1. Seja
\begin{gathered} u=\frac{t+e^{x}}{y-e^{t}}, \\\ x=3 \operatorname{sen}\left(t^{2}+2 t\right), \\\ y=\cos \left(t^{3}\right) . \end{gathered}
Utilizando a regra da cadeia, calcule
\frac{\partial u}{\partial t} .

Passo 1

O enunciado define
u=\frac{t+e^{x}}{y-e^{t}}, \quad x=3\sin\bigl(t^{2}+2t\bigr), \quad y=\cos\bigl(t^{3}\bigr),
e pede \partial u/\partial t. Vamos usar a regra do quociente junto com a regra da cadeia.

Passo 2

Defina o numerador N e o denominador D de u:
N(t)=t+e^{x},\qquad D(t)=y-e^{t}.
Pela regra do quociente,
\frac{\partial u}{\partial t} =\frac{N'\,D - N\,D'}{D^{2}}.

Passo 3

Calcular N'=\frac{d}{dt}(t+e^{x}). Temos
\frac{d}{dt}(t)=1, \quad \frac{d}{dt}(e^{x})=e^{x}\,\frac{dx}{dt},
logo
N'=1 + e^{x}\,\frac{dx}{dt}.

Passo 4

Aplicar regra da cadeia em x=3\sin(t^{2}+2t). Como \frac{d}{dt}\sin(\varphi)=\cos(\varphi)\,\varphi',
\frac{dx}{dt} =3\cos\bigl(t^{2}+2t\bigr)\,\frac{d}{dt}(t^{2}+2t) =3\cos\bigl(t^{2}+2t\bigr)\,(2t+2).

Passo 5

Calcular D'=\frac{d}{dt}(y-e^{t}). Temos
\frac{d}{dt}(e^{t})=e^{t}, \quad \frac{dy}{dt}=\frac{d}{dt}\cos(t^{3}) =-\sin(t^{3})\cdot 3t^{2}=-3t^{2}\sin(t^{3}).
Logo
D'=-3t^{2}\sin(t^{3}) - e^{t}.

Passo 6

Substituindo N', D, N e D' na regra do quociente:
\frac{\partial u}{\partial t} =\frac{\bigl[1 + e^{x}\,(3\cos(t^{2}+2t)\,(2t+2))\bigr]\,(y - e^{t}) -\bigl(t+e^{x}\bigr)\,\bigl(-3t^{2}\sin(t^{3}) - e^{t}\bigr)}{(y - e^{t})^{2}}.

Resposta

\frac{\partial u}{\partial t} =\frac{\bigl(1 + 3e^{x}\cos(t^{2}+2t)\,(2t+2)\bigr)\,(y - e^{t}) + (t + e^{x})\,(3t^{2}\sin(t^{3}) + e^{t})}{\bigl(y - e^{t}\bigr)^{2}}.

Exercícios relacionados