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DERIVADAS PARCIALESDERIVADAS PARCIALES
FUNCIN REAL DE VARIAS FUNCIN REAL DE VARIAS
VARIABLESVARIABLES
CAPTULO IICLCULO VECTORIAL
SESIN 13
DERIVADA PARA UNA FUNCION DE DOS VARIABLES z = f (x, y)
Rosa ique Alvarez 2
xyxfyxxfyxfD
x D-D+
=D
),(),(lim),(0
1
yyxfyyxfyxfD
y D-D+=
D
),(),(lim),(02
Si es que lmite existe
Si es que lmite existe
NOTACION z = f (x, y)
Rosa ique Alvarez 3
xffxfyxfD ==
= 11 ),(
yffyfyxfD ==
= 22 ),(
DERIVADA PARCIAL EN UN PUNTO (x0, y0)
Rosa ique Alvarez 4
xyxfyxxf
yxfDx D
-D+=
D
),(),(lim),( 0000
0001
yyxfyyxf
yxfDy D
-D+=
D
),(),(lim),( 0000
0002
Si es que lmite existe
Si es que lmite existe
Interpretacin Geomtrica
Rosa ique Alvarez 5
Plano: y=y0
Superficie:z= f (x, y)
P=(x0,y0,f(x0, y0 ))
( ) ( )00001 ,, yxfyxfD x=Interpretacin Geomtrica
Rosa ique Alvarez 6
Plano: x=x0
Superficie:z= f (x, y)
P=(x0,y0,f(x0, y0 ))
( ) ( )00002 ,, yxfyxfD y=
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EJEMPLO 1
Rosa ique Alvarez 7
=
+
-=
)0,0(),(0
)0,0(),()(),( 22
22
yx
yxyx
yxyxyxf
18
Calcular: D1 f (0, 0) y D2 f (0, 0).
Solucin
Rosa ique Alvarez 8
0)0,0(
00lim00
lim)0,0(
)0,0()0,(lim)0,0()0,0(lim)0,0(
1
001
001
=
D-=
D
-D=
D-D=
D-D+=
DD
DD
fD
xxxfD
xfxf
xfxffD
xx
xx
18
=
+
-=
)0,0(),(0
)0,0(),()(),( 22
22
yx
yxyx
yxyxyxf
=
+
-=
)0,0(),(0
)0,0(),()(),( 22
22
yx
yxyx
yxyxyxf
0)0,0(
00lim00
lim)0,0(
)0,0(),0(lim)0,0()0,0(lim)0,0(
2
002
002
=
D-
=D
-D=
D-D
=D
-D+=
DD
DD
fD
yyyfD
yfyf
yfyffD
yy
yy
Solucin
Rosa ique Alvarez 9
19
Conclusiones
Rosa ique Alvarez 10
=
+
-=
)0,0(),(0
)0,0(),()(),( 22
22
yx
yxyx
yxyxyxf
0)0,0(
0)0,0(
2
1
=
=
fD
fD
EJEMPLO 2
( )
=
-
--=
yx
yxyx
yyxxyxf
,0
,32),( 2
323
Rosa ique Alvarez 11
Calcular: D1 f (0, 0) y D2 f (0, 0).
Rosa ique Alvarez 12
y=x
Solucin ( )
=
-
--=
yx
yxyx
yyxxyxf
,0
,32),( 2
323
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Solucin
Rosa ique Alvarez 13
202
lim)0,0(
)0,0()0,(lim)0,0()0,0(lim)0,0(
2
3
01
001
=D
-DD
=
D-D
=D
-D+=
D
DD
xxx
fD
xfxf
xfxffD
x
xx
18
( )
=
-
--=
yx
yxyx
yyxxyxf
,0
,32),( 2
323
10
lim)0,0(
)0,0(),0(lim)0,0()0,0(lim)0,0(
2
3
02
002
-=D
-DD-
=
D-D
=D
-D+=
D
DD
yyy
fD
yfyf
yfyffD
y
yy
Solucin
Rosa ique Alvarez 14
19
( )
=
-
--=
yx
yxyx
yyxxyxf
,0
,32),( 2
323
1)0,0(2 -=fD
Conclusiones
Rosa ique Alvarez 15
19
( )
=
-
--=
yx
yxyx
yyxxyxf
,0
,32),( 2
323
2)0,0(1 =fD
EJEMPLO 3
Rosa ique Alvarez 16
a) Calcular: D1 f (0, y) si y 0 ; D1 f (0, 0 )
b) Calcular: D2 f (x,0) si x 0 ; D2 f (0, 0 ).
