Usando as regras de derivação apropriadas, calcule cada uma das derivadas solicitadas.
(b) A terceira derivada de g(x)=\operatorname{sec} (x);

Passo 1

O enunciado pede a terceira derivada de g(x)=\operatorname{sec}(x). Para isso vamos usar as regras de derivação de funções trigonométricas — em particular
  • \displaystyle\frac{d}{dx}\bigl[\operatorname{sec}(x)\bigr]=\operatorname{sec}(x)\tan(x),
  • \displaystyle\frac{d}{dx}\bigl[\tan(x)\bigr]=\operatorname{sec}^2(x),
    e também as regras do produto e da cadeia.

Passo 2

Primeiro, calculamos
g'(x)=\frac{d}{dx}\bigl[\operatorname{sec}(x)\bigr] =\operatorname{sec}(x)\,\tan(x).

Passo 3

Agora, aplicamos a regra do produto a g'(x)=\operatorname{sec}(x)\tan(x):
g''(x)=\frac{d}{dx}\bigl[\operatorname{sec}(x)\bigr]\tan(x) +\operatorname{sec}(x)\,\frac{d}{dx}\bigl[\tan(x)\bigr] =\bigl[\operatorname{sec}(x)\tan(x)\bigr]\tan(x) +\operatorname{sec}(x)\,\operatorname{sec}^2(x).
Isso dá
g''(x)=\operatorname{sec}(x)\,\tan^2(x)+\operatorname{sec}^3(x) =\operatorname{sec}(x)\bigl(\tan^2(x)+\operatorname{sec}^2(x)\bigr) =\operatorname{sec}(x)\bigl(2\operatorname{sec}^2(x)-1\bigr),
já que \tan^2(x)+1=\operatorname{sec}^2(x).

Passo 4

Finalmente, derivamos
g''(x)=2\operatorname{sec}^3(x)-\operatorname{sec}(x).
Pela regra da cadeia,
\frac{d}{dx}\bigl[\operatorname{sec}^3(x)\bigr] =3\operatorname{sec}^2(x)\cdot\frac{d}{dx}\bigl[\operatorname{sec}(x)\bigr] =3\operatorname{sec}^2(x)\bigl[\operatorname{sec}(x)\tan(x)\bigr] =3\operatorname{sec}^3(x)\tan(x).
Logo,
g'''(x) =\frac{d}{dx}\bigl[2\operatorname{sec}^3(x)\bigr] -\frac{d}{dx}\bigl[\operatorname{sec}(x)\bigr] =2\cdot3\operatorname{sec}^3(x)\tan(x) -\operatorname{sec}(x)\tan(x) =6\operatorname{sec}^3(x)\tan(x)-\operatorname{sec}(x)\tan(x).
Podemos fatorar \operatorname{sec}(x)\tan(x):
g'''(x)=\operatorname{sec}(x)\,\tan(x)\bigl(6\operatorname{sec}^2(x)-1\bigr).

Resposta

g'''(x)=\operatorname{sec}(x)\,\tan(x)\bigl(6\operatorname{sec}^2(x)-1\bigr).

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