-2u
d(F,G) = df
df = xeu+v+u xeu+v
du
dg
du
dv
dg
dv
-yeu-v-2v -yeu-v
-2u
du = eu+v
xeu+v
+v
=
y e 2u
+ 2 u e u+v
dx
0
-ye
u-v
– 2u
2u u+v
2xye + 2(u-v)xe + (u+v)ye
u-v
xeu+v+u
xeu+v +v
yeu-v-2v
-yeu-v
-2u
2- f(x,y,u,v)= u+ e u+v
g(x,y,u,v)= v+eu-v
u= Ø(x,y)
dz (1,1)
dx
v= Ø(x,y)
dz (1,1)
dy
dz =
du + dv ,
dz = du + dv
dx
dx
dx
dy dy
dy
du + e u+v du + dv – 1 = 0
dv + eu-v
du – dv = 0
dx
dx
dx
dx
dx
dx
du + e u+v du + dv = 0
dv + eu-v
du – dv -1 = 0
dy
dy
dy
dy
dy
dy
dz = 0+1 = 1
dx
dz = 1+(-2)= -1
dy
3- f(x,y,u,v)= u-v+2x-2y=0
g(x,y,u,v)= 3u3+v3-5×2+y3=0
d(F,G) =
d(u,v)
1
9u2
-1
3v2
= 3v2+9u2
du = 2v2-y2
dy
v2+3u2
dv =
1
1
-2
= -y2+6u2
dy 3v2+ 9u2
9u2
3y2
v2 + 3u2
d2u = – 2v (y2+6u2) + 6u (2v2-y2)2- 2y (v2 + 3u2)
dy
2
2 2 3
(v + 3u )
4- f(x,y,u,v,w)= x+y+u+v+w= 0
g(x,y,u,v,w)= x2-y2+u2-2v2+w2+1= 0
h(x,y,u,v,w)= x3+y3+u4-3v4+8w4+2= 0
En P(1,-1,1,-1,0)
d(F,G,H) =
d(u,y,w)
d(F,G,H) =
d(u,y,w)
d(F,G,H) =
d(u,y,w)
1
2u
4u3
1
2u
4u3
1
2
4
1
-4y
-12v3
1
-2y
3y2
1
2
3
1
2w
32w2
1
2w
32w3
1
0
0
= -2
=8
dv (1,-1)= 1
dy
4
5- f(x,y,z)= 0
z= f(x,y)
df dx + df dy = 0
dx
dy
df dy = – df dx
dy
dx
dy =
dx
– df
dx
df
dy
dy = – df
dx
df
dy
DERIVADA DIRECCIONAL
1- f(x,y,z) = 2×3+7y2+9z2
v = (a,b,c)
x = xo+ as
y = yo+bs
z = zo+cs
gs = 2 (xo+ as)3+7(yo+bs)2+9(zo+cs)2
= 2 (xo3+3x2as+3xoa2s2+a3s3)+7(yo2+2 yobs+b2s2)+9(zo2+2 zocs+c2s2)
= 2 xo3+6 x2as+6 xoa2s2+2 a3s3+7 yo2+14yobs+7 b2s2+9 zo2+18 zocs+9 c2s2
g’(s) = 6ax2+12a2xs+6a3s2+14byo+14b2s+18czo+18c2s
= 6ax2+14byo+18czo
2- f(x,y) = 3x-2y
v = (1/ 2 , 1/ 2 )
x = xo+ 1/
y = yo+ 1/
2 s
2 s
gs = 3(xo+ 1/ 2 s)-2(yo+ 1/ 2 s)
= 3xo+3/ 2 s- 2 yo-2/ 2 s)
g’(s) = 3/ 2 – 2/ 2
= 1/ 2
3- f(x,y) = x2+y2
v = (a,b)
p= (0,0)
x = 0+as
y = 0+bs
gs = (0+as)2+( 0+bs)2
= 02+2 0as+a2s2+ 02+2 0bs+b2s2
g’(s) = 2 0a+2 a2s+2 0b+2 b2s
=0
2
df
(
4- f(x,y) = x y2+ x2y
v=(1,0)
x = xo+ s
y = yo+0
gs = xo+ s (y o +0)2+( x o + s)2 y o +0
g’(s) = (y o +0) + y o +0 .2(x o + s)
2
