Seja z=f(x, y) uma função diferenciável, com x=u+2 v e y=u^2-v. Expresse partial zpartial u e partial zpartial v em termos das derivadas parciais partial fpartial x e partial fpartial y.

Passo 1

O enunciado pede para usar a regra da cadeia em z=f(x,y) com
x = u + 2v,
y = u^2 - v
e expressar partial zpartial u e partial zpartial v em termos de partial fpartial x e partial fpartial y.

Passo 2

Aplicamos a regra da cadeia para partial zpartial u:
partial zpartial u = partial fpartial x,partial xpartial u + partial fpartial y,partial ypartial u
Como partial xpartial u=1 e partial ypartial u=2u, vem
partial zpartial u = partial fpartial xcdot1 + partial fpartial ycdot2u = partial fpartial x + 2u,partial fpartial y

Passo 3

De modo análogo, para partial zpartial v usamos
partial zpartial v = partial fpartial x,partial xpartial v + partial fpartial y,partial ypartial v
Como partial xpartial v=2 e partial ypartial v=-1, obtemos
partial zpartial v = partial fpartial xcdot2 + partial fpartial ycdot(-1) = 2,partial fpartial x - partial fpartial y

Resposta

partial zpartial u = partial fpartial x + 2u,partial fpartial y, partial zpartial v = 2,partial fpartial x - partial fpartial y

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