Use a Regra da Cadeia para achar ∂ z / ∂ s e ∂ z / ∂ t.
z=e^{x+2 y}, \quad x=s / t, \quad y=t / s

Passo 1

O enunciado define
\displaystyle z = e^{\,x + 2y},
\displaystyle x = \frac{s}{t} e
\displaystyle y = \frac{t}{s},
e pede \partial z/\partial s e \partial z/\partial t.
O conteúdo aqui é a regra da cadeia em funções de várias variáveis:
\frac{\partial z}{\partial s} = \frac{\partial z}{\partial x}\,\frac{\partial x}{\partial s} + \frac{\partial z}{\partial y}\,\frac{\partial y}{\partial s}, \quad \frac{\partial z}{\partial t} = \frac{\partial z}{\partial x}\,\frac{\partial x}{\partial t} + \frac{\partial z}{\partial y}\,\frac{\partial y}{\partial t}.

Passo 2

Calculemos as derivadas parciais de z em relação a x e y. Como
\displaystyle z = e^{\,x + 2y}, definindo u = x + 2y, temos \partial z/\partial u = e^u. Logo,
\frac{\partial z}{\partial x} = e^{\,x + 2y}, \qquad \frac{\partial z}{\partial y} = 2\,e^{\,x + 2y}.

Passo 3

Agora as derivadas de x(s,t)=s/t:
\frac{\partial x}{\partial s} = \frac{1}{t}, \qquad \frac{\partial x}{\partial t} = -\frac{s}{t^2}.

Passo 4

E as derivadas de y(s,t)=t/s:
\frac{\partial y}{\partial s} = -\frac{t}{s^2}, \qquad \frac{\partial y}{\partial t} = \frac{1}{s}.

Passo 5

Aplicando a regra da cadeia para \partial z/\partial s:
\frac{\partial z}{\partial s} = e^{\,x + 2y}\cdot\frac{1}{t} + 2\,e^{\,x + 2y}\cdot\Bigl(-\frac{t}{s^2}\Bigr) = e^{\,x + 2y}\,\Bigl(\frac{1}{t} - \frac{2\,t}{s^2}\Bigr).
Substituindo x = s/t e y = t/s, fica
\frac{\partial z}{\partial s} = e^{\,\tfrac{s}{t} + \tfrac{2\,t}{s}} \Bigl(\frac{1}{t} - \frac{2\,t}{s^2}\Bigr).

Passo 6

Aplicando a regra da cadeia para \partial z/\partial t:
\frac{\partial z}{\partial t} = e^{\,x + 2y}\cdot\Bigl(-\frac{s}{t^2}\Bigr) + 2\,e^{\,x + 2y}\cdot\frac{1}{s} = e^{\,x + 2y}\,\Bigl(-\frac{s}{t^2} + \frac{2}{s}\Bigr).
Substituindo x = s/t e y = t/s, fica
\frac{\partial z}{\partial t} = e^{\,\tfrac{s}{t} + \tfrac{2\,t}{s}} \Bigl(-\frac{s}{t^2} + \frac{2}{s}\Bigr).

Resposta

\frac{\partial z}{\partial s} = e^{\,\tfrac{s}{t} + \tfrac{2\,t}{s}} \Bigl(\frac{1}{t} - \frac{2\,t}{s^2}\Bigr), \qquad \frac{\partial z}{\partial t} = e^{\,\tfrac{s}{t} + \tfrac{2\,t}{s}} \Bigl(-\frac{s}{t^2} + \frac{2}{s}\Bigr).

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