Determinar as derivadas parciais partial zpartial u e partial zpartial v usando a regra da cadeia:
z=e^x / y, x=u cdot cos v e y=v sen u

Passo 1

O enunciado dá
z = e^x/y,
x = u cdot cos v e
y = v sin u,
e pede as derivadas parciais partial zpartial u e partial zpartial v.
Para isso vamos usar a regra da cadeia na forma
partial zpartial u = partial zpartial x,partial xpartial u + partial zpartial y,partial ypartial u,
e de modo análogo para partial zpartial v.

Passo 2

Calculamos primeiro partial zpartial x e partial zpartial y. Como
z = e^x/y,
temos
partial zpartial x = 1y,e^x/y, partial zpartial y = -xy^2,e^x/y

Passo 3

Agora achamos partial zpartial u. Primeiro
partial xpartial u = cos v, partial ypartial u = v cos u
Logo
partial zpartial u = 1ye^x/ycdotcos v + l(-xy^2e^x/yr)cdot(vcos u)
Colocando em evidência e^x/y e agrupando,
partial zpartial u = e^x/yl(cos vy -x,vcos uy^2r)

Passo 4

Para partial zpartial v calculamos
partial xpartial v = -u sin v, partial ypartial v = sin u
Assim,
partial zpartial v = 1ye^x/ycdot(-usin v) + l(-xy^2e^x/yr)cdotsin u,
e, fatorando e^x/y,
partial zpartial v = e^x/yl(-usin vy -x,sin uy^2r)

Resposta

partial zpartial u = e^x/yl(cos vy -x,vcos uy^2r), partial zpartial v = e^x/yl(-usin vy -x,sin uy^2r)

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