Utilize a Regra da Cadeia para determinar as derivadas parciais indicadas.
z=x^{2}+x y^{3}, x=u v^{2}+w^{3}, y=u+v e^{2} ;
\frac{\partial z}{\partial u}, \frac{\partial z}{\partial v}, \frac{\partial z}{\partial w} quando u=2, \quad v=1, \quad w=0

Passo 1

O enunciado dá as funções
z = x^{2} + x\,y^{3},\quad x = u\,v^{2} + w^{3},\quad y = u + v\,e^{2}
e pede as derivadas parciais \frac{\partial z}{\partial u}, \frac{\partial z}{\partial v} e \frac{\partial z}{\partial w} no ponto (u,v,w)=(2,1,0). Vamos usar a regra da cadeia:
\frac{\partial z}{\partial u} = z_{x}\,\frac{\partial x}{\partial u} + z_{y}\,\frac{\partial y}{\partial u},
e de modo análogo para v e w.

Passo 2

Calculamos as derivadas parciais de z em relação a x e y:
z_{x} = \frac{\partial}{\partial x}\bigl(x^{2} + x\,y^{3}\bigr) = 2x + y^{3},
z_{y} = \frac{\partial}{\partial y}\bigl(x^{2} + x\,y^{3}\bigr) = 3\,x\,y^{2}.

Passo 3

Agora as derivadas de x e y em relação a u,v,w:
\frac{\partial x}{\partial u} = v^{2}, \quad \frac{\partial x}{\partial v} = 2\,u\,v, \quad \frac{\partial x}{\partial w} = 3\,w^{2},
\frac{\partial y}{\partial u} = 1, \quad \frac{\partial y}{\partial v} = e^{2}, \quad \frac{\partial y}{\partial w} = 0.

Passo 4

No ponto (u,v,w)=(2,1,0) temos
x = 2\cdot1^{2} + 0^{3} = 2, \quad y = 2 + 1\cdot e^{2} = 2 + e^{2},
z_{x}\bigl|_{(2,1,0)} = 2\cdot2 + (2+e^{2})^{3} = 4 + (2+e^{2})^{3},
z_{y}\bigl|_{(2,1,0)} = 3\cdot2\,(2+e^{2})^{2} = 6\,(2+e^{2})^{2},
\frac{\partial x}{\partial u}\Bigl|=1,\quad \frac{\partial x}{\partial v}\Bigl|=2\cdot2\cdot1=4,\quad \frac{\partial x}{\partial w}\Bigl|=0,
\frac{\partial y}{\partial u}\Bigl|=1,\quad \frac{\partial y}{\partial v}\Bigl|=e^{2},\quad \frac{\partial y}{\partial w}\Bigl|=0.

Passo 5

Aplicando a regra da cadeia: \displaystyle\frac{\partial z}{\partial u}\Bigl|_{(2,1,0)} = z_{x}\cdot1 + z_{y}\cdot1 = \bigl[4 + (2+e^{2})^{3}\bigr] + 6\,(2+e^{2})^{2} = 4 + (2+e^{2})^{3} + 6\,(2+e^{2})^{2}, \displaystyle\frac{\partial z}{\partial v}\Bigl|_{(2,1,0)} = z_{x}\cdot4 + z_{y}\cdot e^{2} = 4\bigl[4 + (2+e^{2})^{3}\bigr] + e^{2}\cdot 6\,(2+e^{2})^{2} = 16 + 4\,(2+e^{2})^{3} + 6\,e^{2}\,(2+e^{2})^{2}, \displaystyle\frac{\partial z}{\partial w}\Bigl|_{(2,1,0)} = z_{x}\cdot0 + z_{y}\cdot0 = 0.

Resposta

\displaystyle \frac{\partial z}{\partial u}\Bigl|_{(2,1,0)} = 4 + (2+e^{2})^{3} + 6\,(2+e^{2})^{2}, \displaystyle \frac{\partial z}{\partial v}\Bigl|_{(2,1,0)} = 16 + 4\,(2+e^{2})^{3} + 6\,e^{2}\,(2+e^{2})^{2}, \displaystyle \frac{\partial z}{\partial w}\Bigl|_{(2,1,0)} = 0.

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