Determine as derivadas das seguintes funções:
f(x)=\frac{\sqrt[3]{\operatorname{tg}(x)}}{1+\operatorname{sec} (5 x)}

Passo 1

O enunciado pede a derivada de
f(x)=\frac{\sqrt[3]{\operatorname{tg}(x)}}{1+\operatorname{sec}(5x)}.
Aqui vamos usar a regra do quociente, a regra da cadeia e as derivadas de funções trigonométricas.

Passo 2

Definimos
u(x) = \bigl(\operatorname{tg}(x)\bigr)^{1/3}, \quad v(x) = 1 + \operatorname{sec}(5x),
de modo que
f(x)=\frac{u(x)}{v(x)}.

Passo 3

Pela regra do quociente,
f'(x) =\frac{u'(x)\,v(x)-u(x)\,v'(x)}{\bigl(v(x)\bigr)^2}.

Passo 4

Calculamos u'(x). Como u(x)=(\operatorname{tg} x)^{1/3}, pela regra da cadeia:
u'(x) =\frac{1}{3}\,(\operatorname{tg} x)^{-2/3}\,\frac{d}{dx}\bigl(\operatorname{tg} x\bigr) =\frac{1}{3}\,(\operatorname{tg} x)^{-2/3}\,\operatorname{sec}^2 x.

Passo 5

Agora v'(x). Como v(x)=1+\operatorname{sec}(5x), temos
v'(x) =\frac{d}{dx}\bigl(\operatorname{sec}(5x)\bigr) =\operatorname{sec}(5x)\,\tan(5x)\,\frac{d}{dx}(5x) =5\,\operatorname{sec}(5x)\,\tan(5x).

Passo 6

Substituindo em f'(x):
f'(x) =\frac{\displaystyle \frac{1}{3}(\operatorname{tg} x)^{-2/3}\operatorname{sec}^2 x\,(1+\operatorname{sec}(5x)) -\bigl(\operatorname{tg} x\bigr)^{1/3}\,5\,\operatorname{sec}(5x)\,\tan(5x) }{\displaystyle \bigl(1+\operatorname{sec}(5x)\bigr)^2}.

Passo 7

Podemos deixar a resposta nesse formato ou, se preferir, escrever
f'(x) =\frac{(\operatorname{tg} x)^{-2/3}\operatorname{sec}^2 x\,(1+\operatorname{sec}(5x)) -15\,(\operatorname{tg} x)^{1/3}\,\operatorname{sec}(5x)\,\tan(5x) }{3\,(1+\operatorname{sec}(5x))^2}.

Resposta

f'(x) =\frac{\dfrac{1}{3}(\operatorname{tg} x)^{-2/3}\operatorname{sec}^2 x\,(1+\operatorname{sec}(5x)) -5\,(\operatorname{tg} x)^{1/3}\,\operatorname{sec}(5x)\,\tan(5x) }{(1+\operatorname{sec}(5x))^2}.

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