= y o +2 x o y o +2s y o
= y o2 +2 x o y o
5- f(x,y,z) = xyz
v= (1/3, -2/3,- 2/3)
x = xo+ 1/3s
y = yo-2/3s
z = zo-2/3s
gs = (xo+ 1/3s)( y o -2/3s)( z o -2/3s)
= xo yo – x o 2/3s+1/3s y o -2/9s2(zo-2/3s)
= zo xo y o – z o x o 2/3s+ zo1/3s yo- zo2/9s2 -2/3s xo yo+4/9s2-2/9 yos3+4/27 s3
g’(s) = – z o x o 2/3+ zo1/3 yo-4/9s zo-8/9s-6/9syo+12/27s2
= -2 z o x o + yo zo- 2 xo yo
3
DERIVADA PARCIAL
1- f(x,y) = x2y3
x2(0)+ y32x
= 2xy3
dx
x2 3 y2+ y3(0)
df
dy
3 x2y2
2 – f(x,y) = Sen (
2 x 3
y 2 )
Cos (
2 x 3
y 2 ) . 1/2 (
2 x 3
y 2 )-1/2 . 6×2
df
dx
= 3x2cos(
2 x 3
2 x 3
y 2 )
y 2 )
Cos (
2 x 3
y 2 ).1/2 (
2 x 3
y 2 )-1/2 . 2y
df
y Cos (
2 x 3
y 2 )
dy
(
2 x 3
y 2 )
3- f(x,y) = xy+yx
df
dx
df
dy
= yxy-1 + yxlny
xy lnx + xy x-1
4- f(x,y) = (2y) x + 2 y
si
yxy-1 + yx lny
df
dx
= (2y)x ln 2y
si
xylnx + xyx-1
df
dy
x(2y)x-1 + 2y ln 2
5- f(x,y) = x ln y – y ln x
x(0)+lny(1)
-y(1/x)+lnx(0)
df
dx
= lny – y
x
x(1/y)+(-y)(0)+lnx(-1)
df
dy
x – lnx
y
DERIVADA DIRECCIONAL GRADIENTE
1- f(x, y, z)= x2y3z4
df = 2xy3z4
dx
En el punto (1,1,1) son:
df (1,1,1)= 2
dx
df = 3x2y2z4
dy
df (1,1,1)= 3
dy
df = 4x2y3z3
dz
df (1,1,1)= 4
dz
grad f (1,1,1)= (2,3,4)
2- f(x,y)= 3x2y+cos(xy) en p=(1,1)
df = 6xy -sen (xy) y
dx
df = 3×2 –sen (xy) x
dy
df (1,1) =
dx
5.982
df (1,1)= 2.98
dy
4
-1
grad f (1,1)= (5.982 , 2.98 )
3- f(x,y)= xy en p= (2,2)
df = yxy-1
dx
df (2,2) =
dx
df = xy ln x
dy
df (2,2)= 4 ln 2
dy
grad f (1,1)= (4, 4 ln 2 )
4- f(x,y)=
3 x 2
y 2
df =
-x
df =
-x
dx
3 x
2
y
2
3 x 2
y 2
En el punto (1,1)
df =
-1
df =
-1
1
df (1,1) =
dx
grad f (1,1)= (-1, -1 )
5- f(x, y, z)= ln (x, y, z)
En el punto (1,1,1)
1
df (1,1)= -1
dy
df = 1
df =
1
df = 1
dx
x
dy
y
dz z
df = 1
df =
1
df = 1
dx
1
dy
1
dz 1
df = 1
dx
df =
dy
1
df = 1
dz
grad f (1,1,1)= (1, 1, 1 )
PUNTOS CRITICOS DE UNA FUNCION
1- f(x,y)= 3x+ 8y – 2xy + 4
df = 3 – 2y= 0
-2y =-3
y= 3
)
y= 1
dx
2
df = 8 – 2x = 0
dy
pc= ( 4, 3
2
2- f(x,y)= x2+x+ y2+1
df = 2x+1 = 0
