Lista Derivadas - Calculo 1 - joao bembe - cleo brasil (2)

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UNIVERSIDADE FEDERAL DO OESTE DO PARÁ 
 
Docente: Raimundo Augusto Rego Rodrigues 
Discente: Cléo Marques Brasil – 2021005352 
 João Paulo Santos Bembe – 2021004981 
 
LISTA DE EXERCÍCIO DE DERIVADAS 
 
1.1. Calcule a derivada das funções abaixo utilizando a definição de limite. 
 
lim
ℎ→0
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
 
 
a) 𝑓(𝑥) = sin 𝑥 
 
lim
ℎ→0
sin(𝑥 + ℎ) − sin (𝑥)
ℎ
 
 
lim
ℎ→0
sin(𝑥) . cos (ℎ) + sin(ℎ) . cos(𝑥) − sin(𝑥)
ℎ
 
 
lim
ℎ→0
(
sin(𝑥) . (cos(ℎ) − 1
ℎ
+
sin(ℎ) . cos (𝑥)
ℎ
) 
 
lim
ℎ→0
(
sin(𝑥) . (cos(ℎ) − 1
ℎ
) + lim
ℎ→0
(
sin(ℎ) . cos (𝑥)
ℎ
) 
 
sin(𝑥) . lim
ℎ→0
(
cos(ℎ) − 1
ℎ
) + cos(𝑥) . lim
ℎ→0
(
sin (ℎ)
ℎ
) 
 
sin(𝑥) . lim
ℎ→0
(
cos(ℎ) − 1
ℎ
 . 
cos(ℎ) + 1
cos(ℎ) + 1
) + cos(𝑥) . lim
ℎ→0
(
sin (ℎ)
ℎ
) 
 
sin(𝑥) . lim
ℎ→0
(
cos2(h) − 1
ℎ. (cos(𝑥) + 1)
) + cos(𝑥) . lim
ℎ→0
(
sin (ℎ)
ℎ
) 
 
sin(𝑥) . lim
ℎ→0
(
−sin2(h)
ℎ. (cos(𝑥) + 1)
) + cos(𝑥) . lim
ℎ→0
(
sin (ℎ)
ℎ
) 
 
sin(𝑥) . (− lim
ℎ→0
(
sin(ℎ)
ℎ
)). (lim
ℎ→0
(
sin(ℎ)
cos(ℎ) + 1
)) + cos (𝑥). lim
ℎ→0
(
sin (ℎ)
ℎ
) 
 
sin(𝑥) . (−1). (
0
1 + 1
) + cos(𝑥) . 1 
 
sin(𝑥) . 0 + cos(𝑥) . 1 
 
𝑓(𝑥) = cos(𝑥) 
 
b) 𝑓(𝑥) = ∜𝑥 
 
lim
ℎ→0
(
√𝑥 + ℎ
4
− √𝑥
4
ℎ
) 
 
lim
ℎ→0
(
√𝑥 + ℎ
4
− √𝑥
4
ℎ
.
(√𝑥 + ℎ
4
+ √𝑥
4
). (√𝑥 + ℎ
4
+ √𝑥
4
)²
(√𝑥 + ℎ
4
+ √𝑥
4
). (√𝑥 + ℎ
4
+ √𝑥
4
)²
) 
 
lim
ℎ→0
(
(√𝑥 + ℎ
4
− √𝑥
4
)
4
ℎ. (√𝑥 + ℎ
4
+ √𝑥
4
). (√𝑥 + ℎ
4
+ √𝑥
4
)²
) 
 
lim
ℎ→0
(
𝑥 + ℎ − 𝑥
ℎ. (√𝑥 + ℎ
4
+ √𝑥
4
). (√𝑥 + ℎ
4
+ √𝑥
4
)²
) 
 
lim
ℎ→0
(
ℎ
(√𝑥 + ℎ
4
+ √𝑥
4
). (√𝑥 + ℎ
4
+ √𝑥
4
)²
) 
 
(
1
(√𝑥 + 0
4
+ √𝑥
4
). (√𝑥 + 0
4
+ √𝑥
4
)²
) 
 
 
(
1
(√𝑥
4
+ √𝑥
4
). (√𝑥
4
+ √𝑥
4
)²
) 
 
(
1
(2. √𝑥
4
). (2. √𝑥
4
²)
) 
 
(
1
(2. 𝑥
1
4) . (2. 𝑥
2
4)
) 
 
