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

Passo 1

O enunciado define
P = \sqrt{u^{2} + v^{2} + w^{2}},
onde
u = x e^{y},\quad v = y e^{x},\quad w = e^{x},
e pede as derivadas parciais \frac{\partial P}{\partial x} e \frac{\partial P}{\partial y} no ponto (x,y)=(0,2).
O conteúdo aqui é a Regra da Cadeia em múltiplas variáveis: quando P depende de (u,v,w) e cada uma dessas depende de (x,y), temos
\frac{\partial P}{\partial x} = \frac{\partial P}{\partial u}\,\frac{\partial u}{\partial x} + \frac{\partial P}{\partial v}\,\frac{\partial v}{\partial x} + \frac{\partial P}{\partial w}\,\frac{\partial w}{\partial x},
e análogo para \partial P/\partial y.

Passo 2

Primeiro, note que
P(u,v,w) = \sqrt{u^{2}+v^{2}+w^{2}}
implica
\frac{\partial P}{\partial u} = \frac{u}{\sqrt{u^{2}+v^{2}+w^{2}}}, \quad \frac{\partial P}{\partial v} = \frac{v}{\sqrt{u^{2}+v^{2}+w^{2}}}, \quad \frac{\partial P}{\partial w} = \frac{w}{\sqrt{u^{2}+v^{2}+w^{2}}}.

Passo 3

Agora computamos as derivadas de u,v,w em relação a x e a y:
\frac{\partial u}{\partial x} = e^{y},\quad \frac{\partial u}{\partial y} = x e^{y},
\frac{\partial v}{\partial x} = y e^{x},\quad \frac{\partial v}{\partial y} = e^{x},
\frac{\partial w}{\partial x} = e^{x},\quad \frac{\partial w}{\partial y} = 0.
Pela Regra da Cadeia,
\frac{\partial P}{\partial x} = \frac{u\,u_x + v\,v_x + w\,w_x}{\sqrt{u^{2}+v^{2}+w^{2}}}, \quad \frac{\partial P}{\partial y} = \frac{u\,u_y + v\,v_y + w\,w_y}{\sqrt{u^{2}+v^{2}+w^{2}}}.

Passo 4

Avaliamo tudo em (x,y)=(0,2). Temos
u(0,2)=0\cdot e^{2}=0,\quad v(0,2)=2\cdot e^{0}=2,\quad w(0,2)=e^{0}=1,
logo
P(0,2)=\sqrt{0^{2}+2^{2}+1^{2}}=\sqrt{5}.

Passo 5

No ponto dado,
\frac{\partial P}{\partial x}\Big|_{(0,2)} = \frac{0\cdot e^{2} \;+\; 2\cdot(2e^{0}) \;+\; 1\cdot e^{0}}{\sqrt{5}} = \frac{0 + 4 + 1}{\sqrt{5}} = \frac{5}{\sqrt{5}} = \sqrt{5},
\frac{\partial P}{\partial y}\Big|_{(0,2)} = \frac{0\cdot(0\cdot e^{2}) \;+\; 2\cdot e^{0} \;+\; 1\cdot 0}{\sqrt{5}} = \frac{0 + 2 + 0}{\sqrt{5}} = \frac{2}{\sqrt{5}}.

Resposta

\displaystyle \frac{\partial P}{\partial x}(\,0,2\,)=\sqrt{5} \displaystyle \frac{\partial P}{\partial y}(\,0,2\,)=\frac{2}{\sqrt{5}}

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