dx
df = 2y = 0
dy
pc= (
-1 , 0 )
-2x= -8
2x = -1
2y= 0
x= 4
x= -1
2
y= 0
2
3- f(x,y)= x2 + 2x + y2 – 4y + 10
df = 2x+2 = 0
dx
df = 2y- 4 = 0
dy
pc= (
-1 , -2 )
2x = -2
2y= -4
x= -1
y= -2
4- f(x,y)= 2×3 + 3×2 + 6x +y3 + 3y + 12
df = 6×2+6x +6 = 0
6×2= -6x – 6
x2 = -x -1
raiz negativa
dx
df = 3y2+3y = 0
dy
3y( y + 1) = 0
y = -1
y= 0
No hay puntos criticos
5- f(x,y)= x2y – x2 – 3xy + 3x + 2y -2
df = 2xy – 2x – 3y +3= 0
dx
2x (y -1 ) -3 (y -1) = 0
x= 1
df =
dy
pc= (
x2 – 3x + 2 = 0
1, 1 )
x(x-3) +2 = 0
DERIVADA DIRECCIONAL GRADIENTE
1- f(x, y, z)= x2y3z4
4
dx
x2+y2
x2+y2
df = 2xy3z4
dx
En el punto (1,1,1) son:
df (1,1,1)= 2
dx
df = 3x2y2z4
dy
df (1,1,1)= 3
dy
df = 4x2y3z3
dz
df (1,1,1)= 4
dz
grad f (1,1,1)= (2,3,4)
2- f(x,y)= 3x2y+cos(xy) en p=(1,1)
df = 6xy -sen (xy) y
dx
df = 3×2 –sen (xy) x
dy
df (1,1) =
dx
5.982
df (1,1)= 2.98
dy
grad f (1,1)= (5.982 , 2.98 )
3- f(x,y)= xy en p= (2,2)
df = yxy-1
dx
df (2,2) =
dx
df = xy ln x
dy
df (2,2)= 4 ln 2
dy
grad f (1,1)= (4, 4 ln 2 )
DERIVADAS PARCIALES DE ORDEN SUPERIOR
(Teorema de Schwarz)
1- f(x,y) = x2+y2
si df = 2x , df = 2y
dx
d2f = d ( df ) = d (2x) = 2
dx2 dx dx
d2f = d ( df ) = d (2x) = 0
dy
d2f = d ( df ) = d (2y) = 0
dxdy dx dy dx
d2f = d ( df ) = d (2y) = 2
dydx dy dx
dy
dy2 dy dy
dy
2- f(x,y) = x2e x2+y2
si df = x2e x2+y2 (2x) + 2xe x2+y2 = 2xe
dx
df = x2e x2+y2 (2y) = 2x2ye x2+y2
dy
(x3+x)
d2f = d ( df ) = d 2x (x3+x) = 2xe
dx2 dx dx
dx
(3×2+1)+4 x2e
x2+y2
(x3+x)=
2 e x2+y2(2×4+5×2+)
x2+y2
x2+y2
x2+y2
d2f = d ( df ) = d (2x e x2+y2(x3+x)= 4y e
(x3+x)
dydx dy dx
dy
d2f = d ( df ) = d (2×2 ye x2+y2) = 2×2 ye
dxdy dx dy dx
(2x)+ 4x ye x2+y2= 4y e
(x3+x)
2 2
d f = d ( df ) = d (2x ye
x2+y2 2
) = 2x ye
x2+y2 2
(2y)+ 2x e
x2+y2
2
= 2x e
x2+y2 2
(2y +1)
dy2 dy dy
dy
3- f(x,y) = x3+6x2y4+7xy5+10x3y
si df = 3×2+12xy4+7y5+30x2y
dx
df = 24y3x2+35xy4+10×3
dy
d2f = d ( df ) = d (3×2+12xy4+7y5+30x2y
) = 6x+12y4+60xy
dx2 dx dx
dx