14
( ) ( )
=
+=
0,0,,0
)0,0(),(,2),( 22
yx
yxyx
yxyxf
Rosa ique Alvarez 17
(0, y)0);,0(1 yyfD
)0,0(1 fD
Solucin Solucin
Rosa ique Alvarez 18
0,2),0(
02
lim),0(
0,),0(),0(lim),0(
1
22
01
01
=
D
-+D
D
=
D
-D+=
D
D
yy
yfD
xyxyx
yfD
yx
yfyxfyfD
x
x
( ) ( )
=
+=
0,0,,0
)0,0(),(,2),( 22
yx
yxyx
yxyxf
0)0,0(1 =fD
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Rosa ique Alvarez 19
(x, 0)
0);0,(2 xxfD)0,0(2 fD
Solucin Solucin
Rosa ique Alvarez 20
0,2)0,(
02
lim)0,(
0,)0,()0,(lim)0,(
2
22
02
02
=
D
-D+D
=
D
-D+=
D
D
xx
xfD
yyxyx
xfD
xy
xfyxfxfD
y
y
( ) ( )
=
+=
0,0,,0
)0,0(),(,2),( 22
yx
yxyxyx
yxf
0)0,0(2 =fD
Conclusiones
Rosa ique Alvarez 21
( ) ( )
=
+=
0,0,,0
)0,0(),(,2),( 22
yx
yxyx
yxyxf
0,2),0(1 = yyyfD 0)0,0(1 =fD
0,2)0,(2 = xxxfD 0)0,0(2 =fD
EJEMPLO 4
Rosa ique Alvarez 22
=
+=
)0,0(),(0
)0,0(),(),( 22
22
yx
yxyx
yxyxf
Determine las funciones D1f (x, y) y D2 f (x, y).
GRAFICA
Rosa ique Alvarez 23superficie
=
+=
)0,0(),(,0
)0,0(),(,),( 22
22
yx
yxyx
yxyxf
Solucin:
Rosa ique Alvarez 24
=
+=
)0,0(),(0
)0,0(),(),( 22
22
yx
yxyx
yxyxf
Para (x, y) (0,0)
( ) ( ) ( ) ( )( )222
22222222
22
22
yx
yxx
yxyxx
yx
yxyx
x +
+-
+
=
+
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Solucin:
Rosa ique Alvarez 25
=
+=
)0,0(),(0
)0,0(),(),( 22
22
yx
yxyx
yxyxf
Para (x, y) (0,0)
( )2224
22
22 2
yxxy
yxyx
x +=
+
Solucin:
Rosa ique Alvarez 26
=
+=
)0,0(),(0
)0,0(),(),( 22
22
yx
yxyx
yxyxf
Para (x, y) = (0,0)
0)0,0(
)0,0()0,0(lim)0,0(
1
01
=
D-D+
=D
fD
xfxffD
x
Solucin:
Rosa ique Alvarez 27
( )
=
+=
)0,0(),(0
)0,0(),(2),( 222
4
1
yx
yxyxyx
yxfD
=
+=
)0,0(),(0
)0,0(),(),( 22
22
yx
yxyx
yxyxf
FUNCIN
( )
=
+=
)0,0(),(0
)0,0(),(2),(: 222
4
1
yx
yxyxyx
yxfDS
Rosa ique Alvarez 28
Solucin:
Rosa ique Alvarez 29
=
+=
)0,0(),(0
)0,0(),(),( 22
22
yx
yxyx
yxyxf
Para (x, y) (0,0)
( ) ( ) ( ) ( )( )222
22222222
22
22
yx
yxy
yxyxy
yx
yxyx
y +
+-
+
=
+
Solucin:
Rosa ique Alvarez 30
=
+=
)0,0(),(0
)0,0(),(),( 22
22
yx
yxyx
yxyxf
Para (x, y) (0,0)
( )2224
22
22 2yxyx
yxyx
y +=
+
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Solucin:
Rosa ique Alvarez 31
=
+=
)0,0(),(0
)0,0(),(),( 22
22
yx
yxyx
yxyxf
Para (x, y) = (0,0)
0)0,0(
)0,0()0,0(lim)0,0(
2
02
=
D-D+
=D
fD
yfyffD
y
Solucin:
Rosa ique Alvarez 32
( )
=
+=
)0,0(),(0
)0,0(),(2),( 222
4
2
yx
yxyxyx
yxfD
=
+=
)0,0(),(0
)0,0(),(),( 22
22
yx
yxyx
yxyxf
FUNCIN
( )
=
+=
)0,0(),(0
)0,0(),(2),(: 222
4
2
yx
yxyxyx
yxfDS
Rosa ique Alvarez 33
DERIVADA PARCIAL DE UNA FUNCION DE nVARIABLES
k
nknkk
x
nkk
xxxxfxxxxf
xxxfD
k D-D+
=D
),,(),,,(lim
),,,(
11
0
1
KKKKKK
Rosa ique Alvarez 34
( )nk xxxxfw ,.....,,....,, 21=
DERIVADAS PARCIALES DE ORDEN SUPERIOR PARA f (x, y)
Rosa ique Alvarez 35
xyf
xf
yfffD yx
=
===2
1212
