Utilize a Regra da Cadeia para determinar as derivadas parciais indicadas.
ll P=raizu^2+v^2+w^2, u=x e^y, v=y e^x, w=e^x y ; u=x e^y y, x=alpha^2 beta, y=beta^2 gamma, t=gamma^2 alpha ; partial Ppartial x, partial Ppartial y quando x=0, y=2 & partial upartial alpha, partial upartial beta, partial upartial gamma, quando alpha=-1, beta=2, gamma=1

Passo 1

O enunciado define
P = raizu^2+v^2+w^2, u = x e^y, v = y e^x, w = e^xy,
e pede duas coisas:
  1. partial Ppartial x e partial Ppartial y avaliadas em x=0, y=2.
  2. partial upartial alpha, partial upartial beta, partial upartial gamma para
u = x e^y, x = alpha^2beta, y = beta^2gamma
avaliadas em alpha=-1, beta=2, gamma=1.
O método chave é a Regra da Cadeia para funções compostas.

Passo 2

Para a primeira parte use
partial Ppartial x = partial Ppartial u,partial upartial x + partial Ppartial v,partial vpartial x + partial Ppartial w,partial wpartial x,
e semelhante para partial P/partial y. Como
P = (u^2+v^2+w^2)^1/2 Longrightarrow partial Ppartial u=uP,; partial Ppartial v=vP,; partial Ppartial w=wP,
segue
partial Ppartial x =u,ux+v,vx+w,wxP, partial Ppartial y =u,uy+v,vy+w,wyP

Passo 3

Calcule agora as derivadas parciais de u,v,w em função de x,y:
u = x e^y ;Longrightarrow; ux=e^y,; uy=x e^y=u,
v = y e^x ;Longrightarrow; vx=y e^x,; vy=e^x,
w = e^x y ;Longrightarrow; wx=y e^x y=y,w,; wy=x e^x y=x,w
Substituindo em partial P/partial x e partial P/partial y obtém-se as fórmulas gerais.

Passo 4

Avalie em x=0, y=2:
u(0,2)=0cdot e^2=0, v(0,2)=2cdot e^0=2, w(0,2)=e^0cdot2=1, P=raiz0^2+2^2+1^2=raiz5
Também em (0,2):
ux=e^2,; vx=2e^0=2,; wx=2cdot1=2, uy=0,; vy=e^0=1,; wy=0

Passo 5

Agora calcule os valores numéricos:
partial Ppartial x|(0,2) =u,ux+v,vx+w,wxP =0cdot e^2+2cdot2+1cdot2raiz5 =6raiz5,
partial Ppartial y|(0,2) =u,uy+v,vy+w,wyP =0cdot0+2cdot1+1cdot0raiz5 =2raiz5

Passo 6

Para a segunda parte use u(x,y) com x(alpha,beta), y(beta,gamma). A Regra da Cadeia dá
partial upartial alpha =ux,xalpha+uy,yalpha, partial upartial beta =ux,xbeta+uy,ybeta, partial upartial gamma =ux,xgamma+uy,ygamma
Já sabemos ux=e^y, uy=x e^y=u. De x=alpha^2beta, y=beta^2gamma vem
xalpha=2alphabeta,;xbeta=alpha^2,;xgamma=0, yalpha=0,;ybeta=2betagamma,;ygamma=beta^2

Passo 7

Avalie em alpha=-1, beta=2, gamma=1. Primeiro
y=beta^2gamma=4, e^y=e^4, x=alpha^2beta=1cdot2=2, u=x e^y=2e^4
Logo
partial upartial alpha = e^y,(2alphabeta) =2alphabeta,e^4 =2cdot(-1)cdot2,e^4=-4e^4,
partial upartial beta = e^y,(alpha^2) + u,(2betagamma) =1cdot e^4 + 2e^4cdot(2cdot1) = e^4 +4e^4=5e^4,
partial upartial gamma = u,(beta^2) =2e^4cdot4=8e^4

Resposta

partial Ppartial xg|(0,2)=6raiz5,
partial Ppartial yg|(0,2)=2raiz5,
partial upartial alphag|(-1,2,1)=-4e^4,
partial upartial betag|(-1,2,1)=5e^4,
partial upartial gammag|(-1,2,1)=8e^4

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