Seja g(x, y)=(x^2+y^2) f(u, v) onde u=2 x-y e v=x+2 y. Mostre as seguintes igualdades:
partial gpartial x(x, y) & =2 x f(u, v)+(x^2+y^2)[2 partial fpartial u(u, v)+partial fpartial v(u, v)] partial gpartial y(x, y) & =2 y f(u, v)+(x^2+y^2)[-partial fpartial u(u, v)+2 partial fpartial v(u, v)]

Passo 1

O enunciado define
g(x,y) = l(x^2 + y^2r),fl(u,vr)
com
u = 2x - y, v = x + 2y
Queremos calcular partial gpartial x e partial gpartial y usando regra do produto e regra da cadeia.

Passo 2

Para partial gpartial x, aplicamos a regra do produto:
partial gpartial x = partialpartial xl(x^2 + y^2r),f(u,v) + l(x^2 + y^2r),partialpartial xfl(u,vr)
Como partialpartial x(x^2 + y^2)=2x, fica
partial gpartial x = 2x,f(u,v) + (x^2+y^2),partial fpartial x(u,v)
Agora usamos regra da cadeia para partial fpartial x:
partial fpartial x = partial fpartial u,partial upartial x + partial fpartial v,partial vpartial x
Como partial upartial x=2 e partial vpartial x=1, obtemos
partial fpartial x = 2,partial fpartial u(u,v) + 1cdotpartial fpartial v(u,v)
Substituindo de volta,
partial gpartial x = 2x,f(u,v) + (x^2+y^2)gl[2,partial fpartial u(u,v) + partial fpartial v(u,v)gr]

Passo 3

Para partial gpartial y, de modo análogo:
partial gpartial y = partialpartial y(x^2 + y^2),f(u,v) + (x^2+y^2),partialpartial yf(u,v)
Aqui partialpartial y(x^2+y^2)=2y, então
partial gpartial y = 2y,f(u,v) + (x^2+y^2),partial fpartial y(u,v)
Usando regra da cadeia:
partial fpartial y = partial fpartial u,partial upartial y + partial fpartial v,partial vpartial y
Como partial upartial y=-1 e partial vpartial y=2, segue
partial fpartial y = -1cdotpartial fpartial u(u,v) + 2,partial fpartial v(u,v)
Logo
partial gpartial y = 2y,f(u,v) + (x^2+y^2)gl[-partial fpartial u(u,v) +2,partial fpartial v(u,v)gr]

Resposta

partial gpartial x(x,y) =2x,f(u,v) +(x^2+y^2)l[2,fu(u,v)+fv(u,v)r],
partial gpartial y(x,y) =2y,f(u,v) +(x^2+y^2)l[-fu(u,v)+2,fv(u,v)r]

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