Utilize a Regra da Cadeia para determinar as derivadas parciais indicadas.
P=raizu^2+v^2+u^2, u=x e^y, v=y e^k, w=e^n; partial Ppartial x, partial Ppartial y quando x=0, y=2

Passo 1

O enunciado define
P = raizu^2 + v^2 + w^2,
com
u = x e^y, v = y e^k, w = e^n,
onde k e n são constantes. Pede-se partial Ppartial x e partial Ppartial y no ponto (x,y)=(0,2). Vamos usar a regra da cadeia para funções de várias variáveis.

Passo 2

Aplicamos a regra da cadeia:
partial Ppartial x = partial Ppartial u,partial upartial x + partial Ppartial v,partial vpartial x + partial Ppartial w,partial wpartial x,
e analogamente para partial Ppartial y. Como
P = (u^2+v^2+w^2)^1/2,
temos
partial Ppartial u = uP, partial Ppartial v = vP, partial Ppartial w = wP

Passo 3

Calculamos agora as derivadas de u,v,w:
partial upartial x = e^y, partial vpartial x = 0, partial wpartial x = 0,
partial upartial y = x e^y, partial vpartial y = e^k, partial wpartial y = 0

Passo 4

Substituindo em cada regra da cadeia:
partial Ppartial x = uP,e^y + vPcdot0 + wPcdot0 = u,e^yP,
partial Ppartial y = uP,(x e^y) + vP,e^k + wPcdot0 = u,x e^y + v,e^kP

Passo 5

No ponto (0,2) temos
u(0,2) = 0cdot e^2 = 0, v(0,2) = 2 e^k, w = e^n,
P(0,2) = raiz0 + (2e^k)^2 + (e^n)^2 = raiz4e^2k + e^2n
Logo
partial Ppartial x|(0,2) = 0cdot e^2raiz4e^2k+e^2n = 0,
partial Ppartial y|(0,2) = 0cdot0cdot e^2 + (2e^k),e^kraiz4e^2k+e^2n = 2e^2kraiz4e^2k+e^2n

Resposta

partial Ppartial x(0,2)=0, partial Ppartial y(0,2) =2e^2kraiz4e^2k+e^2n

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