𝑓(𝑥) =
1
4𝑥
3
4
 
 
 
c) 𝑓(𝑥) = cos (𝑥) 
 
lim
ℎ→0
(
cos(𝑥 + ℎ) − cos (𝑥)
ℎ
) 
 
lim
ℎ→0
(
cos(𝑥) . cos(ℎ) − sin(ℎ). sin(𝑥) − cos (𝑥)
ℎ
) 
 
lim
ℎ→0
(
cos(x) . (cos(h) − 1)
ℎ
−
sin(𝑥) . sin (ℎ)
ℎ
) 
 
lim
ℎ→0
(
cos(x) . (cos(h) − 1)
ℎ
− lim
ℎ→0
sin(𝑥) . sin (ℎ)
ℎ
) 
 
cos(x) . lim
ℎ→0
(
(cos(h) − 1)
ℎ
− sin(𝑥) . lim
ℎ→0
sin (ℎ)
ℎ
) 
 
cos(𝑥) . 0 − sin(𝑥) . 1 
 
𝑓(𝑥) = −sin (𝑥) 
 
d) 𝑓(𝑥) = 𝑥𝑛 
 
lim
ℎ→0
(
(𝑥 + ℎ)𝑛 + 𝑥𝑛
ℎ
) 
 
lim
ℎ→0
(
𝑥𝑛 + (1). 𝑥𝑛−1. ℎ + (2). 𝑥𝑛−2. ℎ2…+ (𝑛). 𝑥0. ℎ𝑛 − 𝑥𝑛
ℎ
) 
 
lim
ℎ→0
(
(1). 𝑥𝑛−1. ℎ + (2). 𝑥𝑛−2. ℎ2…+ (𝑛). 𝑥.0 . ℎ𝑛
ℎ
) 
 
lim
ℎ→0
(
ℎ((1). 𝑥𝑛−1 + (2). 𝑥𝑛−2…+ (𝑛). 𝑥0. ℎ𝑛−1)
ℎ
) 
 
lim
ℎ→0
((1). 𝑥𝑛−1 + (2). 𝑥𝑛−2…+ (𝑛)𝑥0. ℎ𝑛−1) 
 
lim
ℎ→0
((1). 𝑥𝑛−1) 
 
𝑓(𝑥) = 𝑛. 𝑥𝑛−1 
 
1.2. Derive: 
 
a) 𝑓(𝑥) =
𝑥
𝑥2+1
 
 
𝑥. (𝑥2 + 1) − (𝑥. (𝑥2 + 1))
(𝑥2 + 1)2
 
 
1. (𝑥2 + 1) − (𝑥. (𝑥2 + 1))
(𝑥2 + 1)2
 
 
𝑥2 + 1 − (𝑥. (2. 𝑥 + 0))
(𝑥2 + 1)2
 
 
𝑥2 + 1 − 2𝑥2
(𝑥2 + 1)2
 
 
𝑓(𝑥) =
−𝑥2 + 1
(𝑥2 + 1)2
 
 
b) 
𝑥2−1
𝑥+1
 
 
(𝑥2 − 1). (𝑥 − 1) − ((𝑥2 − 1). (𝑥 + 1))
(𝑥 + 1)2
 
 
(2𝑥 + 0). (𝑥 − 1) − ((𝑥2 − 1). (1 + 0))
(𝑥 + 1)2
 
 
2𝑥2 + 2𝑥 + 𝑥2 + 1
(𝑥 + 1)2
 
 
𝑥2 + 2𝑥 + 1
(𝑥 + 1)2
 
 
𝑥2 + 2𝑥 + 1
𝑥2 + 2𝑥 + 1
 
 
𝑓(𝑥) = 1 
 
c) 𝑓(𝑥) =
3𝑥2+3
5𝑥−3
 
 
(3𝑥2 + 3). (5𝑥 − 3) − ((3𝑥2 + 3). (5𝑥 − 3))
(5𝑥 − 3)2
 
 
(6𝑥 + 0). (5𝑥 − 3) − ((3𝑥2 + 3). (5.1 − 0))
(5𝑥 − 3)2
 
 
30𝑥2 − 18𝑥 − 15𝑥2 − 15
(5𝑥 − 3)2
 
 
15𝑥2 − 18𝑥 − 15
(5𝑥 − 3)2
 
 
 
d) 𝑓(𝑥) =
√𝑥
𝑥+1
 
 
(√𝑥). (𝑥 + 1)((√𝑥. (𝑥 + 1)))
( 𝑥 + 1)2
 
 
(
1
2√𝑥
) . (𝑥 + 1) − ((√𝑥). (𝑥 + 1)) 
 
