Use a Regra da Cadeia para determinar as derivadas parciais partial zpartial u e partial zpartial v das funções:
z=x / y, para x=u e^v e y=1+u e^-v.

Passo 1

O enunciado define
z = dfracxy
com
x = u,e^v e y = 1 + u,e^-v.
Queremos calcular as derivadas parciais dfracpartial zpartial u e dfracpartial zpartial v.
O conteúdo usado será a Regra da Cadeia aliada à Regra do Quociente:
partialpartial wl(xyr) = y,partial xpartial w-x,partial ypartial wy^2

Passo 2

Calculamos primeiro as derivadas de x e y em relação a u e v:
partial xpartial u = e^v, partial xpartial v = u,e^v,
partial ypartial u = e^-v, partial ypartial v = -u,e^-v

Passo 3

Usando a Regra do Quociente para dfracpartial zpartial u:
partial zpartial u = ypartial xpartial u-xpartial ypartial uy^2 = l(1+u,e^-vr)e^v - u,e^v,e^-vl(1+u,e^-vr)^2
No numerador:
(1+u,e^-v)e^v - u,e^ve^-v = e^v + u - u = e^v
Logo,
partial zpartial u = e^vl(1+u,e^-vr)^2

Passo 4

Agora dfracpartial zpartial v:
partial zpartial v = ypartial xpartial v-xpartial ypartial vy^2 = l(1+u,e^-vr)(u,e^v) - u,e^v,(-u,e^-v)l(1+u,e^-vr)^2
No numerador:
(1+u,e^-v)u,e^v + u,e^v,u,e^-v = u,e^v + u^2 + u^2 = u,e^v + 2,u^2
Portanto,
partial zpartial v = u,e^v + 2,u^2l(1+u,e^-vr)^2 = ul(e^v+2ur)l(1+u,e^-vr)^2

Resposta

partial zpartial u = e^vl(1+u,e^-vr)^2, partial zpartial v = u,e^v + 2,u^2l(1+u,e^-vr)^2

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