asdasd

1. Curso de Cálculo
Diferencial e
Integral
Ing. José M Gómez Martínez
Ingeniero Petrolero
Cédula Profesional 11997514

2. Ing. José Gómez
𝐹 𝑥 = 𝑌 = 2
Y X
2 2
2 4
2 6
2 8
2 10
2 12
2 14
𝐷𝑂𝑀𝐼𝑁𝑂 (X)
(-infinito , +infinito)
𝑅𝐴𝑁𝐺𝑂 (Y)
CONTRADOMINO
2

3. Ing. José Gómez
𝐹 𝑥 = −3
𝑌 = −3
Y X
-3 2
-3 1
-3 6
-3 8
-3 10
-3 12
-3 14

4. Ing. José Gómez
𝐹 𝑥 = 𝑌 = −2
Y X
-2 2
-2 4
-2 6
-2 8
-2 10
-2 12
-2 14
𝐹 𝑥 = −2
𝐷𝑂𝑀𝐼𝑁𝑂 (X)
(-infinito , +infinito)
𝑅𝐴𝑁𝐺𝑂 (Y)
CONTRADOMINO
−2

5. Ing. José Gómez
𝐹 𝑥 = 𝑌 = −2𝑋
Y X
-4 2
-2 4
-2 6
-2 8
-2 10
-2 12
-2 14
𝐹 𝑥 = −2𝑋
𝐹 2 = −2(2)
𝐹 2 = −2(+2)
𝐹 2 = −4

6. Ing. José Gómez
𝐹 𝑦 = 𝑥 = 0
x y
0 2
0 4
0 6
0 8
0 10
0 12
0 14

7. Ing. José Gómez
𝐹 𝑥 = 𝑌 = 3/4
Y X
3/4 2
3/4 4
3/4 6
3/4 8
3/4 10
3/4 12
3/4 14
𝐹 𝑥 = 3/4
𝐷𝑂𝑀𝐼𝑁𝑂 (X)
(-infinito , +infinito)
𝑅𝐴𝑁𝐺𝑂 (Y)
CONTRADOMINO
0.75

8. Ing. José Gómez
𝐹 𝑥 = 𝑌 = 𝑋
Y X
10 10
15 15
21 21
25 25
-50 -50
0 0
-3 -3
𝐹 10 = 10
𝐷𝑂𝑀𝐼𝑁𝑂 (X)
(-)
𝑅𝐴𝑁𝐺𝑂 (Y)
CONTRADOMINO
𝐹 15 = 15

9. Ing. José Gómez
10 = 3𝑥 + 2
10 − 2 = 3𝑥
8 = 3𝑥
8/3 = 𝑥

10. Ing. José Gómez
−5𝑥 − 0 𝑦 = −40
𝑆𝑢𝑚𝑎 𝑦 𝑅𝑒𝑠𝑡𝑎
+
−1 ∗ (15 𝑥 + 10 𝑦 = 180)
+
− −
−
−5𝑥 = −40
40

11. Ing. José Gómez
𝑆𝑢𝑚𝑎 𝑦 𝑅𝑒𝑠𝑡𝑎
+
𝑥 = −
40
−5
= 8
10𝑥 + 10𝑦 = 140
10(8) + 10𝑦 = 140
80 + 10𝑦 = 140
10𝑦 = 140 − 80
10𝑦 = 60
𝑦 =
60
10
= 6

12. Ing. José Gómez
𝑆𝑢𝑚𝑎 𝑦 𝑅𝑒𝑠𝑡𝑎
+
𝑥 = −
40
−5
= 8
10(8) + 10(6) = 140
𝑦 =
60
10
= 6
80 + 60 = 140
140 = 140

13. Ing. José Gómez
𝑆𝑢𝑠𝑡𝑖𝑡𝑢𝑐𝑖𝑜𝑛
𝑦 = 180 − 5𝑥 /15
10𝑥 + 10𝑦 = 140
10𝑥 + 10(
180 − 5𝑥
15
) = 140
10𝑥 + 10(12 −
𝑥
3
) = 140
10𝑥 + 120 −
10
𝑥 = 140

