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}.
    (g) z=\operatorname{arcsen}(3 x+y), x=s^{2}, y=\operatorname{sen}(s t)

Passo 1

O enunciado dá
z = \operatorname{arcsen}(3x + y), \quad x = s^{2}, \quad y = \operatorname{sen}(s t),
e pede as derivadas parciais \frac{\partial z}{\partial s} e \frac{\partial z}{\partial t}.
O conteúdo necessário é a regra da cadeia para função de duas 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

Calculamos primeiro as derivadas de z em relação a x e y. Como
z = \operatorname{arcsen}(u) \quad\text{com}\quad u = 3x + y,
temos
\frac{d}{du}\bigl(\operatorname{arcsen}(u)\bigr) = \frac{1}{\sqrt{1 - u^{2}}}
e
\frac{\partial u}{\partial x} = 3, \quad \frac{\partial u}{\partial y} = 1.
Logo
\frac{\partial z}{\partial x} = \frac{3}{\sqrt{1 - (3x+y)^{2}}}, \qquad \frac{\partial z}{\partial y} = \frac{1}{\sqrt{1 - (3x+y)^{2}}}.

Passo 3

Agora as derivadas de x e y em função de s,t:
x = s^{2} \implies \frac{\partial x}{\partial s} = 2s, \quad \frac{\partial x}{\partial t} = 0.
y = \operatorname{sen}(s t) \implies \frac{\partial y}{\partial s} = t\cos(s t), \quad \frac{\partial y}{\partial t} = s\cos(s t).

Passo 4

Aplicando a regra da cadeia:
\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} = \frac{3\,(2s) + 1\,(t\cos(s t))}{\sqrt{1 - (3x+y)^{2}}},
\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{0 + s\cos(s t)}{\sqrt{1 - (3x+y)^{2}}}.
Substituindo x = s^{2} e y = \operatorname{sen}(s t) no denominador, temos as respostas finais.

Resposta

\frac{\partial z}{\partial s} = \frac{6s + t\cos(s t)}{\sqrt{1 - \bigl(3s^{2} + \operatorname{sen}(s t)\bigr)^{2}}}, \qquad \frac{\partial z}{\partial t} = \frac{s\cos(s t)}{\sqrt{1 - \bigl(3s^{2} + \operatorname{sen}(s t)\bigr)^{2}}}.

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