d2f = d ( df ) = d (24y3x2+35xy4+10×3) = 48xy3+35y4+30x
dxdy dx dy dx
d2f = d ( df ) = d (3×2+12xy4+7y5+30x2y
) = 6x+12y4+60xy
dydx dy dx
dy
d2f = d ( df ) = d (24y3x2+35xy4+10×3) = 72x2y2+140xy3
dy2 dy dy
dy
4- f(x,y) = x+y Verificar que satisfaga lo siguiente: d2f + d2f = 0
2 2
df = y -2xy-x
x2+y2
dx2 dy2
dx
(x2+y2)2
d2f = d ( df ) = 2×3-2y3+6x2y-6xy2
dx2 dx dx
df = x2-2xy-y2
(x2+y2)3
dy
(x2+y2)2
d2f = 2y3-2×3+6y2x-6xy2
dy2
(x2+y2)3
d2f + d2f = 2×3-2y3+6x2y-6xy2 + 2y3-2×3+6y2x-6xy2 = 0
dx2
dy2
(x2+y2)3
(x2+y2)3
5- f(x,y) = xy
df = yxy-1
dy
d2f = xy-2(y2-y)
df = xy lnx
dx
d2f = xyln2x
dy2
d2f = xy-1(ylnx+1)
dx dy
dx2
d2f = xy-1(ylnx+1)
dydx
FUNCIONES DIFERENCIABLES
1- f(x,y) =
xy2
?f= (x+?x) (y+?y)2= x+?x (y2+2y ?y + (?y)2) – xy2=
xy2+2xy?y+x(?y)2+ ?x y2+ ?x2y ?y+ ?x(?y)2- xy2=
2xy?y+x(?y)2+ ?x y2+ ?x2y ?y+ ?x(?y)2
Si es diferenciable
2-f(x,y) = x2+y2
?f= (x+?x)2+(y+ ?y)2 –( x2+y2) =
x2+2x ?x+( ?x)2+ y2+2y ?y + (?y)2)- ( x2+y2)=
2x ?x+( ?x)2+2y ?y + (?y)2
Si es diferenciable
3- f(x,y) = e-(x2+y2)
df = -2xe-(x2+y2)
dx
df = -2y e-(x2+y2)
dy
Si es diferenciable
4- f(x,y,z) = cos (x+y2+z3)
df = -sen (x+y2+z3)
dx
df = -2ysen (x+y2+z3)
dy
df = -3z2 sen (x+y2+z3)
dz
Si es diferenciable
5- f(x,y) = 3x
?f= 3(x+?x)-3x =
3x+3 ?x-3x = 3 ?x
Si es diferenciable
DIVERGENCIA Y ROTACIONAL
Y LAPLACIANO
1. f (xyz) = x i + xy j + k
? xF=
i
j
k
d
dx
x
d
dy
xy
d
dz
1
=(0-0)i- (0-0)j + (y-0)k
? x F = yk
2. f (xyz) = -wy i + wx j
rot =
i
j
k
d
dx
-wy
d
dy
wx
d
dz
0
= 2wk
rot = 2w
3. F = x 2y i + z j + xyz k
div F = d (x 2y) + d (z) + d(xyz) = 2xy + 0 + xy = 3xy
dx
dy
dz
2xy + xy =
4. F = F1i + F2 j + F3 k
? 2 f = ? . (? f) = d2f + d2f + d2f
dx 2
3xy
dy 2 dz 2
? 2 f = ? 2 F1i +? 2 F2 j + ? 2F3 k
5. F = 3 x 2y i + 5xz3j – y2k
div F = d (x 2y) + d (z) + d(xyz) = 5xy + 0 + xy = 6xy
dx
dy
dz
5xy + xy = 6xy
ECUACIONES DEL PLANO OSCULADOR, NORMAL
Y RECTIFICANTE
1- F (s) =
cos s , sen s , s
p= f ( 2 ?) = (0 ,1, ?)