14. Ing. José Gómez
𝑆𝑢𝑠𝑡𝑖𝑡𝑢𝑐𝑖𝑜𝑛
30
3
𝑥 −
10
3
𝑥 = 140 − 120
20
3
𝑥 = 20
𝑥 =
20
1
20
3
=
60
20
= 3

15. Ing. José Gómez
𝑆𝑢𝑠𝑡𝑖𝑡𝑢𝑐𝑖𝑜𝑛
𝑥 =
20
1
20
3
=
60
20
= 3
𝑦 = 180 − 5𝑥 /15
𝑦 = 180 − 5(3) /15
𝑦 = 180 − 15 /15
𝑦 = 165 /15
𝑦 = 11

16. Ing. José Gómez
𝑆𝑢𝑠𝑡𝑖𝑡𝑢𝑐𝑖𝑜𝑛
𝑥 = 3
10𝑥 + 10𝑦 = 140
𝑦 = 11
10(3) + 10(11) = 140
30 + 110 = 140
140 = 140

17. Ing. José Gómez
𝐼𝑔𝑢𝑎𝑙𝑎𝑐𝑖ó𝑛
𝑥 = (180 − 10𝑦)/15
𝑥 = (140 − 10𝑦)/10
𝑥1 = 𝑥2
(140 − 10𝑦)
10
=
(180 − 10𝑦)
15
15 ∗ (140 − 10𝑦)
=
10 ∗ (180 − 10𝑦)

18. Ing. José Gómez
𝐼𝑔𝑢𝑎𝑙𝑎𝑐𝑖ó𝑛
15 140 − 10𝑦 = 10(180 − 10𝑦)
2100 − 150𝑦 = 1800 − 100𝑦
100𝑦 − 150𝑦 = 1800 − 2100
−50𝑦 = −300
𝑦 = −
300
−50
= 6

19. Ing. José Gómez
𝐼𝑔𝑢𝑎𝑙𝑎𝑐𝑖ó𝑛
𝑦 = −
300
−50
= 6
𝑥 = (180 − 10𝑦)/15
𝑥 = (180 − 10(6))/15
𝑥 = (180 − 60)/15
𝑥 =
120
15
= 8

20. Ing. José Gómez
𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡𝑒𝑠
𝑋 =
140
180
10
10
10
15
10
10
1400 −1800
100 −150
=
−400
−50
= 8

21. Ing. José Gómez
𝐷𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑛𝑡𝑒𝑠
𝑌 =
10
15
140
180
10
15
10
10
1800 − 2100
100 −150
=
−300
−50
= 6

22. Ing. José Gómez
𝑔 𝑜 𝑓 𝑥 = 𝑔 𝑓 𝑥
𝑔 𝑜 𝑓 𝑥 = 𝑓 𝑔 𝑥

23. Ing. José Gómez
2.- 𝑓 𝑓(𝑥) =
(3𝑥+2 )+3
(3𝑥+2 )+1

24. Ing. José Gómez
+5𝑋2
EXPONENTE
LITERAL
INCOGNITA
VARIABLE
CONSTANTE
SIGNO
TÉRMINO

25. Ing. José Gómez
5𝐴 − 5𝐴 = 0
(+5𝐴) − (+5𝐴) = 0
+5𝐴 − (+5𝐴) = 0
+5𝐴 − 5𝐴 = 0
0 = 0

26. Ing. José Gómez
𝑣 =
𝑑
𝑡
𝑣𝑡 = 𝑑
𝑡 =
𝑑
𝑣

27. Ing. José Gómez
𝑓 𝑥 = 3𝑥 + 2
𝒇 𝒈(𝒙) = 𝟑
𝒙 + 𝟑
𝟐𝒙 + 𝟏
+ 𝟐
𝑓 𝑔(𝑥) = 3
𝑥 + 3
2𝑥 + 1
+ 2
2.- 𝑓 𝑓(𝑥) =
(3𝑥+2 )+3
(3𝑥+2 )+1

