Para cada um dos itens abaixo, calcule F^{\prime}(t).
F(t)=\left(\operatorname{sec} ^{3}\left(\sqrt[4]{2^{t+1}}\right), \operatorname{cossec}^{4}\left(3^{t^{2}+1}\right)\right)

Passo 1

O enunciado nos dá a função vetorial
F(t)=\bigl(f(t),\,g(t)\bigr)
com
f(t)=\operatorname{sec}^{3}\bigl(\sqrt[4]{2^{\,t+1}}\bigr) \quad\text{e}\quad g(t)=\csc^{4}\bigl(3^{\,t^{2}+1}\bigr).
Para calcular F'(t), vamos usar a regra da cadeia para funções compostas, a regra da potência e as derivadas conhecidas
\frac{d}{dx}\operatorname{sec} x=\operatorname{sec} x\tan x, \quad \frac{d}{dx}\csc x=-\csc x\cot x, \quad \frac{d}{dx}a^{u(x)}=a^{u(x)}\ln(a)\,u'(x).

Passo 2

Definimos
u(t)=\sqrt[4]{2^{\,t+1}}=2^{\frac{t+1}{4}}.
Aplicando a derivada de 2^{u},
u'(t) =2^{\frac{t+1}{4}}\ln(2)\,\frac{d}{dt}\!\Bigl(\tfrac{t+1}{4}\Bigr) =2^{\frac{t+1}{4}}\ln(2)\,\frac{1}{4} =\frac{\ln(2)}{4}\,2^{\frac{t+1}{4}}.

Passo 3

Agora derivamos
f(t)=\operatorname{sec}^{3}\bigl(u(t)\bigr).
Pela regra da potência e da cadeia,
f'(t) =3\,\operatorname{sec}^{2}\bigl(u(t)\bigr)\,\frac{d}{dt}\bigl[\operatorname{sec}(u(t))\bigr] =3\,\operatorname{sec}^{2}(u)\,\bigl[\operatorname{sec}(u)\tan(u)\bigr]\,u'(t) =3\,u'(t)\,\operatorname{sec}^{3}(u)\,\tan(u).
Substituindo u e u',
f'(t) =\frac{3\ln(2)}{4}\,2^{\frac{t+1}{4}} \;\operatorname{sec}^{3}\!\bigl(\sqrt[4]{2^{\,t+1}}\bigr) \;\tan\!\bigl(\sqrt[4]{2^{\,t+1}}\bigr).

Passo 4

Definimos
v(t)=3^{\,t^{2}+1}.
Derivando v por t,
v'(t) =3^{\,t^{2}+1}\ln(3)\,\frac{d}{dt}(t^{2}+1) =3^{\,t^{2}+1}\ln(3)\cdot2t =2t\ln(3)\,3^{\,t^{2}+1}.

Passo 5

Derivamos
g(t)=\csc^{4}\bigl(v(t)\bigr).
Pela regra da potência e da cadeia,
g'(t) =4\,\csc^{3}(v)\,\frac{d}{dt}\bigl[\csc(v)\bigr] =4\,\csc^{3}(v)\,[-\csc(v)\cot(v)]\,v'(t) =-4\,v'(t)\,\csc^{4}(v)\,\cot(v).
Substituindo v e v',
g'(t) =-4\bigl(2t\ln(3)\,3^{\,t^{2}+1}\bigr) \;\csc^{4}\!\bigl(3^{\,t^{2}+1}\bigr) \;\cot\!\bigl(3^{\,t^{2}+1}\bigr) =-8t\ln(3)\,3^{\,t^{2}+1} \;\csc^{4}\!\bigl(3^{\,t^{2}+1}\bigr) \;\cot\!\bigl(3^{\,t^{2}+1}\bigr).

Resposta

F'(t) =\bigl(f'(t),\,g'(t)\bigr) =\!\Bigl(\, \frac{3\ln(2)}{4}\,2^{\tfrac{t+1}{4}} \operatorname{sec}^{3}\!\bigl(\sqrt[4]{2^{\,t+1}}\bigr) \tan\!\bigl(\sqrt[4]{2^{\,t+1}}\bigr), \;-8t\ln(3)\,3^{\,t^{2}+1} \csc^{4}\!\bigl(3^{\,t^{2}+1}\bigr) \cot\!\bigl(3^{\,t^{2}+1}\bigr) \Bigr).

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