Exercício 3.13 (Resolvido) Calcule, com detalhamento, as derivadas de
f(x)=\operatorname{tg}\left(\cos \left(\operatorname{sen}^{4}(x)\right)\right)

Passo 1

O enunciado pede a derivada de
f(x)=\tan\!\bigl(\cos\bigl(\sin^4(x)\bigr)\bigr).
Conteúdo: regra da cadeia aplicada três vezes, usando que \frac{d}{du}\tan(u)=\operatorname{sec}^2(u), \frac{d}{du}\cos(u)=-\sin(u) e \frac{d}{dx}\sin^4(x)=4\sin^3(x)\cos(x).

Passo 2

Aplicamos a regra da cadeia à função externa \tan(u) com u(x)=\cos(\sin^4(x)). Logo,
f'(x) =\operatorname{sec}^2\bigl(u(x)\bigr)\cdot u'(x) =\operatorname{sec}^2\!\bigl(\cos(\sin^4(x))\bigr)\cdot \frac{d}{dx}\bigl[\cos(\sin^4(x))\bigr].

Passo 3

Agora derivamos u(x)=\cos(v(x)) com v(x)=\sin^4(x). Pela regra da cadeia,
\frac{d}{dx}\bigl[\cos(v(x))\bigr] =-\sin\bigl(v(x)\bigr)\cdot v'(x) =-\sin\!\bigl(\sin^4(x)\bigr)\cdot \frac{d}{dx}\bigl[\sin^4(x)\bigr].

Passo 4

Derivada de v(x)=\sin^4(x) é
v'(x)=4\sin^3(x)\cos(x).

Passo 5

Juntando tudo, temos
f'(x) =\operatorname{sec}^2\!\bigl(\cos(\sin^4(x))\bigr) \;\times\; \Bigl[-\sin\!\bigl(\sin^4(x)\bigr)\Bigr] \;\times\; \bigl[4\sin^3(x)\cos(x)\bigr].
Ou seja,
f'(x) =-4\,\sin^3(x)\,\cos(x)\,\sin\!\bigl(\sin^4(x)\bigr)\, \operatorname{sec}^2\!\bigl(\cos(\sin^4(x))\bigr).

Resposta

f'(x) =-4\,\sin^3(x)\,\cos(x)\,\sin\!\bigl(\sin^4(x)\bigr)\, \operatorname{sec}^2\!\bigl(\cos(\sin^4(x))\bigr).

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