28. Ing. José Gómez
𝑓 𝑔(𝑥) = 3
𝑥 + 3
2𝑥 + 1
+ 2
𝑓 𝑓(𝑥) =
(3𝑥 + 2 ) + 3
(3𝑥 + 2 ) + 1

29. Ing. José Gómez
𝑎
𝑎
= 1
𝑎
1
= 𝑎
𝒂
𝟎
= 𝑬𝑹𝑹𝑶𝑹 O INDETERMINACION
0
𝑎
= 0
𝐥𝐢𝐦
𝒙→𝒂
𝒇
𝟏𝟎
𝒙
= 𝒂
𝐥𝐢𝐦
𝒙→𝟎
𝒇
𝟏𝟎
𝟎
= 𝒆𝒓𝒓𝒐𝒓

30. Ing. José Gómez
3
𝑥 + 3
2𝑥 + 1
3
1
𝑥 + 3
2𝑥 + 1
𝑥
2
X
𝑥
2
3 ∗ (𝑥 + 3)
1 ∗ (2𝑥 + 1)

31. Ing. José Gómez
3 ∗ (𝑥 + 3)
1 ∗ (2𝑥 + 1)
3𝑥 + 9
2𝑥 + 1

32. Ing. José Gómez
𝑓 𝑔(𝑥) = 3
𝑥 + 3
2𝑥 + 1
+ 2
𝑓 𝑔(𝑥) =
3𝑥 + 9
2𝑥 + 1
+ 2

33. Ing. José Gómez
𝒇 𝒈(𝒙) =
𝟑𝒙 + 𝟗
𝟐𝒙 + 𝟏
+ 𝟐
F(g(x)) X
-3
-2
-1
0
1
2
3
𝑓 𝑔(−3) =
3𝑥 + 9
2𝑥 + 1
+ 2
𝑓 𝑔(−3) =
3(−3) + 9
2(−3) + 1
+ 2
𝑓 𝑔(−3) =
9 + 9
−6 + 1
+ 2
𝑓 𝑔(−3) =
18
−5
+ 2
𝑓 𝑔(−3) = −
18
5
+ 2

34. Ing. José Gómez
𝒇 𝒈(𝒙) =
𝟑𝒙 + 𝟗
𝟐𝒙 + 𝟏
+ 𝟐
F(g(x)) X
-8/5 -3
-2
-1
0
1
2
3
𝑓 𝑔(−3) = −
18
5
+ 2
𝑥
𝑎
+/−
𝑦
𝑎
𝑓 𝑔(−3) = −
18
5
+
2
1
2
1
∗
5
5
=
10
5
𝑓 𝑔(−3) = −
18
5
+
10
5
𝑓 𝑔(−3) = −
8
5

35. Ing. José Gómez
𝒇 𝒈(𝒙) =
𝟑𝒙 + 𝟗
𝟐𝒙 + 𝟏
+ 𝟐
F(g(x)) X
-8/5 -3
-2
-1
11 0
1
2
3
𝑓 𝑔(0) =
3(0) + 9
2(0) + 1
+ 2
𝑓 𝑔(0) =
0 + 9
0 + 1
+ 2
𝑓 𝑔(0) =
9
1
+ 2
𝑓 𝑔(0) = 9 + 2
𝑓 𝑔(0) = 11

36. Ing. José Gómez
𝑓 𝑥 = 𝑌 = 2
Y X
2 1
𝐥𝐢𝐦
𝒙→𝒂
𝒇 𝒙 = 𝒂
lim
𝑥→1
𝑓 1 = 1
Y = f(x) X
1 1

37. Ing. José Gómez
𝐹 𝑥 = 𝑌 = 2
Y X
2 2
2 4
2 6
2 8
2 10
2 12
2 14
𝐷𝑂𝑀𝐼𝑁𝑂 (X)
(-infinito , +infinito)
𝑅𝐴𝑁𝐺𝑂 (Y)
CONTRADOMINO
2

