− 1)2 =
5
3
x2/3(x− 1)− x5/3
(x− 1)2 =
=
5
3
x5/3 − 5
3
x2/3 − x5/3
(x− 1)2 =
2
3
x5/3 − 5
3
x2/3
(x− 1)2 ;
(h) f ′(x) = 7(2x2−x+1)6(2x2−x+1)′ = 7(2x2−x+1)6(4x−1) = (2x2−x+1)6(28x−7);
(i) como f(x) =
5
√
12x− x2 = (12x− x2)1/5, temos f ′(x) = 1
5
(12x− x2)−4/5(12− 2x) =
=
12− 2x
5 5
√
(12x− x2)4 ;
(j) como f(x) = x2 − 1
x− 1 +
√
x2 − 1 = x2 − (x− 1)−1 + (x2 − 1)1/2, temos
f ′(x) = 2x− (−1)(x− 1)−2 + 1
2
(x2 − 1)−1/2 · 2x = 2x+ 1
(x− 1)2 +
2x
2
√
x2 − 1;
(k) f ′(x) = 5
(
2x+ 1
x− 2
)4(
2x+ 1
x− 2
)′
= 5
(
2x+ 1
x− 2
)4 −5
(x− 2)2 =
−25(2x+ 1)4
(x− 2)6 ;
4
(l) f ′(x) = 4(3x+ 1)3 · 3 · (4x− 1)3 + (3x+ 1)4 · 3(4x− 1)2 · 4 =
= 12(3x+ 1)3(4x− 1)3 + 12(3x+ 1)4(4x− 1)2;
(m) f ′(x) = − sen(x2− 1)(x2− 1)′+2 sen(x)( sen(x))′ = −2x sen(x2− 1)+2 sen(x) cos(x);
(n) f ′(x) = (cos(x))′ sen(x) + cos(x)( sen(x))′ = − sen(x) sen(x) + cos(x) cos(x) =
= cos2(x)− sen2(x);
(o) como f(x) =
sen(x)
x
= x−1 sen(x), temos f ′(x) = −1 · x−2 sen(x) + x−1 cos(x) =
− sen(x)
x2
+
cos(x)
x
Este item tambe´m pode ser resolvido com a regra do quociente:
f ′(x) =
cos(x) · x− sen(x) · 1
x2
=
x cos(x)− sen(x)
x2
.
Note que esta e´ a mesma expressa˜o obtida anteriormente pois, colocando no mesmo
denominador, temos:
− sen(x)
x2
+
cos(x)
x
=
− sen(x) + x cos(x)
x2
=
x cos(x)− sen(x)
x2
;
(p) f ′(x)=
(1 + sen(x))′(1− sen(x))− (1 + sen(x))(1− sen(x))′
(1− sen(x))2 =
=
cos(x)(1− sen(x))− (1 + sen(x))(− cos(x))
(1− sen(x))2 =
2 cos(x)
(1− sen(x))2 ;
(q) f ′(x) = (5x3 − x)′e5x3−x = (15x2 − 1)e5x3−x;
(r) como f(x) =
ln(x)
x
= x−1 ln(x), temos
f ′(x) = −1 · x−2 ln(x) + x−1 1
x
=
− ln(x)
x2
+
1
x2
=
1− ln(x)
x2
Este item tambe´m pode ser resolvido com a regra do quociente:
f ′(x) =
(ln(x))′x− ln(x)x′
x2
=
1
x
· x− ln(x) · 1
x2
=
1− ln(x)
x2
;
(s) como f(x) = e−1/x + ln(4 + 5x) = ex
−1
+ ln(4 + 5x), temos
f ′(x) = (−1)x−2ex−1 + 5
4 + 5x
=
−e1/x
x2
+
5
4 + 5x
;
(t) como f(x) =
3
√
e4x + 5− 1
ln(x)
= (e4x + 5)1/3 − (ln(x))−1, temos
f ′(x) =
1
3
(e4x + 5)−2/3(e4x + 5)′ − (−1)(ln(x))−2(ln(x))′ =
=
1
3
1
3
√
(e4x + 5)2
· 4e4x + 1
(ln(x))2
· 1
x
=
4e4x
3 3
√
(e4x + 5)2
+
1
x(ln(x))2
5