(
1
2√𝑥
) . (𝑥 + 1) − ((√𝑥). (𝑥 + 1))
(𝑥 + 1)2
 
 
(
1
2√𝑥
) − √𝑥
(𝑥 + 1)2
 
 
𝑥 + 1 − 2√𝑥2
2√𝑥
(𝑥 + 1)2
 
 
1 − 𝑥
2√𝑥
.
1
(𝑥 + 1)2
 
 
𝑓(𝑥) =
1 − 𝑥
2√𝑥. (𝑥 + 1)2
 
 
e) 𝑓(𝑥) = 5𝑥 +
√𝑥
𝑥+1
 
 
5.1 +
1 − 𝑥
2√𝑥. (𝑥 + 1)^2
 
 
𝑓(𝑥) = 5 +
1 − 𝑥
2√𝑥. (𝑥 + 1)2
 
 
f) 𝑓(𝑥) = √𝑥 +
3
𝑥2+2
 
 
(
1
2√𝑥
) + (
(3)(𝑥3 + 2). ((3). (𝑥3 + 2))
(𝑥3 + 2)2
) 
 
(
1
2√𝑥
) + (
(0)(𝑥3 + 2). ((3). (𝑥3 + 2))
(𝑥3 + 2)2
) 
 
(
1
2√𝑥
) + (
−((3)(3𝑥2 + 0))
(𝑥3 + 2)2
) 
 
 
𝑓(𝑥) =
1
2√𝑥
−
9𝑥2
(𝑥3 + 2)
 
 
g) 𝑓(𝑥) =
√𝑥
3
+𝑥
√𝑥
 
 
(
(√𝑥
3
+ 𝑥). (√𝑥). ((√𝑥
3
+ 𝑥). (√𝑥))
(√𝑥)
2 ) 
 
(
(𝑥
1
3 + 𝑥) . (√𝑥) − ((√𝑥
3
+ 𝑥). (𝑥
1
2))
(√𝑥)
2 ) 
 
(
(
1
3∛𝑥2
+ 1) . (√𝑥) − ((√𝑥
3
+ 𝑥). (
1
2√𝑥
))
(√𝑥)
2 ) 
 
(
 
 
(
√𝑥
3√𝑥2
3 ) . (√𝑥) −
√𝑥
3
+ 𝑥
2√𝑥
(√𝑥)
2
)
 
 
 
 
1 + (3𝑥
1
6. 𝑥
1
2)
3 √x
6 −
√𝑥
3
+ 𝑥
2√𝑥
𝑥
 
 
1 + 3𝑥
2
6
3 √x
6 −
√𝑥
3
+ 𝑥
2√𝑥
𝑥
 
 
(
(1 + 3𝑥
2
6) . (2𝑥
1
2) − (3𝑥
1
6) . (3𝑥
1
3 + 𝑥)
(3𝑥
1
6) . (2𝑥
1
2)
𝑥
) 
 
(
6𝑥
7
6 + 2√𝑥 − (3√𝑥 + 3𝑥
7
6)
6𝑥
2
3
𝑥
) 
 
3𝑥
7
6 − √𝑥
6𝑥
2
3
𝑥
 
 
 
(
3𝑥
7
6 − √𝑥
6𝑥
7
3
.
1
𝑥
) 
 
√𝑥. (3𝑥
2
3 − 1)
√𝑥. (6𝑥
7
3)
 
 
(
3𝑥
2
3 − 1
6𝑥
7
3
) 
 
h) 𝑓(𝑥) =
𝑥+ √𝑥
4
𝑥2+3
 
 
(
(𝑥 + 𝑥
1
4) . (𝑥2 + 3) − ((𝑥 + ∜𝑥). (𝑥2 + 3))
(𝑥2 + 3)2
) 
 
(
(1 +
1
4𝑥
3
4
) (𝑥2 + 3) − ((𝑥 + √𝑥
4
). (2𝑥 + 0))
(𝑥2 + 3)2
) 
 
(
𝑥2 + 3
4𝑥
3
4
+ 𝑥2 + 3 − 2𝑥2 − 2√𝑥
4
(𝑥2 + 3)2
) 
 
