Questão.(2,0 pts) Calcule.
F^{\prime}\left(\frac{\pi}{6}\right) se F(x)=\frac{\operatorname{sen}(G(x))}{\cos 2 x}, onde G é diferenciável, G\left(\frac{\pi}{6}\right)=\frac{\pi}{3} e G^{\prime}\left(\frac{\pi}{6}\right)=-3.

Passo 1

O enunciado dá
F(x)=\dfrac{\operatorname{sen}\bigl(G(x)\bigr)}{\cos 2x}
e pede o valor de F'\bigl(\tfrac{\pi}{6}\bigr), sabendo que G é diferenciável,
G\bigl(\tfrac{\pi}{6}\bigr)=\tfrac{\pi}{3},\qquad G'\bigl(\tfrac{\pi}{6}\bigr)=-3.
Vamos usar a regra do quociente e a regra da cadeia para derivar.

Passo 2

Defina
u(x)=\operatorname{sen}\bigl(G(x)\bigr), \quad v(x)=\cos 2x.
Pela regra do quociente,
F'(x) =\frac{u'(x)\,v(x)-u(x)\,v'(x)}{v(x)^{2}}.

Passo 3

Calcule u'(x) e v'(x):
Usando que a derivada de \operatorname{sen}(G(x)) é \cos(G(x))\cdot G'(x), temos
u'(x)=\cos\bigl(G(x)\bigr)\,G'(x).
E como \dfrac{d}{dx}\cos 2x=-\sen2x\cdot2,
v'(x)=-2\,\sen2x.

Passo 4

Substituindo em F'(x):
F'(x) =\frac{\cos\bigl(G(x)\bigr)\,G'(x)\,\cos2x \;-\;\operatorname{sen}\bigl(G(x)\bigr)\,(-2\sen2x)} {\cos^{2}2x} =\frac{\cos\bigl(G(x)\bigr)\,G'(x)\,\cos2x +2\,\operatorname{sen}\bigl(G(x)\bigr)\,\sen2x} {\cos^{2}2x}.

Passo 5

Agora avalie em x=\tfrac{\pi}{6}. Sabemos
\;G(\tfrac{\pi}{6})=\tfrac{\pi}{3}\implies\cos(G)=\tfrac12,\\ \operatorname{sen}(G)=\tfrac{\sqrt3}{2},
\;\cos2x=\cos\tfrac{\pi}{3}=\tfrac12,\\ \sen2x=\tfrac{\sqrt3}{2},
e G'(\tfrac{\pi}{6})=-3.
Então o numerador fica
\cos(G)\,G'\,\cos2x +2\,\operatorname{sen}(G)\,\sen2x =\Bigl(\tfrac12\Bigr)(-3)\Bigl(\tfrac12\Bigr) +2\Bigl(\tfrac{\sqrt3}{2}\Bigr)\Bigl(\tfrac{\sqrt3}{2}\Bigr) =-\tfrac{3}{4}+2\cdot\tfrac{3}{4} =-\tfrac{3}{4}+\tfrac{3}{2} =\tfrac{3}{4},
e o denominador é
\cos^{2}2x=\Bigl(\tfrac12\Bigr)^{2}=\tfrac{1}{4}.
Logo
F'\bigl(\tfrac{\pi}{6}\bigr) =\frac{\tfrac{3}{4}}{\tfrac{1}{4}}=3.

Resposta

3

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