Ejercicios Preparativos para el Examen
1. Calcular las derivadas parciales y las segundas derivadas de las siguientes funciones
(mostrar el desarrollo paso a paso):
3 2
I. 𝑓(𝑥, 𝑦) = 𝑥 − 2𝑥 + 𝑦
Solución:
∂
∂𝑥
𝑓(𝑥, 𝑦) =
∂
∂𝑥 (𝑥3 − 2𝑥2 + 𝑦)
=
∂
∂𝑥 (𝑥3) + ∂𝑥∂ (− 2𝑥2) + ∂
∂𝑥
(𝑦)
=
∂
∂𝑥 (𝑥3) − 2 ∂𝑥∂ (𝑥2) + ∂
∂𝑥
(𝑦)
2
= 3𝑥 − 4𝑥 + 0
2
= 3𝑥 − 4𝑥.
2
∂
∂𝑥
2 𝑓(𝑥, 𝑦) =
∂
∂𝑥 ( ∂
∂𝑥
𝑓(𝑥, 𝑦) )
=
∂
∂𝑥 (3𝑥2 − 4𝑥)
=
∂
∂𝑥 (3𝑥2) + ∂
∂𝑥
(− 4𝑥)
= 3
∂
∂𝑥 (𝑥2) − 4 ∂𝑥∂ (𝑥)
= 6𝑥 − 4.
∂
∂𝑦
𝑓(𝑥, 𝑦) =
∂
∂𝑦 (𝑥3 − 2𝑥2 + 𝑦)
=
∂
∂𝑦 (𝑥3) + ∂𝑦∂ (− 2𝑥2) + ∂
∂𝑦
(𝑦)
, =
∂
∂𝑦 (𝑥3) − 2 ∂𝑦∂ (𝑥2) + ∂
∂𝑦
(𝑦)
= 0+ 0+ 1
= 1.
2
∂𝑦
∂
2 𝑓(𝑥, 𝑦) =
∂
∂𝑦 ( ∂
∂𝑦
𝑓(𝑥, 𝑦) )
∂
= ∂𝑦
(1)
= 0.
3
𝑥
II. 𝑔(𝑥, 𝑦) = 2𝑥𝑦
+ 𝑙𝑛(𝑥𝑦)
Solución:
( )
3
∂ ∂ 𝑥
∂𝑥
𝑔(𝑥, 𝑦) = ∂𝑥 2𝑥𝑦
+ 𝑙𝑛(𝑥𝑦)
( )+
3
∂ 𝑥 ∂
= ∂𝑥 2𝑥𝑦 ∂𝑥
(𝑙𝑛(𝑥𝑦))
( )+
2
∂ 𝑥 ∂
= ∂𝑥 2𝑦 ∂𝑥
(𝑙𝑛(𝑥𝑦))
=
1 ∂
2𝑦 ∂𝑥 (𝑥2) + ∂
∂𝑥
(𝑙𝑛(𝑥) + 𝑙𝑛(𝑦))
=
1 ∂
2𝑦 ∂𝑥 (𝑥2) + ∂
∂𝑥
(𝑙𝑛(𝑥)) +
∂
∂𝑥
(𝑙𝑛(𝑦))
1 1
= 2𝑦
(2𝑥) + 𝑥
+ 0
𝑥 1
= 𝑦
+ 𝑥
.
2
∂𝑥
∂
2 𝑔(𝑥, 𝑦) =
∂
∂𝑥 ( ∂
∂𝑥
𝑔(𝑥, 𝑦) )
, =
∂
∂𝑥 ( 𝑥
𝑦
+
1
𝑥 )
=
∂
∂𝑥 ( )+ ( )
𝑥
𝑦
∂
∂𝑥
1
𝑥
=
1 ∂
𝑦 ∂𝑥
(𝑥) +
∂
∂𝑥 ( ) 1
𝑥
1 1
= 𝑦
− 2 .
𝑥
( )
3
∂ ∂ 𝑥
∂𝑦
𝑔(𝑥, 𝑦) = ∂𝑦 2𝑥𝑦
+ 𝑙𝑛(𝑥𝑦)
( )+
3
∂ 𝑥 ∂
= ∂𝑦 2𝑥𝑦 ∂𝑦
(𝑙𝑛(𝑥𝑦))
( )+
2
∂ 𝑥 ∂
= ∂𝑦 2𝑦 ∂𝑦
(𝑙𝑛(𝑥𝑦))
2
=
𝑥
2
∂
∂𝑦 ( )+ 1
𝑦
∂
∂𝑦
(𝑙𝑛(𝑥) + 𝑙𝑛(𝑦))
2
=
𝑥
2
∂
∂𝑦 ( )+ 1
𝑦
∂
∂𝑦
(𝑙𝑛(𝑥)) +
∂
∂𝑦
(𝑙𝑛(𝑦))
(− ) + 0 +
2
𝑥 1 1
= 2 2 𝑦
𝑦
2
𝑥 1
=− 2 + 𝑦
.
