Determine partial hpartial u, partial hpartial v onde h(u, v)=f(x(u, v), y(u, v)) nos seguintes casos:
f(x, y)=x^2+y^2x^2-y^2, x(u, v)=e^-u-v, y(u, v)=e^u v

Passo 1

Estamos com
h(u,v)=fl(x(u,v),,y(u,v)r)
em que
f(x,y)=x^2+y^2x^2-y^2, x(u,v)=e^-u-v, y(u,v)=e^,u,v
O que precisamos é aplicar a regra da cadeia para obter partial hpartial u e partial hpartial v.

Passo 2

Calculemos primeiro as derivadas parciais de f. Defina
N(x,y)=x^2+y^2, D(x,y)=x^2-y^2, f=ND
Então
partial fpartial x =NxD - N,DxD^2 =2x(x^2-y^2)-2x(x^2+y^2)(x^2-y^2)^2 =-4x,y^2(x^2-y^2)^2,
partial fpartial y =NyD - N,DyD^2 =2y(x^2-y^2)+2y(x^2+y^2)(x^2-y^2)^2 =4x^2,y(x^2-y^2)^2

Passo 3

Agora as derivadas de x(u,v) e y(u,v):
xu=partialpartial ue^-u-v=-e^-u-v, xv=-e^-u-v,
yu=partialpartial ue^uv=v,e^uv, yv=u,e^uv

Passo 4

Pela regra da cadeia:
partial hpartial u =partial fpartial x,xu +partial fpartial y,yu, partial hpartial v =partial fpartial x,xv +partial fpartial y,yv

Passo 5

Substituindo partial f/partial x, partial f/partial y, xu, xv, yu, yv e ainda x=e^-u-v, y=e^uv, note que
x,y^2e^-u-v =e^-u-v,(e^uv)^2e^-u-v =e^-2u-2v+2uv, x^2y,e^uv =e^-2u-2v,e^2uv =e^-2u-2v+2uv
Também x^2-y^2=e^-2u-2v-e^2uv. Assim
partial hpartial u =4,e^-2u-2v+2uvl(e^-2u-2v-e^2uvr)^2 ,(1+v),
partial hpartial v =4,e^-2u-2v+2uvl(e^-2u-2v-e^2uvr)^2 ,(1+u)

Resposta

partial hpartial u = 4,(1+v),e^-2u-2v+2uvl(e^-2u-2v-e^2uvr)^2
partial hpartial v = 4,(1+u),e^-2u-2v+2uvl(e^-2u-2v-e^2uvr)^2

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