Calculo

Derivación de funciones implícitas
Sea Fx,y=0
1. Método por Diferenciación

∂F∂xdx+∂F∂ydy=0

Si y=gx o
∂F∂xdxdx+∂F∂ydydx=0
dydx=-∂F∂x∂F∂y

Si x=hy
∂F∂xdxdy+∂F∂ydydy=0
dxdy=-∂F∂y∂F∂x

dxdydydx=1
2. Método por Derivación Implícita
Sea Fx,y=0

a. Derivando con respecto a x:
∂F∂xdxdx+∂F∂ydydx=0
dydx=-∂F∂x∂F∂y
b. Derivando con respecto a y:∂F∂xdxdy+∂F∂ydydy=0
dxdy=-∂F∂y∂F∂x
Sea Fx,y,z=0
∂F∂xdx+∂F∂ydy+∂F∂zdz=0
1. Método por Diferenciación

Si x=gy,z
∂F∂xdxdy+∂F∂ydydy+∂F∂zdzdy=0
dzdy=0
∂x∂y=-∂F∂y∂F∂x

∂F∂xdxdz+∂F∂ydydz+∂F∂zdzdz=0
dydz=0
∂x∂z=-∂F∂z∂F∂x
Si y=hx,z
∂F∂xdxdx+∂F∂ydydx+∂F∂zdzdx=0
dzdx=0
∂y∂x=-∂F∂x∂F∂y
Si z=jx,y
∂F∂xdxdx+∂F∂ydydx+∂F∂zdzdx=0
dydx=0
∂z∂x=-∂F∂x∂F∂z

∂f∂xdxdz+∂f∂ydydz+∂f∂zdzdz=0
dxdz=0∂y∂z=-∂f∂z∂f∂y
z=k(x,y)
∂f∂xdxdy+∂f∂ydydy+∂f∂zdzdy=0
dxdy=0
∂z∂y=-∂f∂y∂f∂z

Sistemas de Funciones Implícitas
Fx,y,u,v=0Gx,y,u,v=0
x=xu,v
y=yu,v
1. Método por Diferenciación
∂F∂xdx+∂F∂ydy+∂F∂udu+∂F∂vdv=0
∂F∂xdx+∂F∂ydy=-∂F∂udu-∂F∂vdv
∂G∂xdx+∂G∂ydy+∂G∂udu+∂G∂vdv=0
∂G∂xdx+∂G∂ydy=-∂G∂udu-∂G∂vdv
dx=∂x∂udu+∂x∂vdv
dy=∂y∂udu+∂y∂vdv
Sea D el determinante Jacobiano:J=D=∂F∂x∂F∂y∂G∂x∂G∂y=∂F,G∂x,y
D1=-∂F∂udu-∂F∂vdv∂F∂y-∂G∂udu-∂G∂vdv∂G∂y=-∂F∂u∂F∂y∂G∂u∂G∂ydu-∂F∂v∂F∂y∂G∂v∂G∂ydv
D1=-∂F,G∂u,ydu-∂F,G∂v,ydv
D2=∂F∂x-∂F∂udu-∂F∂vdv∂G∂x-∂G∂udu-∂G∂vdv=-∂F∂x∂F∂u∂G∂x∂G∂udu-∂F∂x∂F∂v∂G∂x∂G∂vdv
D2=-∂F,G∂x,udu-∂F,G∂x,vdv
dx=D1D=-∂F,G∂u,y∂F,G∂x,ydu-∂F,G∂v,y∂F,G∂x,ydv
∂x∂u=-∂F,G∂u,y∂F,G∂x,y
∂x∂v=-∂F,G∂v,y∂F,G∂x,y
dy=D2D=-∂F,G∂x,u∂F,G∂x,ydu-∂F,G∂x,v∂F,G∂x,ydv
∂y∂u=-∂F,G∂x,u∂F,G∂x,y∂y∂v=-∂F,G∂x,v∂F,G∂x,y
2. Método por Derivación Implícita
Fx,y,u,vGx,y,u,v
x=xu,v ; y=yu,v
Derivación con respecto a u:
∂F∂x∂x∂u+∂F∂y∂y∂u+∂F∂u∂u∂u+∂F∂v∂v∂u=0
∂v∂u=0
∂F∂x∂x∂u+∂F∂y∂y∂u+∂F∂u=0
∂F∂x∂x∂u+∂F∂y∂y∂u=-∂F∂u
∂G∂x∂x∂u+∂G∂y∂y∂u+∂G∂u∂u∂u+∂G∂v∂v∂u=0
∂v∂u=0
∂G∂x∂x∂u+∂G∂y∂y∂u+∂G∂u=0
∂G∂x∂x∂u+∂G∂y∂y∂u=-∂G∂u
Derivación con respecto a v:∂F∂x∂x∂v+∂F∂y∂y∂v+∂F∂u∂u∂v+∂F∂v∂v∂v=0
∂u∂v=0
∂F∂x∂x∂v+∂F∂y∂y∂v+∂F∂v=0
∂F∂x∂x∂v+∂F∂y∂y∂v=-∂F∂v
∂G∂x∂x∂v+∂G∂y∂y∂v+∂G∂u∂u∂v+∂G∂v∂v∂v=0
∂u∂v=0
∂G∂x∂x∂v+∂G∂y∂y∂v+∂G∂v=0
∂G∂x∂x∂v+∂G∂y∂y∂v=-∂G∂v
D=∂F∂x∂F∂y∂G∂x∂G∂y=∂F,G∂x,y
∂x∂u=-∂F,G∂u,y∂F,G∂x,y
∂y∂u=-∂F,G∂x,u∂F,G∂x,y
∂x∂v=-∂F,G∂v,y∂F,G∂x,y
∂y∂v=-∂F,G∂x,v∂F,G∂x,y
Ejercicios
Dadas las funciones:
* x=u+v
* y=u2+v2
* z=u3+v3
Donde z=fx,y. Hallar ∂z∂x y ∂z∂y.
*Método de Diferenciación

