Calcule as derivadas partial zpartial u e partial zpartial v.
z=e^x y, x=u(u+v), y=v(u+v);

Passo 1

O enunciado define
z = e^xy, x = u(u+v), y = v(u+v)
e pede as derivadas parciais partial zpartial u e partial zpartial v.
O conteúdo que vamos usar é a regra da cadeia para derivadas de funções compostas:
partial zpartial u = e^xy,partial (xy)partial u, partial zpartial v = e^xy,partial (xy)partial v

Passo 2

Calculemos primeiro partial (xy)partial u. Pela regra do produto,
partial (xy)partial u = x,partial ypartial u + y,partial xpartial u
Agora vamos achar partial xpartial u e partial ypartial u.

Passo 3

Como x = u(u+v) e y = v(u+v), temos
partial xpartial u = partialpartial u[,u^2 + uv,] = 2u + v, partial ypartial u = partialpartial u[,uv + v^2,] = v
Substituindo em partial(xy)partial u,
partial(xy)partial u = xcdot v + ycdot(2u+v)
Portanto,
partial zpartial u = e^xyl(x,v + y,(2u+v)r)

Passo 4

Agora calculamos partial (xy)partial v. Similarmente,
partial(xy)partial v = x,partial ypartial v + y,partial xpartial v
E
partial xpartial v = partialpartial v[,u^2+uv,] = u, partial ypartial v = partialpartial v[,uv+v^2,] = u+2v
Logo,
partial(xy)partial v = x,(u+2v) + y,u
e
partial zpartial v = e^xyl(x,(u+2v) + y,ur)

Passo 5

Se quisermos, podemos substituir x = u(u+v) e y = v(u+v) e simplificar:
x,v + y,(2u+v) = u(u+v),v + v(u+v),(2u+v) = v(u+v)l(u+2u+vr) = v(u+v)(3u+v),
x,(u+2v) + y,u = u(u+v),(u+2v) + v(u+v),u = u(u+v)l(u+2v+vr) = u(u+v)(u+3v)
Assim,
partial zpartial u = v(u+v)(3u+v),e^xy, partial zpartial v = u(u+v)(u+3v),e^xy

Resposta

partial zpartial u = e^xyl(x,v + y,(2u+v)r) = v(u+v)(3u+v),e^xy,
partial zpartial v = e^xyl(x,(u+2v) + y,ur) = u(u+v)(u+3v),e^xy

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