Derivada de orden dos
2
2
1111 xf
xf
xfffD xx
=
===
DERIVADAS PARCIALES DE ORDEN SUPERIOR PARA f (x, y)
Rosa ique Alvarez 36
2
3
2
2
112112 xyf
xf
yfffD xxy
=
===
Derivadas de orden tres
3
3
2
2
222222 yf
yf
yfffD yyy
=
===
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DEFINICION DE DERIVADA PARCIAL DE ORDEN SUPERIOR
Rosa ique Alvarez 37
yyxfDyyxfDyxfD
yxfDyx
fy
yxfD
y
funcin
D-D+=
=
=
D
),(),(lim),(
),(),(
11012
112
),( yxfz =
SI EL LMITE EXISTE
DEFINICION DE DERIVADA PARCIAL DE ORDEN SUPERIOR
Rosa ique Alvarez 38
xyxfDyxxfDyxfD
yxfDxy
fx
yxfD
x
funcin
D-D+=
=
=
D
),(),(lim),(
),(),(
22021
221
),( yxfz =
SI EL LMITE EXISTE
EJEMPLO 5:
=
+
-=
)0,0(),(;0
)0,0(),(;)(),( 22
22
yx
yxyx
yxyxyxf
Rosa ique Alvarez 39
Calcular las derivadas de orden dos: D12 f (0, 0) y D21 f (0,
0).
Solucin: calculo de D1 f (x, y)
Rosa ique Alvarez 40
( )2224224
22
22 )4()(
)0,0(),(Para
yxyyxxy
yxyxyx
x
yx
+
-+=
+
-
0)0,0()0,0(lim)0,0(
)0,0(),(Para
01=
D-D+
=
=
D xfxffD
yx
x
=
+
-=
)0,0(),(;0
)0,0(),(;)(),( 22
22
yx
yxyx
yxyxyxf
Solucin
Rosa ique Alvarez 41
=
+
-=
)0,0(),(;0
)0,0(),(;)(),( 22
22
yx
yxyx
yxyxyxf
( )
=
+
-+=
)0,0(),(;0
)0,0(),(;)4(),( 222
4224
1
yx
yxyx
yyxxyyxfD
( )
=
+
-+=
)0,0(),(;0
)0,0(),(;)4(),( 222
4224
1
yx
yxyx
yyxxyyxfD
Rosa ique Alvarez 42
yfDyfDfD
y D-D+
=D
)0,0()0,0(lim)0,0( 11021
Solucin: calculo de D12 f (0, 0)
10
lim)0,0(4
5
021-=
D
-DD-
=D y
yy
fDy
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( )
=
+
--
=
)0,0(),(;0
)0,0(),(;)4(
),( 222
4224
2
yx
yxyx
yyxxxyxfD
Rosa ique Alvarez 43
1)0,0(12 =fD
Solucin: calculo de D21 f (0, 0) Conclusiones
Rosa ique Alvarez 44
=
+
-=
)0,0(),(;0
)0,0(),(;)(),( 22
22
yx
yxyx
yxyxyxf
1)0,0(21 -=fD 1)0,0(12 =fD
0)0,0(1 =fD 0)0,0(2 =fD
EJEMPLO 6:
( ) ( )
=
+=
0,0,,0
)0,0(),(,2),( 22
yx
yxyx
yxyxf
Rosa ique Alvarez 45
10
Calcular D12 f (0,0) y D21 f (0,0) si es que existen
Solucin
( )( ) ( )
=
+
-=
0,0,,0
)0,0(),(,22
),( 22223
1
yx
yxyx
yxyyxfD
Rosa ique Alvarez 46
( )existeno2lim)0,0()0,0(lim)0,0( 20
110
12yy
fDyfDfDyy D
=D
-D+=
DD
Solucin
( )( ) ( )
=
+
-=
0,0,,0
)0,0(),(,22
),( 222
23
2
yx
yxyxxyx
yxfD
Rosa ique Alvarez 47
( )existeno2lim
)0,0()0,0(lim)0,0(
2022
021
xxfDxfD
fDxx D
=D
-D+=
DD
Conclusiones
Rosa ique Alvarez 48
( ) ( )
=
+=
0,0,,0
)0,0(),(,2),( 22
yx
yxyx
yxyxf
existenno)0,0(y)0,0(
0)0,0(,0)0,0(
2112
21
fDfD
fDfD ==
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TEOREMA 1
Rosa ique Alvarez 49
),(),( 0000 yxfyxf xyyx =
Suponga que f es una funcin en lasvariables x e y, que est
definida en el discoabierto B((x0, y0), ) y que fx , fy, fx y y f y
xestn definidas en B. Adems, suponga quefx y y fy x son continuas
en B. Entonces
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