Use a regra da cadeia para calcular \frac{\partial z}{\partial u} e \frac{\partial z}{\partial v}, onde:
z=e^{x y} t g(y), onde x=x(u, v)=u+2 v e y=y(u, v)=\frac{u}{v}

Passo 1

O enunciado dá
z = e^x y,tg(y), x = x(u,v) = u + 2v, y = y(u,v) = uv,
e pede partial z/partial u e partial z/partial v. Vamos usar a regra da cadeia
partial zpartial u = zx,partial xpartial u + zy,partial ypartial u, partial zpartial v = zx,partial xpartial v + zy,partial ypartial v

Passo 2

Calcular as derivadas parciais de z em relação a x e y.
A função é z(x,y)=e^x y,tg(y).
zx = partialpartial xl(e^xyr),tg(y) = y,e^x y,tg(y),
zy = partialpartial yl(e^xyr),tg(y) + e^x y,ddyl(tg(y)r) = x,e^x y,tg(y) + e^x y,sec^2(y)

Passo 3

Calcular as derivadas parciais de x(u,v) e y(u,v).
partial xpartial u = 1, partial xpartial v = 2,
partial ypartial u = partialpartial ul(uvr) = 1v, partial ypartial v = partialpartial vl(uvr) = -uv^2

Passo 4

Aplicando a regra da cadeia:
partial zpartial u = zxcdot1 + zycdot1v = y,e^x y,tg(y) + 1vl(x,e^x y,tg(y) + e^x y,sec^2(y)r),
partial zpartial v = zxcdot2 + zycdotl(-uv^2r) = 2,y,e^x y,tg(y) - uv^2l(x,e^x y,tg(y) + e^x y,sec^2(y)r)

Resposta

partial zpartial u = y,e^x y,tg(y) + x,e^x y,tg(y) + e^x y,sec^2(y)v,
partial zpartial v = 2,y,e^x y,tg(y) - ul(x,e^x y,tg(y) + e^x y,sec^2(y)r)v^2

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