2𝑦
2
∂
∂𝑦
2 𝑔(𝑥, 𝑦) =
∂
∂𝑦 ( ∂
∂𝑦
𝑔(𝑥, 𝑦) )
( )
2
∂ 𝑥 1
= ∂𝑦
− 2 + 𝑦
2𝑦
( )+
2
=
∂
∂𝑦
−
𝑥
2𝑦
2
∂
∂𝑦 ( ) 1
𝑦
( )+
2
=−
𝑥
2
∂
∂𝑦 𝑦
1
2
∂
∂𝑦 ( ) 1
𝑦
1. Calcular las derivadas parciales y las segundas derivadas de las siguientes funciones
(mostrar el desarrollo paso a paso):
3 2
I. 𝑓(𝑥, 𝑦) = 𝑥 − 2𝑥 + 𝑦
Solución:
∂
∂𝑥
𝑓(𝑥, 𝑦) =
∂
∂𝑥 (𝑥3 − 2𝑥2 + 𝑦)
=
∂
∂𝑥 (𝑥3) + ∂𝑥∂ (− 2𝑥2) + ∂
∂𝑥
(𝑦)
=
∂
∂𝑥 (𝑥3) − 2 ∂𝑥∂ (𝑥2) + ∂
∂𝑥
(𝑦)
2
= 3𝑥 − 4𝑥 + 0
2
= 3𝑥 − 4𝑥.
2
∂
∂𝑥
2 𝑓(𝑥, 𝑦) =
∂
∂𝑥 ( ∂
∂𝑥
𝑓(𝑥, 𝑦) )
=
∂
∂𝑥 (3𝑥2 − 4𝑥)
=
∂
∂𝑥 (3𝑥2) + ∂
∂𝑥
(− 4𝑥)
= 3
∂
∂𝑥 (𝑥2) − 4 ∂𝑥∂ (𝑥)
= 6𝑥 − 4.
∂
∂𝑦
𝑓(𝑥, 𝑦) =
∂
∂𝑦 (𝑥3 − 2𝑥2 + 𝑦)
=
∂
∂𝑦 (𝑥3) + ∂𝑦∂ (− 2𝑥2) + ∂
∂𝑦
(𝑦)
, =
∂
∂𝑦 (𝑥3) − 2 ∂𝑦∂ (𝑥2) + ∂
∂𝑦
(𝑦)
= 0+ 0+ 1
= 1.
2
∂𝑦
∂
2 𝑓(𝑥, 𝑦) =
∂
∂𝑦 ( ∂
∂𝑦
𝑓(𝑥, 𝑦) )
∂
= ∂𝑦
(1)
= 0.
3
𝑥
II. 𝑔(𝑥, 𝑦) = 2𝑥𝑦
+ 𝑙𝑛(𝑥𝑦)
Solución:
( )
3
∂ ∂ 𝑥
∂𝑥
𝑔(𝑥, 𝑦) = ∂𝑥 2𝑥𝑦
+ 𝑙𝑛(𝑥𝑦)
( )+
3
∂ 𝑥 ∂
= ∂𝑥 2𝑥𝑦 ∂𝑥
(𝑙𝑛(𝑥𝑦))
( )+
2
∂ 𝑥 ∂
= ∂𝑥 2𝑦 ∂𝑥
(𝑙𝑛(𝑥𝑦))
=
1 ∂
2𝑦 ∂𝑥 (𝑥2) + ∂
∂𝑥
(𝑙𝑛(𝑥) + 𝑙𝑛(𝑦))
=
1 ∂
2𝑦 ∂𝑥 (𝑥2) + ∂
∂𝑥
(𝑙𝑛(𝑥)) +
∂
∂𝑥
(𝑙𝑛(𝑦))
1 1
= 2𝑦
(2𝑥) + 𝑥
+ 0
𝑥 1
= 𝑦
+ 𝑥
.
2
∂𝑥
∂
2 𝑔(𝑥, 𝑦) =
∂
∂𝑥 ( ∂
∂𝑥
𝑔(𝑥, 𝑦) )
, =
∂
∂𝑥 ( 𝑥
𝑦
+
1
𝑥 )
=
∂
∂𝑥 ( )+ ( )
𝑥
𝑦
∂
∂𝑥
1
𝑥
=
1 ∂
𝑦 ∂𝑥
(𝑥) +
∂
∂𝑥 ( ) 1
𝑥
1 1
= 𝑦
− 2 .
𝑥
( )
3
∂ ∂ 𝑥
∂𝑦
𝑔(𝑥, 𝑦) = ∂𝑦 2𝑥𝑦
+ 𝑙𝑛(𝑥𝑦)
( )+
3
∂ 𝑥 ∂
= ∂𝑦 2𝑥𝑦 ∂𝑦
(𝑙𝑛(𝑥𝑦))
( )+
2
∂ 𝑥 ∂
= ∂𝑦 2𝑦 ∂𝑦
(𝑙𝑛(𝑥𝑦))
2
=
𝑥
2
∂
∂𝑦 ( )+ 1
𝑦
∂
∂𝑦
(𝑙𝑛(𝑥) + 𝑙𝑛(𝑦))
2
=
𝑥
2
∂
∂𝑦 ( )+ 1
𝑦
∂
∂𝑦
(𝑙𝑛(𝑥)) +
∂
∂𝑦
(𝑙𝑛(𝑦))
(− ) + 0 +
2
𝑥 1 1
= 2 2 𝑦
𝑦
2
𝑥 1
=− 2 + 𝑦
.
2𝑦
2
∂
∂𝑦
2 𝑔(𝑥, 𝑦) =
∂
∂𝑦 ( ∂
∂𝑦
𝑔(𝑥, 𝑦) )
( )
2
∂ 𝑥 1
= ∂𝑦
− 2 + 𝑦
2𝑦
( )+
2
=
∂
∂𝑦
−
𝑥
2𝑦
2
∂
∂𝑦 ( ) 1
𝑦
( )+
2
=−
𝑥
2
∂
∂𝑦 𝑦
1
2
∂
∂𝑦 ( ) 1
𝑦
