Se h(u, v)=g(u^2+v^2, 2 u-3 v) onde g=g(x, y) é uma função diferenciável tais que partial gpartial x(2,-1)=3, partial gpartial y(2,-1)=4. Calcule partial hpartial u(1,1), partial hpartial v(1,1).

Passo 1

O enunciado define a função composta
h(u,v)=gl(x(u,v),,y(u,v)r)
onde
x(u,v)=u^2+v^2, y(u,v)=2u-3v,
e sabemos que g é diferenciável com
partial gpartial x(2,-1)=3 e
partial gpartial y(2,-1)=4.
Para encontrar partial hpartial u e partial hpartial v em (1,1), vamos usar a regra da cadeia.

Passo 2

Calcular os valores de x e y no ponto (u,v)=(1,1):
x(1,1)=1^2+1^2=2, y(1,1)=2cdot1-3cdot1=-1
Isso confirma que usaremos partial gpartial x(2,-1) e partial gpartial y(2,-1).

Passo 3

Pela regra da cadeia para derivadas parciais,
partial hpartial u =partial gpartial x,partial xpartial u +partial gpartial y,partial ypartial u, partial hpartial v =partial gpartial x,partial xpartial v +partial gpartial y,partial ypartial v

Passo 4

Calculamos as derivadas de x e y:
partial xpartial u=2u, partial xpartial v=2v, partial ypartial u=2, partial ypartial v=-3
Em (1,1) temos 2u=2, 2v=2. Usando partial gpartial x(2,-1)=3 e partial gpartial y(2,-1)=4, obtemos
partial hpartial u(1,1) =3cdot2 + 4cdot2 =6+8=14, partial hpartial v(1,1) =3cdot2 + 4cdot(-3) =6-12=-6

Resposta

partial hpartial u(1,1)=14, partial hpartial v(1,1)=-6.

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