Determine:
\frac{d R}{d x}, \frac{d R}{d y}, quando x=3, y=-1, onde R=u e^{v-w^{2}}, u=x+2 y, v=x-y, w=x+y.

Passo 1

O enunciado pede as derivadas parciais de
R = u\,e^{v - w^2}
com
u = x + 2y,\quad v = x - y,\quad w = x + y
avaliadas em (x,y)=(3,-1). Vamos usar regra do produto e regra da cadeia para achar \frac{\partial R}{\partial x} e \frac{\partial R}{\partial y}.

Passo 2

Defina g = v - w^2. Então
R = u\,e^{g}.
Usando a regra do produto e a cadeia:
\frac{\partial R}{\partial x} = \frac{\partial u}{\partial x}\,e^{g} + u\,e^{g}\,\frac{\partial g}{\partial x}, \qquad \frac{\partial R}{\partial y} = \frac{\partial u}{\partial y}\,e^{g} + u\,e^{g}\,\frac{\partial g}{\partial y}.
Agora,
\frac{\partial u}{\partial x}=1,\quad \frac{\partial u}{\partial y}=2,
\frac{\partial v}{\partial x}=1,\quad \frac{\partial v}{\partial y}=-1,
\frac{\partial w}{\partial x}=1,\quad \frac{\partial w}{\partial y}=1.
Logo,
\frac{\partial g}{\partial x} = \frac{\partial v}{\partial x} - 2w\frac{\partial w}{\partial x} = 1 - 2w,
\frac{\partial g}{\partial y} = \frac{\partial v}{\partial y} - 2w\frac{\partial w}{\partial y} = -1 - 2w.
Substituindo em cada derivada:
\frac{\partial R}{\partial x} = e^{g} + u\,e^{g}\,(1 - 2w) = e^{g}\bigl[1 + u(1 - 2w)\bigr],
\frac{\partial R}{\partial y} = 2\,e^{g} + u\,e^{g}\,(-1 - 2w) = e^{g}\bigl[2 + u(-1 - 2w)\bigr].

Passo 3

Avalie as variáveis em (x,y)=(3,-1):
u = 3 + 2(-1) = 1,\quad v = 3 - (-1) = 4,\quad w = 3 + (-1) = 2,
g = v - w^2 = 4 - 4 = 0, \qquad e^{g} = e^{0} = 1.

Passo 4

Substituindo estes valores nas expressões de \frac{\partial R}{\partial x} e \frac{\partial R}{\partial y}:
\frac{\partial R}{\partial x}(3,-1) = 1\bigl[1 + 1\cdot(1 - 2\cdot2)\bigr] = 1\bigl[1 + (1 - 4)\bigr] = 1\cdot(-2) = -2,
\frac{\partial R}{\partial y}(3,-1) = 1\bigl[2 + 1\cdot(-1 - 2\cdot2)\bigr] = 1\bigl[2 + (-1 - 4)\bigr] = 1\cdot(-3) = -3.

Resposta

\displaystyle \frac{\partial R}{\partial x}(3,-1) = -2,\quad \frac{\partial R}{\partial y}(3,-1) = -3.

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