Se z=f(x, y) com x=u+v e y=u-v, demonstre que
partial zpartial u+partial zpartial v=2 partial fpartial x

Passo 1

A gente tem a função z = f(x, y) com as mudanças de variável x = u + v e y = u - v, e o objetivo é usar derivadas parciais e a regra da cadeia para mostrar que partial zpartial u + partial zpartial v = 2 partial fpartial x.

Passo 2

Aplicamos a regra da cadeia para partial zpartial u:
partial zpartial u = partial fpartial x cdot partial xpartial u + partial fpartial y cdot partial ypartial u
Como x = u+v e y = u-v, temos partial xpartial u = 1 e partial ypartial u = 1. Substituindo dá
partial zpartial u = partial fpartial x + partial fpartial y

Passo 3

Agora fazemos o mesmo para partial zpartial v:
partial zpartial v = partial fpartial x cdot partial xpartial v + partial fpartial y cdot partial ypartial v
Aqui partial xpartial v = 1 e partial ypartial v = -1, então
partial zpartial v = partial fpartial x - partial fpartial y

Passo 4

Basta somar as duas expressões para cancelar o termo partial fpartial y:
partial zpartial u + partial zpartial v = l(partial fpartial x + partial fpartial yr) + l(partial fpartial x - partial fpartial yr) = 2,partial fpartial x

Resposta

2 partial fpartial x

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