Exercício 9: (Derivada implícita) Determine a derivada:
-e^{-x} \operatorname{sec} x^{2}\left[1-2 x \operatorname{tg} x^{2}\right]

Passo 1

O enunciado dá a função
f(x) = -e^{-x}\,\operatorname{sec}\bigl(x^{2}\bigr)\,\bigl[1-2x\tan\bigl(x^{2}\bigr)\bigr]
e pede f'(x). Vamos usar regra do produto (para três fatores) e regra da cadeia (para e^{-x}, \operatorname{sec}(x^{2}) e \tan(x^{2})).

Passo 2

Identifiquemos os três fatores:
u(x) = -e^{-x},\quad v(x) = \operatorname{sec}\bigl(x^{2}\bigr),\quad w(x) = 1 - 2x\,\tan\bigl(x^{2}\bigr).
Então f(x)=u(x)\,v(x)\,w(x).

Passo 3

Calculamos u'(x). Como u(x)=-e^{-x} e \tfrac{d}{dx}e^{-x}=-e^{-x}, vem
u'(x) = -\bigl(-e^{-x}\bigr) = e^{-x}.

Passo 4

Agora v'(x). Sabe-se que \tfrac{d}{du}\operatorname{sec} u = \operatorname{sec} u\tan u e, por cadeia, com u=x^{2} e u'=2x,
v'(x) = \operatorname{sec}\bigl(x^{2}\bigr)\,\tan\bigl(x^{2}\bigr)\cdot 2x = 2x\,\operatorname{sec}\bigl(x^{2}\bigr)\,\tan\bigl(x^{2}\bigr).

Passo 5

Finalmente w'(x). Escrevendo w(x)=1 - 2x\,T(x) onde T(x)=\tan(x^{2}), temos
w'(x) = -2\,T(x) \;-\;2x\,T'(x).
E T'(x)=\operatorname{sec}^{2}(x^{2})\cdot2x. Logo
w'(x) = -2\,\tan\bigl(x^{2}\bigr)\;-\;2x\bigl[2x\,\operatorname{sec}^{2}\bigl(x^{2}\bigr)\bigr] = -2\tan\bigl(x^{2}\bigr)\;-\;4x^{2}\operatorname{sec}^{2}\bigl(x^{2}\bigr).

Passo 6

Aplicando a regra do produto para três fatores:
f'(x)=u'\,v\,w \;+\;u\,v'\,w \;+\;u\,v\,w'.
Substituindo cada derivada,
\begin{aligned} f'(x)=&\;e^{-x}\,\operatorname{sec}(x^{2})\bigl[1-2x\tan(x^{2})\bigr]\\\ &\;+\bigl(-e^{-x}\bigr)\,\bigl[2x\,\operatorname{sec}(x^{2})\tan(x^{2})\bigr]\bigl[1-2x\tan(x^{2})\bigr]\\\ &\;+\bigl(-e^{-x}\bigr)\,\operatorname{sec}(x^{2})\bigl[-2\tan(x^{2})-4x^{2}\operatorname{sec}^{2}(x^{2})\bigr]. \end{aligned}

Passo 7

Fatorando e^{-x}\operatorname{sec}(x^{2}),
f'(x)=e^{-x}\,\operatorname{sec}(x^{2}) \bigl\{ [1-2x\tan(x^{2})] -2x\tan(x^{2})[1-2x\tan(x^{2})] +[2\tan(x^{2})+4x^{2}\operatorname{sec}^{2}(x^{2})] \bigr\}.

Passo 8

Somando os termos dentro das chaves:
\begin{aligned} &1-2x\tan+(-2x\tan+4x^{2}\tan^{2})+(2\tan+4x^{2}\operatorname{sec}^{2})\\\ &\quad=1 +\bigl(-4x\tan+2\tan\bigr) +4x^{2}\tan^{2}+4x^{2}\operatorname{sec}^{2}\\\ &\quad=1+2\tan(x^{2})(1-2x)+4x^{2}\bigl[\tan^{2}(x^{2})+\operatorname{sec}^{2}(x^{2})\bigr]. \end{aligned}

Resposta

f'(x)=e^{-x}\,\operatorname{sec}\bigl(x^{2}\bigr)\Bigl\{1+2\tan\bigl(x^{2}\bigr)(1-2x) +4x^{2}\bigl[\tan^{2}\bigl(x^{2}\bigr)+\operatorname{sec}^{2}\bigl(x^{2}\bigr)\bigr]\Bigr\}.

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