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}.
    (e) z=\operatorname{cosh} \left(\frac{y}{x}\right), x=3 t^{2} s, y=6 t e^{3}

Passo 1

O enunciado define
z = \operatorname{cosh}\!\bigl(\tfrac{y}{x}\bigr), \quad x = 3\,t^{2}s, \quad y = 6\,t\,e^{3},
e pede \frac{\partial z}{\partial t} e \frac{\partial z}{\partial s}.
Vamos usar a regra da cadeia para funções de duas variáveis:
\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},
\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}.

Passo 2

Calculemos \frac{\partial z}{\partial x} e \frac{\partial z}{\partial y}.
Como z = \operatorname{cosh}(u) com u = y/x, temos
\frac{dz}{du} = \operatorname{sinh}(u),
\frac{\partial u}{\partial x} = \frac{\partial}{\partial x}\Bigl(\tfrac{y}{x}\Bigr) = -\,\frac{y}{x^{2}}, \qquad \frac{\partial u}{\partial y} = \frac{1}{x}.
Logo, por cadeia,
\frac{\partial z}{\partial x} = \operatorname{sinh}\!\bigl(\tfrac{y}{x}\bigr)\,\Bigl(-\frac{y}{x^{2}}\Bigr), \quad \frac{\partial z}{\partial y} = \operatorname{sinh}\!\bigl(\tfrac{y}{x}\bigr)\,\frac{1}{x}.

Passo 3

Agora \frac{\partial x}{\partial t},\frac{\partial x}{\partial s},\frac{\partial y}{\partial t},\frac{\partial y}{\partial s}.
Como x = 3\,t^{2}s e y=6\,t\,e^{3},
\frac{\partial x}{\partial t} = 6\,t\,s, \quad \frac{\partial x}{\partial s} = 3\,t^{2}, \quad \frac{\partial y}{\partial t} = 6\,e^{3}, \quad \frac{\partial y}{\partial s} = 0.

Passo 4

Aplique a regra da cadeia:
\frac{\partial z}{\partial t} = \Bigl(-\frac{y}{x^{2}}\operatorname{sinh}\!\tfrac{y}{x}\Bigr)\,(6\,t\,s) + \Bigl(\frac{1}{x}\operatorname{sinh}\!\tfrac{y}{x}\Bigr)\,(6\,e^{3}),
\frac{\partial z}{\partial s} = \Bigl(-\frac{y}{x^{2}}\operatorname{sinh}\!\tfrac{y}{x}\Bigr)\,(3\,t^{2}) + \Bigl(\frac{1}{x}\operatorname{sinh}\!\tfrac{y}{x}\Bigr)\,(0).

Resposta

\frac{\partial z}{\partial t} = \operatorname{sinh}\!\bigl(\tfrac{y}{x}\bigr)\,\Bigl(-\frac{6\,t\,s\,y}{x^{2}} + \frac{6\,e^{3}}{x}\Bigr), \qquad \frac{\partial z}{\partial s} = -\,\frac{3\,t^{2}\,y}{x^{2}}\;\operatorname{sinh}\!\bigl(\tfrac{y}{x}\bigr).

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