Usando a regra da cadeia e a derivada de funçōes trigonométricas inversas, calcule a primeíra derivada de:
g(x)=\operatorname{Arctg}\left(\operatorname{sen}\left(\frac{x^{3}+3 x^{3}}{3}\right)\right)

Passo 1

O enunciado pede a derivada da função
g(x)=\Arctg\Bigl(\operatorname{sen}\bigl(\tfrac{x^3+3x^3}{3}\bigr)\Bigr),
ou seja,
g(x)=\arctan\bigl(\sin\bigl(\tfrac{x^3+3x^3}{3}\bigr)\bigr).
Para resolver vamos usar a regra da cadeia e as fórmulas
\frac{d}{dx}\bigl[\arctan(u)\bigr]=\frac{u'}{1+u^2} \quad\text{e}\quad \frac{d}{dx}[\sin(v)]=\cos(v)\,v'.

Passo 2

Primeiro simplificamos o argumento interno:
\frac{x^3+3x^3}{3}=\frac{4x^3}{3}.
Assim podemos reescrever
g(x)=\arctan\bigl(\sin\bigl(\tfrac{4}{3}x^3\bigr)\bigr).

Passo 3

Aplicando a derivada de \arctan(u) com u(x)=\sin\bigl(\tfrac{4}{3}x^3\bigr), obtemos
g'(x) =\frac{d}{dx}\Bigl[\arctan\bigl(u(x)\bigr)\Bigr] =\frac{u'(x)}{1+\bigl(u(x)\bigr)^2} =\frac{u'(x)}{1+\sin^2\bigl(\tfrac{4}{3}x^3\bigr)}.

Passo 4

Agora calculamos u'(x) com u(x)=\sin\bigl(v(x)\bigr) e v(x)=\tfrac{4}{3}x^3. Pela regra da cadeia:
u'(x) =\cos\bigl(v(x)\bigr)\,v'(x) =\cos\bigl(\tfrac{4}{3}x^3\bigr)\times\frac{d}{dx}\!\bigl(\tfrac{4}{3}x^3\bigr) =\cos\bigl(\tfrac{4}{3}x^3\bigr)\times4x^2.

Passo 5

Substituindo u'(x) em g'(x):
g'(x) =\frac{4x^2\cos\bigl(\tfrac{4}{3}x^3\bigr)}{1+\sin^2\bigl(\tfrac{4}{3}x^3\bigr)}.

Resposta

g'(x) =\frac{4x^2\cos\!\bigl(\tfrac{4}{3}x^3\bigr)}{1+\sin^2\!\bigl(\tfrac{4}{3}x^3\bigr)}.

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