38. Ing. José Gómez
𝐥𝐢𝐦
𝒙→𝟑
𝟕𝟕
Y X
77 1
2
77 3
4
5
6
7
𝐥𝐢𝐦
𝒙→𝟑
𝟕𝟕 =77
𝐥𝐢𝐦
𝒙→𝟏
𝟕𝟕 = 𝟕𝟕

39. Ing. José Gómez
𝐥𝐢𝐦
𝒙→𝟓
(𝟑𝒙 − 𝟕)
Y X
1
2
3
4
8 5
6
7
𝐥𝐢𝐦
𝒙→𝟓
𝟑𝒙 − 𝟕 = 𝐥𝐢𝐦
𝒙→𝟓
𝟑(𝟓) − 𝟕 =
𝐥𝐢𝐦
𝒙→𝟓
𝟑(𝟓) − 𝟕 = 𝐥𝐢𝐦
𝒙→𝟓
𝟏𝟓 − 𝟕 =
𝐥𝐢𝐦
𝒙→𝟓
𝟏𝟓 − 𝟕 = 𝐥𝐢𝐦
𝒙→𝟓
= 𝟖

40. Ing. José Gómez
𝐥𝐢𝐦
𝒙→𝟐
(𝒙𝟐 + 𝟐𝒙 − 𝟏)
Y X
1
7 2
ERROR -3
4
5
6
7
𝐥𝐢𝐦
𝒙→𝟐
( 𝟐 𝟐 + 𝟐(𝟐) − 𝟏)
𝐥𝐢𝐦
𝒙→𝟐
(𝟒 + 𝟒 − 𝟏)
𝐥𝐢𝐦
𝒙→𝟐
𝟕 = 𝟕

42. Ing. José Gómez
𝐥𝐢𝐦
𝒙→𝟑
𝟒𝒙 − 𝟓
𝟓𝒙 − 𝟏
Y X
5 0
2
1/2 3
4
5
6
7
𝐥𝐢𝐦
𝒙→𝟑
𝟒(𝟑) − 𝟓
𝟓(𝟑) − 𝟏
𝐥𝐢𝐦
𝒙→𝟑
𝟏𝟐 − 𝟓
𝟏𝟓 − 𝟏
𝐥𝐢𝐦
𝒙→𝟑
𝟕
𝟏𝟒
𝐥𝐢𝐦
𝒙→𝟑
𝟕
𝟏𝟒
=
𝟏
𝟐
𝐥𝐢𝐦
𝒙→𝟎
−𝟓
−𝟏
= 𝟓

43. 𝐥𝐢𝐦
𝒙→𝟑
𝟐𝑿𝟑 − 𝟓𝑿𝟐 − 𝟐𝑿 − 𝟑
𝟒𝑿𝟑 − 𝟏𝟑𝑿𝟐 + 𝟒𝑿 − 𝟑
𝐥𝐢𝐦
𝒙→𝟑
𝟐(𝟑)𝟑−𝟓 𝟑 𝟐 − 𝟐(𝟑) − 𝟑
𝟒(𝟑)𝟑−𝟏𝟑(𝟑)𝟐+𝟒(𝟑) − 𝟑
𝐥𝐢𝐦
𝒙→𝟑
𝟐(𝟐𝟕) − 𝟓(𝟗) − 𝟔 − 𝟑
𝟒(𝟐𝟕) − 𝟏𝟑(𝟗) + 𝟏𝟐 − 𝟑
𝐥𝐢𝐦
𝒙→𝟑
𝟓𝟒 − 𝟒𝟓 − 𝟔 − 𝟑
𝟏𝟎𝟖 − 𝟏𝟏𝟕 + 𝟏𝟐 − 𝟑
𝐥𝐢𝐦
𝒙→𝟑
𝟎
𝟎
= 𝑰𝑵𝑫𝑬𝑻𝑬𝑹𝑴𝑰𝑵𝑨𝑪𝑰𝑶𝑵