* x=u+v⟶dx=du+dv Ecuación 1
* y=u2+v2⟶dy=2udu+2vdv Ecuación 2
* z=u3+v3⟶dz=3u2du+3v2dv Ecuación 3
De Ecuación 1: -2udx=-2udu-2udv Ecuación 4
Ecuación 4 + Ecuación 2:
dy-2udx=2vdv-2udv
dy-2udx=2v-2udv
dv=dy-2udx2v-2u
De Ecuación 1: -2vdx=-2vdu-2vdv Ecuación 5
Ecuación 5 + Ecuación 2:
dy-2vdx=2udu-2vdu
dy-2vdx=2u-2vdu
du=dy-2vdx2u-2vdu=2vdx-dy2v-2u
du y dv en Ecuación 3:
dz=3u22vdx-dy2v-2u+3v2dy-2udx2v-2u
dz=3u22vdx-dy+3v2dy-2udx2v-2u
dz=6u2vdx-3u2dy+3v2dy-6uv2dx2v-u
dz=6u2v-6uv2dx+3v2-3u2dy2v-u
dz=6u2v-6uv22v-udx+3v2-3u22v-udy
dz=6uvu-v2v-udx+3v2-u22v-udy
dz=-6uvv-u2v-udx+3v-uv+u2v-udy
dz=-3uvdx+32v+udy
∂z∂x=-3uv
∂z∂y=32u+v
* Método por Derivación
∂z∂x=∂z∂u∂u∂x+∂z∂v∂v∂x⟶∂z∂x=3u2∂u∂x+3v2∂v∂x∂z∂y=∂z∂u∂u∂y+∂z∂v∂v∂y⟶∂z∂y=3u2∂u∂y+3v2∂v∂y
* x=u+v⟶u+v-x=0 Fu,v,x,y=0
* y=u2+v2⟶u2+v2-y=0 Gu,v,x,y=0
∂u∂x=-∂F,G∂x,v∂F,G∂u,v⟶∂u∂x=-∂F∂x∂F∂v∂G∂x∂G∂v∂F∂u∂F∂v∂G∂u∂G∂v
∂u∂x=--1102v112u2v=--2v2v-2u⟶∂u∂x=vv-u
∂v∂x=-∂F,G∂u,x∂F,G∂u,v⟶∂v∂x=-∂F∂u∂F∂x∂G∂u∂G∂x∂F∂u∂F∂v∂G∂u∂G∂v
∂v∂x=-1-12u0112u2v=-2u2v-2u⟶∂v∂x=-uv-u
∂u∂y=-∂F,G∂y,v∂F,G∂u,v⟶∂u∂y=-∂F∂y∂F∂v∂G∂y∂G∂v∂F∂u∂F∂v∂G∂u∂G∂v
∂u∂y=-01-12v112u2v=-12v-2u⟶∂u∂y=-12v-u∂v∂y=-∂F,G∂u,y∂F,G∂u,v⟶∂v∂y=-∂F∂u∂F∂y∂G∂u∂G∂y∂F∂u∂F∂v∂G∂u∂G∂v
∂v∂y=-102u-1112u2v=--12v-2u⟶∂v∂y=12v-u
∂z∂x=3u2∂u∂x+3v2∂v∂x
∂z∂x=3u2vv-u+3v2-uv-u
∂z∂x=3u2v-3v2uv-u
∂z∂x=-3uv-u+vv-u
∂z∂x=-3uv
∂z∂y=3u2∂u∂y+3v2∂v∂y
∂z∂y=3u2-12v-u+3v212v-u
∂z∂y=-3u2+3v22v-u
∂z∂y=3v2-u22v-u
∂z∂y=3v-uv+u2v-u
∂z∂y=32u+v
* Método por...