2
2
2
s
s
0
1
,
T (s) = f’’(s) =
-1 sen s
, 1 cos s ,
2
2
2
2
2
f’’ (s) =
-1 cos s ,
-1 sen s
, 0
2
2
2
2
k ( s) = ½
N (s) = 1 f’’ (s) = 1
-1 cos s ,
-1 sen s
, 0
k ( s)
½
2
2
2
2
=
– cos s ,
2
– sen s
2
, 0
B (s) = T (s) x N (s) =
det
i
j
k
-1 sen s
1 cos s
2
2
2
2
2
– cos s
2
– sen s
2
=
1 sen s
, -1 cos s ,
1
2
2
2
2
2
T ( 2 ?)=
0 , -1 ,
2
2
N ( 2 ?) = ( 1, 0, 0)
B(
2 ? ) =
0 , 1
2
2
1
0 ( x-0) + 1
( y-1) + 1
( z- ?) = 0
y+ z = ? +1
Osculador
2
2
0 ( x-0) – 1
( y-1) + 1
( z- ?) = 0
-y+ z = ? +1
Normal
2
2
1 ( x-0) + 0 ( y-1) + 0 ( z- ?) = 0
2 3
2- F (t) = ( t , t , t )
u = f’(t) = (1, 2t, 3t2 ) ,
x = 0
p= f (2) = ( 2, 4, 8 )
f’’’(t) =( 0, 2, 6t)
Rectificante
v= f’(t) x f´´(t) = det
i
j
k
1
2t
3t2
= ( 6t2 – 6t, 2)
3 4 3
0
0
0
2
6t
w = v x u = det
i
j
k
6t
1
2
– 6t
2t
2
3t2
= ( 18t -4t, 3 – 18 t , 12 t + 6t)
24 ( x-2) – 12 ( y – 4) + 2 ( z – 8 ) = 0
12x – 6y + z = 8
1 ( x-2) + 4 ( y – 4) + 12 ( z – 8 ) = 0
x + 4y + 12 z = 114
-152 ( x-2) – 286 ( y – 4)+ 108 ( z – 8 ) = 0
76x + 143 y – 54z= 292
Osculador
Normal
Rectificante
3- f ( t) = ( 2 cost,
2 sent , 2 )
p= (1 , 1, 2 )
f’(t) x f´´(t) = (- 2 sent,
2 cost , 0 ) x (- 2 cost, – 2 sent , 0)
= det
i
– 2 sent
j
2 cost
k
= ( 0, 0, 2)
– 2 cost
0 ( x-1) + 0 ( y-1) + 2 ( z-2) = 0
– 2 sent
Osculador
4 – si T ( – 3/5, 0, 4/5) ; p ( 0, 3 , 2 ?)
-3 ( x-0) + 0 ( y-3) + 4 (z- 2 ?) = 0
5
5
-3 x- + 0 + 4 z- 8 ? = 0
5
5
5
– 3x + 4z – 8 ? = 0
3x – 4z + 8 ? = 0
4x + 3z – 6 ? =0
Normal
Osculador
5- x= t – cost
y= 3 + sen 2t
z= 1 + cos 3 t
p= t= ?
2
lim
x’ = 1 + sen t = 2
y’= 3 + sen2t = -2
z’= 1 + cos 3t = 3
x= t – cos t = ? – cos ?
=
?
2
2
2
y= 3 + sen 2t = 3
z= 1 + cos 3t = 1
2x- ? – 2y + 6 + 3z – 3 = 0
2x- 2y + 3z + 3 – ? = 0
FUNCIONES VECTORIALES
1- F(t) = (t2 +1, 2t, t)
Normal
lim
(t2 + 1) = 1 +1=2 , lim
(2t) = 2 , lim
(2t) = 2
t
1
t
1
t
1
lim
t
(t) = 1
1
lim
t
1
=
2i + 2j + k
2- F(t) = (t3 +1, t2 – 2 t +1 )
lim (t3 +1)= 2
( t2 – 2 t +1) = 0
t
1
t
1
lim = 2i
t
1
3- F(t) = (t , t2 , sent )
t
lim
(t , t2 , sent ) = (0, 0 1)
t
0
t
lim
t
(t) = 0
0
t
lim
0
(t2) = 0
t
lim
0
(t) =
t
sent
=1
lim
t
0
=
k
4- F(t) = (3t-1, t)
lim t
(3t-1)
= 3- 1 = 2
t
lim t
t
lim t
t
1
1
1
(t) = 1
= 2i , j
5- F(t) = (1- cos t , 1- cos2t )
t2
t2
lim t
t
lim t
t
1
1
=
=
1 .
t
1 .
t
t
t
lim t
1
lim t
1
1- cost
t
1- cos2t
t
=
=
1.0=0
1. .01 = .01
lim t
t
1
=
.01 j
MATRIZ HESIANA
1. f(x,y)= 2 (x-1)2 + 3(y-2)2
df = 4 (x-1) = 0
dx
x= 1
df = 6(y-2) = 0
dy
y= 2
d2f
d2f
4
0
dx2
dydx
=
0
6
d2f
dxdy
(1,2 )
d2f
dy2
mínimo local
2 – f(x,y,z)= senx + sen y + sen z – sen (x+y+z)
en P(? , ? , ? )
2
2 2
H (? , ? , ?
