1. Usando a regra da cadeia para z=f(x, y) e w=f(x, y, z), calcule: \frac{\partial z}{\partial t}, \frac{\partial z}{\partial s} e \frac{\partial w}{\partial t}, \frac{\partial w}{\partial s} e \frac{\partial w}{\partial r}.
    (h) w=x e^{y}, x=\operatorname{arctg}(r s t), y=\ln (3 r s+5 s t)

Passo 1

O enunciado dá a função
w = x e^{y},
em que
x = \arctan(r s t), \qquad y = \ln\bigl(3 r s + 5 s t\bigr).
O objetivo é usar a regra da cadeia para obter \frac{\partial w}{\partial r}, \frac{\partial w}{\partial s} e \frac{\partial w}{\partial t}. Precisamos das derivadas parciais de w em relação a x e y, e das derivadas de x e y em relação a r,s,t.

Passo 2

Calculamos primeiro
\frac{\partial w}{\partial x} = e^{y}, \qquad \frac{\partial w}{\partial y} = x e^{y}.
Isto vem de derivar w=x e^{y} tratando y ou x como constante.

Passo 3

Agora derivamos
x = \arctan(r s t).
Como \displaystyle\frac{d}{du}\arctan(u)=\frac1{1+u^2}, temos
\frac{\partial x}{\partial r} = \frac{1}{1+(r s t)^{2}}\,(s t), \quad \frac{\partial x}{\partial s} = \frac{1}{1+(r s t)^{2}}\,(r t), \quad \frac{\partial x}{\partial t} = \frac{1}{1+(r s t)^{2}}\,(r s).

Passo 4

Em seguida,
y = \ln\bigl(3 r s + 5 s t\bigr).
Como \displaystyle\frac{d}{du}\ln(u)=\frac1u, definindo u=3rs+5st, obtemos
\frac{\partial y}{\partial r} = \frac{3s}{3rs+5st}, \quad \frac{\partial y}{\partial s} = \frac{3r+5t}{3rs+5st}, \quad \frac{\partial y}{\partial t} = \frac{5s}{3rs+5st}.

Passo 5

Aplicando a regra da cadeia
\frac{\partial w}{\partial u} = \frac{\partial w}{\partial x}\,\frac{\partial x}{\partial u} + \frac{\partial w}{\partial y}\,\frac{\partial y}{\partial u},
para u=r,s,t, e lembrando que \frac{\partial w}{\partial x}=e^{y} e \frac{\partial w}{\partial y}=x e^{y}, temos:
\frac{\partial w}{\partial r} = e^{y}\,\frac{s t}{1+(r s t)^{2}} + x\,e^{y}\,\frac{3s}{3rs+5st},
\frac{\partial w}{\partial s} = e^{y}\,\frac{r t}{1+(r s t)^{2}} + x\,e^{y}\,\frac{3r+5t}{3rs+5st},
\frac{\partial w}{\partial t} = e^{y}\,\frac{r s}{1+(r s t)^{2}} + x\,e^{y}\,\frac{5s}{3rs+5st}.

Resposta

\frac{\partial w}{\partial r} = e^{y}\,\frac{s t}{1+(r s t)^{2}} + x\,e^{y}\,\frac{3s}{3rs+5st},
\frac{\partial w}{\partial s} = e^{y}\,\frac{r t}{1+(r s t)^{2}} + x\,e^{y}\,\frac{3r+5t}{3rs+5st},
\frac{\partial w}{\partial t} = e^{y}\,\frac{r s}{1+(r s t)^{2}} + x\,e^{y}\,\frac{5s}{3rs+5st}.

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