(
𝑥2 + 3 + 12𝑥
3
4 + 4𝑥
11
4 − 8𝑥
11
4 − 8𝑥
7
4. √𝑥
4
4𝑥
3
4
(𝑥2 + 3)² 
) 
 
𝑓(𝑥) = (
𝑥2 + 3 + 12𝑥
3
4 − 4𝑥
11
4 − 8𝑥
7
4. √𝑥
4
4𝑥
3
4. (𝑥2 + 3)2
) 
 
1.3. Continue derivando. 
 
a) 𝑓(𝑥) = sin(3𝑥) 
 
lim
ℎ→0
(
𝑠𝑖𝑛3(𝑥 + ℎ) − sin (3𝑥)
ℎ
) 
 
lim
ℎ→0
(
sin (3𝑥 + 3ℎ) − sin (3𝑥)
ℎ
) 
 
lim
ℎ→0
(
sin(3𝑥) . cos 3(ℎ) + sin(3ℎ) . cos(3𝑥) − sin(3𝑥)
ℎ
) 
 
 
lim
ℎ→0
(
sin(3𝑥). (cos 3 (3ℎ) − 1)
ℎ
+
sin (3ℎ). cos(3𝑥)
ℎ
) 
 
lim
ℎ→0
(
sin(3𝑥). (cos 3 (3ℎ) − 1)
ℎ
+ lim
ℎ→0
sin (3ℎ). cos(3𝑥)
ℎ
) 
 
sin(3𝑥). lim
ℎ→0
(
cos(3ℎ) − 1
ℎ
) + cos(3𝑥) . lim
ℎ→0
(
sin(3ℎ)
ℎ
) 
 
 
sin(3𝑥). lim
ℎ→0
(
cos(3ℎ) − 1
ℎ
+
cos(3ℎ) + 1
cos(3ℎ) + 1
) + cos(3𝑥) . lim
ℎ→0
(
sin(3ℎ)
ℎ
) 
 
sin(3𝑥). lim
ℎ→0
(
cos2(3ℎ) − 1
ℎ. (cos(3ℎ) + 1
) + cos(3𝑥). lim
ℎ→0
(
sin(3ℎ)
ℎ
) 
 
sin(3𝑥). lim
ℎ→0
(
−sin(3ℎ)
ℎ. (cos(3ℎ) − 1
) + cos(3𝑥) . lim
ℎ→0
(
sin (3ℎ)
ℎ
) 
 
sin(3𝑥) . (−lim
ℎ→0
(
sin(3ℎ)
ℎ
)) . (lim
ℎ→0
(
sin(3ℎ)
cos(3ℎ) + 1
)) + cos(3𝑥) . lim
ℎ→0
(
sin(3ℎ)
ℎ
) 
 
sin(3𝑥). (−3). (
0
3 + 1
) + cos(3𝑥). 3 
 
sin(3𝑥). 0 + cos(3𝑥) . 3 
 
𝑓(𝑥) = 3. cos(3𝑥) 
 
b) 𝑓(𝑥) = 𝑒𝑥
2
 
 
𝑒𝑥
2
. 𝑥2 
 
 
𝑒𝑥
2
. 2𝑥 
 
𝑓(𝑥) = 2𝑥. 𝑒𝑥
2
 
 
 
c) 𝑓(𝑥) = (3𝑥2 − 2𝑥)10 
 
10(3𝑥2 − 2𝑥)9. (3𝑥2 − 2𝑥) 
 
10(3𝑥2 − 2𝑥)9. (2.3𝑥 − 2) 
 
10(3𝑥2 − 2𝑥)9. (6𝑥 − 2) 
 
10. 𝑥9. 2(3𝑥 − 2)9. (3𝑥 − 1) 
 
𝑓(𝑥) = 20𝑥9(3𝑥 − 1). (3𝑥 − 2)9 
 
d) 𝑓(𝑥) = √𝑥2 − 1 
 
𝑓(𝑥) = (𝑥2 − 1)
1
2 
 
1
2
. (𝑥2 − 1)
1
2
−1. (𝑥2 − 1) 
 
1
2. √𝑥2 − 1
. (2𝑥) 
 
2𝑥
2. √𝑥2 − 1
 
 
𝑓(𝑥) =
𝑥
√𝑥2 − 1
 
 
e) 𝑓(𝑥) = sin(𝑥). cos(𝑥) 
 