44. CONDICIONES Y REGLAS
ALGEBRAICAS
DIVISIÓN
MULTIPLICACIÓN
POTENCIA
RADICACIÓN
𝑿
𝒀
𝑿𝒀
𝑿𝒀
𝒀
𝑿

45. 𝑎
𝑎
= 1
𝑎
1
= 𝑎
𝒂
𝟎
= 𝑬𝑹𝑹𝑶𝑹 O INDETERMINACION
0
𝑎
= 0
a(a + b) = a2 + ab
𝑎2
+ 𝑎𝑏
𝑎
=
𝑥3
𝑥3
+
𝑎𝑏
𝑎
=
𝑎 ∗ 𝑎
𝑎
+
𝑎 ∗ 𝑏
𝑎
𝑎 ∗ 𝑎
𝑎
+
𝑎 ∗ 𝑏
𝑎
= 1 ∗ 𝑎 + 1 ∗ 𝑏 = 𝑎 + 𝑏
𝑎 𝑎 + 𝑏 = 𝑎2 + 𝑎𝑏

46. 𝑎 ∗ 1 = 𝑎
1𝑎 + 1𝑎 = 2𝑎
1𝑎 ∗ 1𝑎 = 1 ∗ 1 ∗ 𝑎1+1
= 1𝑎2
= 𝑎2
𝑎22
= (𝑎2)2 = 𝑎2 ∗ 𝑎2 = 𝑎2+2=4= 𝑎4
𝑎1 = 𝑎
𝑎0
= 1
𝑎−1 =
1
𝑎
𝒂𝟏/𝟐
=
𝟐
𝒂𝟏

47. 2
+1 = 1
2
−1 = 𝑖𝑛𝑑𝑒𝑡𝑒𝑟𝑚𝑖𝑛𝑎𝑑𝑜
2
0 = 0
2
1 ∗ 𝑖
2
−1 = 𝑖
1 ∗ 𝑖 = i
2
+11= 1
1
2
2
𝑥1 = 𝑥
1
2
3
𝑥1 = 𝑥
1
3

48. 2
𝑥1 ∗
2
𝑥1 =
2
𝑥1
2
= 𝑥
1
2
2
= 𝑥
1
2
∗
2
1 = 𝑥
2
2 = 𝑥1 = 𝑥

50. 𝑥 − 1
(𝑥 − 1)
1
( 𝑥 + 1)
∗
1
1
( 𝑥 + 1)
∗
1
𝑥 + 1

51. 1
𝑥 + 1
1
1 + 1
=
1
1 + 1
=
1
2
=
1
2
=0.5
0.5 =0.5

52. 𝑓 𝑥 = 𝑥5
dx
𝑓′ 𝑥
= 𝑑(5 ∗ 𝑥5−1
)/𝑑𝑥 = 5𝑥4
𝑓 𝑥 = 5𝑥1
dx
𝑓′ 𝑥 = 5 ∗ 𝑥1
dx
𝑓′ 𝑥 = 5 ∗ 1 ∗ 𝑥1−1
𝑓′ 𝑥 = 5 ∗ 1 ∗ 𝑥0
𝑓′ 𝑥
= 5 ∗ 1 ∗ 1
𝑓′ 𝑥
= 5

53. 𝑓 𝑥 = 2𝑥4
+ 𝑥3
− 𝑥2
+ 4
𝑓′ 𝑥 = 2𝑥4
𝑑𝑥 + 𝑥3
𝑑𝑥 − 𝑥2
𝑑𝑥 + 4𝑑𝑥
𝑓′ 𝑥 = 4 ∗ 2𝑥4−1
+ 3 ∗ 𝑥3−1
− 2 ∗ 𝑥2−1
+ 0
𝑓′ 𝑥 = 8𝑥3
+ 3𝑥2
− 2𝑥1
+ 0
𝑓′ 𝑥 = 8𝑥3
+ 3𝑥2
− 2𝑥1

