Calcule as derivadas partial zpartial u e partial zpartial v.
z=e^x sin y, x=u v^2, y=u^2 v;

Passo 1

O enunciado define
z = e^x,sin y
com
x = u,v^2 e y = u^2v
e pede as derivadas parciais tfracpartial zpartial u e tfracpartial zpartial v.
O conteúdo usado será a regra da cadeia para funções de várias variáveis: precisamos de tfracpartial zpartial x, tfracpartial zpartial y, tfracpartial xpartial u, tfracpartial xpartial v, tfracpartial ypartial u e tfracpartial ypartial v.

Passo 2

Calculamos primeiro as derivadas de z em relação a x e y, tratando z como função de duas variáveis x,y:
partial zpartial x = partialpartial xl(e^xsin yr) = e^xsin y,
partial zpartial y = partialpartial yl(e^xsin yr) = e^xcos y

Passo 3

Agora derivamos x(u,v) e y(u,v):
partial xpartial u = partialpartial ul(u,v^2r) = v^2, partial xpartial v = partialpartial vl(u,v^2r) = 2u,v,
partial ypartial u = partialpartial ul(u^2vr) = 2u,v, partial ypartial v = partialpartial vl(u^2vr) = u^2

Passo 4

Aplicando a regra da cadeia:
partial zpartial u = partial zpartial x,partial xpartial u + partial zpartial y,partial ypartial u,
partial zpartial v = partial zpartial x,partial xpartial v + partial zpartial y,partial ypartial v
Substituindo os resultados dos passos anteriores:
partial zpartial u = e^xsin ycdot v^2 + e^xcos ycdot (2u,v) = e^xl(v^2sin y + 2u,vcos yr),
partial zpartial v = e^xsin ycdot (2u,v) + e^xcos ycdot u^2 = e^xl(2u,vsin y + u^2cos yr)
Finalmente, voltamos a x=u,v^2 e y=u^2v:

Resposta

partial zpartial u = e^u v^2l(v^2sin(u^2v) + 2u,vcos(u^2v)r),
partial zpartial v = e^u v^2l(2u,vsin(u^2v) + u^2cos(u^2v)r)

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