(sin(𝑥). cos (𝑥)) 
 
sin(𝑥) . cos(𝑥) + sin(𝑥) . cos(𝑥) 
 
cos(𝑥). cos(𝑥) + sin(𝑥) . (− sin(𝑥)) 
 
𝑓(𝑥) = cos2(𝑥) − sin2(𝑥) 
 
f) 𝑓(𝑥) = 𝑒sin(𝑥) 
 
𝑒sin(𝑥). sin(𝑥) 
 
𝑒sin(𝑥). cos (𝑥) 
 
1.4. Sabendo que lim
𝜃→0
sin (𝜃)
𝜃
= 1 mostre que lim
ℎ→0
cos(ℎ)−1
ℎ
= 0. 
 
 
lim
ℎ→0
cos(ℎ) − 1
ℎ
 
 
lim
ℎ→0
(
cos(ℎ) − 1
ℎ
.
cos(ℎ) + 1
cos(ℎ) + 1
 
 
lim
ℎ→0
(
cos2(ℎ) − 1
ℎ(cos(𝑥) + 1)
) 
 
lim
ℎ→0
(
−𝑠𝑖𝑛2(ℎ)
ℎ(cos(𝑥) + 1)
) 
 
(−lim
ℎ→0
(
sin(ℎ)
ℎ
)) . (lim
ℎ→0
(
sin(ℎ)
cos(ℎ) + 1
)) 
 
(−1). (
0
1 + 1
) 
 
0 
 
lim
ℎ→0
(
cos(ℎ) − 1
ℎ
) = 0 
 
1.5. Demonstre as seguintes propriedades: 
 
a) (𝑓(𝑥) + 𝑔(𝑥))
′
= 𝑓′(𝑥) + 𝑔′(𝑥) 
 
(𝑓(𝑥) + 𝑔(𝑥)) = 𝐹 
 
lim
ℎ→0
𝐹(𝑥 + ℎ) − 𝐹(𝑥)
ℎ
 
 
lim
ℎ→0
𝑓(𝑥 + ℎ) + 𝑔(𝑥 + ℎ) − (𝑓(𝑥) + 𝑔(𝑥))
ℎ
 
 
lim
ℎ→0
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
+
𝑔(𝑥 + ℎ) − 𝑔(𝑥)
ℎ
 
 
lim
ℎ→0
𝑓(𝑥 + ℎ) − 𝑓(𝑥)
ℎ
+ lim
ℎ→0
𝑔(𝑥 + ℎ) − 𝑔(𝑥)
ℎ
 
 
𝑓′(𝑥) + 𝑔′(𝑥) 
 
b) (𝑓 (𝑥) ⋅ 𝑔(𝑥 ))′ = 𝑓 ′(𝑥) ⋅ 𝑔(𝑥 ) + 𝑓 (𝑥) ⋅ 𝑔 ′(𝑥 ) 
 
lim
ℎ→0
𝑓 (𝑥 + ℎ) ⋅ 𝑔(𝑥 + ℎ) − 𝑓 (𝑥) ⋅ 𝑔(𝑥 )
ℎ
 
 
lim
ℎ→0
𝑓 (𝑥 + ℎ) ⋅ 𝑔 (𝑥 + ℎ) − 𝑔(𝑥 + ℎ) ⋅ 𝑓 (𝑥) + 𝑔(𝑥 + ℎ) ⋅ 𝑓 (𝑥) − 𝑓 (𝑥) ⋅ 𝑔(𝑥 )
ℎ
 
 
lim
ℎ→0𝑔(𝑥 + ℎ) ⋅ (𝑓 (𝑥 + ℎ) − 𝑓 (𝑥 )) + 𝑓 (𝑥) ⋅ (𝑔(𝑥 + ℎ) − 𝑔(𝑥))
ℎ
 
 
 
lim
ℎ→0
𝑔(𝑥 + ℎ). lim
ℎ→0
(f ( x + h) − f (x ))
ℎ
+ lim
ℎ→0
 𝑓(𝑥 + ℎ). lim
ℎ→0
(𝑔( 𝑥 + ℎ) − 𝑔 (𝑥))
1
 
 
(𝑓 (𝑥) ⋅ 𝑔(𝑥 ))′ = 𝑔(𝑥 ) ⋅ 𝑓 ′(𝑥 ) + 𝑓 (𝑥) ⋅ 𝑔 ′(𝑥 )