Utilize a Regra da Cadeia para determinar as derivadas parciais indicadas.
M=x e^y-z^2, x=2 u v, y=u-v, z=u+v ;
partial Mpartial u, partial Mpartial v quando u=3, v=-1

Passo 1

Temos a função M(x,y,z)=x e^y - z^2 com as substituições x=2uv, y=u-v e z=u+v. Precisamos de partial Mpartial u e partial Mpartial v no ponto (u,v)=(3,-1).

Passo 2

Calculamos primeiro as derivadas parciais de M em relação a x, y e z:
partial Mpartial x=e^y-z^2, partial Mpartial y=x e^y-z^2, partial Mpartial z=-2xz e^y-z^2

Passo 3

Agora derivamos x,y,z em relação a u e v:
partial xpartial u=2v, ;partial xpartial v=2u, ;partial ypartial u=1, partial ypartial v=-1, partial zpartial u=1, partial zpartial v=1

Passo 4

Aplicando a regra da cadeia:
partial Mpartial u=partial Mpartial xpartial xpartial u+partial Mpartial ypartial ypartial u+partial Mpartial zpartial zpartial u,
partial Mpartial v=partial Mpartial xpartial xpartial v+partial Mpartial ypartial ypartial v+partial Mpartial zpartial zpartial v

Passo 5

Avançando para o ponto (u,v)=(3,-1):
x=2cdot3cdot(-1)=-6, ;y=3-(-1)=4, ;z=3+(-1)=2, ;y-z^2=4-4=0, ;e^y-z^2=1

Passo 6

Substituindo e simplificando:
partial Mpartial u=1l[2(-1)+(-6)-2cdot(-6)cdot2r]=-2-6+24=16,
partial Mpartial v=1l[2cdot3-(-6)-2cdot(-6)cdot2r]=6+6+24=36

Resposta

partial Mpartial u=16 e partial Mpartial v=36.

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