Para cada um dos itens abaixo, calcule F^{\prime}(t).
F(t)=\left(\operatorname{tg}\left(\frac{1}{(t+2)^{2}}\right), \operatorname{sec} \left(\operatorname{cotg}\left(e^{t+1}\right)\right)\right)

Passo 1

O enunciado define
F(t)=\bigl(\operatorname{tg}\bigl((t+2)^{-2}\bigr),\;\operatorname{sec}\bigl(\operatorname{cotg}(e^{t+1})\bigr)\bigr).
Queremos F'(t), ou seja, as derivadas das duas componentes. Vamos usar a regra da cadeia e as fórmulas:
  1. \dfrac{d}{dx}\,\operatorname{tg}(x)=\operatorname{sec}^2(x).
  2. \dfrac{d}{dx}\,\operatorname{sec}(x)=\operatorname{sec}(x)\tan(x).
  3. \dfrac{d}{dx}\,\operatorname{cotg}(x)=-\csc^2(x).
  4. \dfrac{d}{dt}e^{t+1}=e^{t+1}.

Passo 2

Primeira componente:
f_1(t)=\operatorname{tg}\bigl(u(t)\bigr), \quad u(t)=(t+2)^{-2}.
Aplicando regra da cadeia:
f_1'(t)=\operatorname{sec}^2\bigl(u(t)\bigr)\,u'(t).
Agora
u'(t)=\frac{d}{dt}(t+2)^{-2}=-2\,(t+2)^{-3}.
Logo
f_1'(t)=-2\,(t+2)^{-3}\,\operatorname{sec}^2\!\bigl((t+2)^{-2}\bigr).

Passo 3

Segunda componente:
f_2(t)=\operatorname{sec}\bigl(w(t)\bigr), \quad w(t)=\operatorname{cotg}\bigl(z(t)\bigr), \quad z(t)=e^{t+1}.
De novo, regra da cadeia em camadas:
f_2'(t)=\operatorname{sec}\bigl(w(t)\bigr)\,\tan\bigl(w(t)\bigr)\,w'(t).
Em seguida,
w'(t)=\frac{d}{dt}\operatorname{cotg}(z) =-\csc^2(z)\,z'(t) =-\csc^2\bigl(e^{t+1}\bigr)\,e^{t+1}.
Assim
f_2'(t)=-e^{t+1}\,\operatorname{sec}\bigl(\operatorname{cotg}(e^{t+1})\bigr)\,\tan\bigl(\operatorname{cotg}(e^{t+1})\bigr)\,\csc^2\bigl(e^{t+1}\bigr).

Resposta

F'(t)=\Bigl( -2\,(t+2)^{-3}\,\operatorname{sec}^2\!\bigl((t+2)^{-2}\bigr),\; -e^{t+1}\,\operatorname{sec}\bigl(\operatorname{cotg}(e^{t+1})\bigr)\,\tan\bigl(\operatorname{cotg}(e^{t+1})\bigr)\,\csc^2\bigl(e^{t+1}\bigr) \Bigr).

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