Ache a derivadas parciais indicadas usando a regra da cadeia
z=x^{2}+x y, x=u^{2}+v^{2}, y=3 u-2 v ; \frac{\partial z}{\partial u}, \frac{\partial z}{\partial v}

Passo 1

O enunciado dá z = x^{2} + x\,y, x = u^{2} + v^{2} e y = 3u - 2v,
e pede \partial z/\partial u e \partial z/\partial v usando a regra da cadeia em variáveis múltiplas. Vamos precisar de \partial z/\partial x, \partial z/\partial y, \partial x/\partial u, \partial x/\partial v, \partial y/\partial u e \partial y/\partial v.

Passo 2

Calculemos as derivadas parciais intermediárias: \displaystyle \frac{\partial z}{\partial x} = 2x + y e \displaystyle \frac{\partial z}{\partial y} = x.
Como x = u^{2}+v^{2} e y=3u-2v, temos \displaystyle \frac{\partial x}{\partial u} = 2u,\quad \frac{\partial x}{\partial v} = 2v,\quad \frac{\partial y}{\partial u} = 3,\quad \frac{\partial y}{\partial v} = -2.

Passo 3

Aplicando a regra da cadeia:
\frac{\partial z}{\partial u} = \frac{\partial z}{\partial x}\,\frac{\partial x}{\partial u} + \frac{\partial z}{\partial y}\,\frac{\partial y}{\partial u},
\frac{\partial z}{\partial v} = \frac{\partial z}{\partial x}\,\frac{\partial x}{\partial v} + \frac{\partial z}{\partial y}\,\frac{\partial y}{\partial v}.

Passo 4

Substituindo os valores:
\frac{\partial z}{\partial u} = (2x + y)\,(2u) + x\,(3) = 2u\bigl(2(u^{2}+v^{2}) + (3u-2v)\bigr) + 3(u^{2}+v^{2}),
\frac{\partial z}{\partial v} = (2x + y)\,(2v) + x\,(-2) = 2v\bigl(2(u^{2}+v^{2}) + (3u-2v)\bigr) - 2(u^{2}+v^{2}).
Agora basta expandir e agrupar termos:
\frac{\partial z}{\partial u} = 4u^{3} + 4u\,v^{2} + 6u^{2} - 4u\,v + 3u^{2} + 3v^{2} = 4u^{3} + 4u\,v^{2} + 9u^{2} - 4u\,v + 3v^{2},
\frac{\partial z}{\partial v} = 4u^{2}v + 4v^{3} + 6u\,v - 4v^{2} - 2u^{2} - 2v^{2} = 4u^{2}v + 4v^{3} + 6u\,v - 2u^{2} - 6v^{2}.

Resposta

\displaystyle \frac{\partial z}{\partial u} = 4u^{3} + 4u\,v^{2} + 9u^{2} - 4u\,v + 3v^{2} \displaystyle \frac{\partial z}{\partial v} = 4u^{2}v + 4v^{3} + 6u\,v - 2u^{2} - 6v^{2}

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