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TERCERA EVALUACIÓN
DERIVADAS – SIMPLIFICACIÓN

MAT I
TEMA XV

EXPONENTES NEGATIVOS
f ( x) =

3
(5 x 2 − 1) 4

f ( x ) = 3 ⋅ (5 x 2 − 1) − 4 → f ' ( x ) = −12 ⋅ (5 x 2 − 1) − 510 x =
f (x) =

(5 x 2 − 1) 4
3 ⋅ (5 x 2 − 1) 4 − 3 x ⋅ 4 ⋅ (5 x 2 − 1)310 x

=

(5 x 2 − 1)8
EXPONENTES RACIONALES

(

(5 x 2 − 1)3 15 x 2 − 3 − 120 x 2
(5 x 2 − 1)8

) = − 3 − 105 x 2
(5 x 2 −1)5

3
4

5x2 −1

(

−1
4

)

2

f ( x) = 3 ⋅ 5 x − 1

f ( x) =

(5 x 2 − 1)5

3x

→ f ' ( x) =

f ( x) =

− 120 x

−5

(

)

−3
→ f ' ( x) =
3 ⋅ 5 x 2 − 1 4 10x =
4

− 30 x

(

)

2

5
4

4 ⋅ 5x −1

− 30 x

=
4⋅

4

(5 x 2 − 1)5

3x
4

5x2 −1
4

→ f ' ( x) =

3 ⋅ 5 x 2 − 1 − 3x

(

−3
4 10 x

)

1
5x2 −1
4

4=

(5x − 1)
12 ⋅ (5 x 2 − 1)− 30 x 2
30 x 2 − 12
=
=
5
5
4
4
4 ⋅ (5 x 2 − 1)
4 ⋅ (5 x 2 − 1)
4

2

2

30 x 2

3 ⋅ 5x2 −1 −

(5 x 2 − 1)3 =
4
(5 x 2 − 1)2
4⋅

4EJERCICIOS para PRACTICAR

1 f ( x) =
2 f ( x) =

2

(7 x 2 − 1)

3

2x +1

(2 x − 1)

3

→ sol :
→ sol :

− 84 x

(7 x 2 − 1)

4

3 f ( x) =

− 8x − 8

(7 x − 1)
2

4

MATI – 3 EV – Derivadas - Simplificación 1 de 2

4 f ( x) =

2
3

7x2 −1
2x + 1

3

2x − 1

− 28 x

→ sol :

→ sol :

3⋅

3

(7 x 2 − 1)4

8x − 8
3 ⋅ 3 (2 x − 1)4

TERCERAEVALUACIÓN
DERIVADAS – SIMPLIFICACIÓN

MAT I

FUNCIONES ARCO - SENO
x−2
f ( x ) = Arcsen
x+2
1
1 ⋅ ( x + 2) − 1 ⋅ ( x − 2) x + 2
4
4
=
⋅
=
⋅
→ f ' ( x) =
2
2
2
8 x (x + 2 )
( x +2) 8 x
(x + 2 )
 x−2
1− 

 x+ 2

2
2
 1 −  x − 2  = 1 − (x − 2 ) =



 x + 2
(x + 2 )2

f ( x ) = Arcsen

(x + 2 )2 − (x − 2 )2
(x + 2 )2

x2 + 4x + 4 − x 2 + 4x −4

=

(x + 2 )2

x−2
x+2
−1

1  x − 2  2 1 ⋅ ( x + 2 ) − 1 ⋅ ( x − 2)
⋅
=
→ f ' ( x) =

(x + 2 )2
 x−2 2 x+2
1− 

 x+ 2
1


x+2 x−2
4
 x−2
−
=
=
=
 1−...