) =
-2
-1
-1
2
2
2
-1
-1
-2
-1
-1
-2
A1 = -2
A 2 =
-2
-1
= 3
A 3 = -4
-1
-2
H(p) – ?I) = -?3 – 6 ?2 – 9? – 4 = – (? + 1 )2 (? + 4 )
? = -1
? = -4
máximo local de p = 4
3- Se f: R4 – {( 0, 0, 0, 0)}
f ( x, y, z, u ) = x + y + z +u + 1
x y z u
dada por
df = 1 – y = 0
df = 1 – z = 0
df = 1 – u = 0
df = 1 – 1 = 0
dx
x2
dy
x y2
dz y z2
du z u2
H (x, y, z, u ) =
2y
x3
-1 0
x2
0
0
-1
x2
-1
2z
y3
-1
y2
2u -1
0
y2
z3 z2
0
0
-1
2
z2 u3
En el punto (1, 1,1 ,1 )
H ( 1, 1,1 ,1 ) =
2 -1 0 0
-1 2 -1 0
0 -1 2 -1
0 0 -1 2
A1 = 2
A2 = 2 -1 = 3
A3=
2 -1 0 = 2
A4 = 2 -1 0 0
=5
-1 2
-1 2 -1 -1 2 -1 0
0 -1 2
0 -1 2 -1
0 0 -1 2
mínimo local en ( 1, 1,1 ,1 ) = 5
x
4- f(x,y,z)= e-x2 + e –y2 + z2
df = -2xe-x2
dx
df = -2y-y2
dy
df = 2z
dz
H=
(4×2-2)e-x2
0
0
0
(4y2-2)e-y2
0
0
0
2
H ( 0,0, 0)=
?1 = -2
-2
0
0
? 2 = 2
0
-2
0
0
0
2
Tiene punto de silla
5- f(x,y)= ax2 +by4
df = 2ax
dx
df = 4by3
dy
H (x,y) =
2a
0
0
12by2
H (0,0) =
2a
0
0
0
? = 0
Cualquier cosa puede pasar
LIMITES
1) lim
(x,y)
3x2y =
(0,0) x4+y2
si y(x=0)
lim
y 0
si y=x
0 = 0
y2 0
lim
0
3x2x = lim
x4+x 2
3 x3 =
x2(x2+1)
lim 3x
=
x2+1
0= 0
0+1
si y=x2
lim
x
0
3x2x 2 = lim
x4+x 4
3 x4 =
x2(x2+x2)
lim 3×2
2×2
=
3
2
)
)
)
2
2
x
x
=
2) lim
sen(x2+y2) =
(x,y)
(0,0
x2+y2
De acuerdo a una propiedad de limites de funciones trigonometricas especifica por definición que :
lim
x 0
lim
(x,y)
3) lim
(x,y)
sen x = 1
x
sen(x2+y2) = 1
(0,0
x2 =
(0,0
por lo tanto
x2+y2
( x2+y2 )
si y(x=0)
lim
(x,y)
si y=x
0 = 0 = 1
0 0+y2
lim
0
x2 = lim x2 =
x + x x 0 2 x2
0 =
0
indeterminacion
si y=x2
lim
x
0
x2 = lim
x2 +x4
x2
=
x2(x+x2)
0 =
0
indeterminacion
4) lim
(x,y)
x4y =
(0,0) x4+y4
si y(x=0)
lim
y 0
si y=x
0 = 0
x4
lim
0
x4x = lim x5 =
x4+x 4 2×4
0 =
0
indeterminacion
5) lim
x
0=
0
x 1
x 1
1
indeterminacion
saco el conjugado
x 1 x 1
.
x 1 x 1
x1/2 +1
implica que x=1
DOMINIOS
1) f(x,y)=
1 X 2 Y 2 =
1- X2 –y2> 0
despejo el uno
X2 +y2 x < x+y =>x2 0
y> 0
y< 0
3) f(x,y,z)= ln(1-x2-y2+z)
El dominio del logaritmo debe ser mayor que cero.
entoces:
2 2
1-x -y +z > 0
-x2-y2+z > -1
x2+ y2+z
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