54. 𝑓 𝑥 = (4𝑋 + 1)(10𝑥2
− 5)
𝑓 𝑥 = (𝑢)(𝑣)dx
𝒇′ 𝒙
= 𝒖
𝒅 𝒗
𝒅𝒙
+ (𝒗)
𝒅 𝒖
𝒅𝒙
𝑓′ 𝑥
= 4𝑋 + 1
𝑑 10𝑥2−5
𝑑𝑥
+ (10𝑥2
− 5)
𝑑 4𝑋+1
𝑑𝑥
𝑓′ 𝑥
= 4𝑋 + 1 (20𝑥 − 0) + (10𝑥2
− 5) (4 + 0)
𝒇′ 𝒙
= 𝟒𝑿 + 𝟏 (𝟐𝟎𝒙) + (𝟏𝟎𝒙𝟐
− 𝟓) (𝟒)

55. 𝒇′ 𝒙
= 𝒖
𝒅 𝒗
𝒅𝒙
+ (𝒗)
𝒅 𝒖
𝒅𝒙
𝑓′ 𝑥
= 4𝑋 + 1
𝑑 10𝑥2−5
𝑑𝑥
+ (10𝑥2
− 5)
𝑑 4𝑋+1
𝑑𝑥
𝑓′ 𝑥
= 4𝑋 + 1 (20𝑥 − 0) + (10𝑥2
− 5) (4 + 0)
𝒇′ 𝒙
= 𝟒𝑿 + 𝟏 (𝟐𝟎𝒙) + (𝟏𝟎𝒙𝟐
− 𝟓) (𝟒)
𝑓′ 𝑥
= 4𝑋 + 1 (
𝑑 10𝑥2
𝑑𝑥
−
𝑑 5
𝑑𝑥
) + (10𝑥2
− 5) (
𝑑 4𝑋
𝑑𝑥
+
𝑑 1
𝑑𝑥
)

56. 𝑓 𝑥 =
𝑥3
+ 2
3
𝑓 𝑥 =
𝑥3
3
+
2
3
𝑓 𝑥 =
1
3
(
𝑥3
1
) +
2
3
𝑓′ 𝑥 =
1
3
(
𝑥3
1
dx)+
2
3
𝑑𝑥
𝑓′ 𝑥 =
1
3
(3 ∗ 𝑥3−1dx)+
2
3
𝑑𝑥
𝑓′ 𝑥 =
1
3
(3𝑥2)+ 0
𝑓′ 𝑥 =
1
3
(
3𝑥2
1
)
𝑓′ 𝑥 = (
3𝑥2
3
)=𝑥2

57. 𝑓 𝑥 = 𝑥5
− 5𝑥4
𝑑𝑥
𝒇′ 𝒙 = 𝟓𝒙𝟒
− 𝟐𝟎𝒙𝟑
𝒅𝒙
𝒇′′ 𝒙 = 𝟐𝟎𝒙𝟑
− 𝟔𝟎𝒙𝟐
𝒅𝒙
𝒇′′′ 𝒙 = 𝟔𝟎𝒙𝟐
− 𝟏𝟐𝟎𝒙𝟏
𝒅𝒙
𝒇′′′′ 𝒙 = 𝟏𝟐𝟎𝒙𝟏
− 𝟏𝟐𝟎𝒅𝒙
𝒇′′′′′ 𝒙 = 𝟏𝟐𝟎 − 𝟎𝒅𝒙
𝒇′′′′′′ 𝒙 = 𝟎𝒅𝒙

58. 5𝑥4
= 5 𝑥 𝑥 𝑥 𝑥 = 5 −1 −1 −1 −1
𝟓 −𝟏 −𝟏 −𝟏 −𝟏 = +𝟓
20𝑥3
= 20 𝑥 𝑥 𝑥 = 20 −1 −1 −1
20 −1 −1 −1 = −20
+5 − −20 = +5 + 20 = +𝟐𝟓

59. 𝑒𝑥
𝑑𝑥
𝑒𝑥
𝑑𝑥 = 𝑒𝑢
𝑑𝑢 = 𝑒𝑢
+ 𝐶
𝑒𝑢
𝑑𝑢 = 𝑒𝑢
+ 𝐶
𝒆𝒙
𝒅𝒙 = 𝒆𝒙
+ 𝑪

60. (3𝑥4
+𝑥2
+ 2)𝑑𝑥
3𝑥4
𝑑𝑥 + 𝑥2
𝑑𝑥 + 2 𝑑𝑥
3 𝑥4
𝑑𝑥 + 𝑥2
𝑑𝑥 + 2 𝑑𝑥
3 ∗
𝑥4+1
4 + 1
+
𝑥2+1
2 + 1
+ 2(𝑥)
3 ∗
𝑥5
5
+
𝑥3
3
+ 2𝑥
𝟑𝒙𝟓
𝟓
+
𝒙𝟑
𝟑
+ 𝟐𝒙 + 𝑪

61. 𝒙 + 𝒙
𝑥1
+ 𝑥
𝑥1
+
2
𝑥1
𝒙𝟏
+ 𝒙
𝟏
𝟐
𝒙𝟏
+ 𝒙
𝟏
𝟐)𝑑𝑥
𝒙𝟏
𝒅𝒙 + 𝒙
𝟏
𝟐 𝑑𝑥
𝑥2
2
+
𝑥
1
2
+1
1
2
+ 1
𝑥2
2
+
𝑥
1
2
+
2
2
=
3
2
1
2
+
2
2
=
3
2
𝑥2
2
+
𝑥
3
2
1
3
2
𝑥2
2
+
2 ∗ 𝑥
3
2
3 ∗ 1
𝒙𝟐
𝟐
+
𝟐𝒙
𝟑
𝟐
𝟑
+ 𝑪

62. 𝑥5
𝑥2
=
𝑥 ∗ 𝑥 ∗ 𝑥 ∗ 𝑥 ∗ 𝑥
𝑥 ∗ 𝑥
=
𝑥 ∗ 𝑥
𝑥 ∗ 𝑥
∗ 𝑥 ∗ 𝑥 ∗ 𝑥 = 1 ∗ 𝑥3
= 𝑥3
𝑥5−2
= 𝑥3

64. 𝑓′
𝑥 = 3𝑥2
− 3

65. 𝒙𝟐
𝒙
𝑥2
𝒙
𝒙 =
𝟐
𝒙𝟏 = 𝒙
𝟏
𝟐
𝑥2
𝑥
1
2
𝒙𝟐
∗ 𝒙−
𝟏
𝟐
𝒙𝟐
∗ 𝒙−
𝟏
𝟐 = 𝒙
𝟐+ −
𝟏
𝟐 = 𝒙𝟐−
𝟏
𝟐 = 𝒙
𝟑
𝟐 =
𝟐
𝒙𝟑

66. 𝟐
𝒙𝟑
Dominio
(x > o = 0)
(XeR+:. X > o = 0)
(0,+infinito)
Rango
(y > o = 0)
(YeR+:. Y > o = 0)
(0,+infinito)

67. 𝒇 𝒙 = 𝒙
𝟑
𝟐𝒅𝒙
𝒇′ 𝒙 =
𝟑
𝟐
∗ 𝒙
𝟑
𝟐
−𝟏
𝒇′ 𝒙 =
𝟑
𝟐
∗ 𝒙
𝟑
𝟐
−
𝟐
𝟐
𝒇′ 𝒙 =
𝟑
𝟐
𝒙
𝟏
𝟐
𝒇′
𝒙 =
𝟑
𝟐
𝟐
𝒙𝟏
𝒇′
𝒙 =
𝟑
𝟐
𝟐
𝒙𝟏
Dominio
(x > o = 0)
(XeR+:. X > o = 0)
(0,+infinito)

68. 𝑥
3
2𝑑𝑥
𝑥
3
2
+ 1
3
2
+1
𝑥
3
2
+ 1 =
5
2
3
2
+
2
2
=
5
2
𝑥
5
2
5
2
𝑥
5
2
1
5
2
=
2𝑥
5
2
5
=
𝟐
𝟓
𝒙
𝟓
𝟐 + 𝑪
Integral F(X)=
𝟐
𝟓
𝒙
𝟓
𝟐 + 𝑪

69. Curso de Cálculo
Diferencial e
Integral
GRACIAS