Calcule f^{\prime}(x) para as funções f abaixo:
f(x)=\frac{\sqrt[3]{x} \operatorname{sen} x}{x^{2} \cos \left(x^{2}\right)}

Passo 1

O enunciado pede a derivada de
f(x)=\frac{\sqrt[3]{x}\,\operatorname{sen} x}{x^{2}\cos\bigl(x^{2}\bigr)}.
Aqui vamos usar as regras de derivação de potência, produto, cadeia (para \cos(x^2)) e quociente. Definimos
u(x)=x^{\tfrac13}\operatorname{sen} x,\quad v(x)=x^{2}\cos\bigl(x^{2}\bigr),
de modo que f(x)=u(x)/v(x).

Passo 2

Calculamos u'(x) usando a regra do produto:
u(x)=x^{\tfrac13}\operatorname{sen} x \quad\Longrightarrow\quad u'(x)=\bigl(x^{\tfrac13}\bigr)'\operatorname{sen} x + x^{\tfrac13}\bigl(\operatorname{sen} x\bigr)'.
Como \bigl(x^{\tfrac13}\bigr)'=\tfrac13 x^{-\tfrac23} e \bigl(\operatorname{sen} x\bigr)'=\cos x, obtemos
u'(x)=\tfrac13\,x^{-\tfrac23}\,\operatorname{sen} x \;+\; x^{\tfrac13}\cos x.

Passo 3

Agora v'(x). Aplicando produto e cadeia a
v(x)=x^{2}\cos\bigl(x^{2}\bigr),
temos
v'(x)=2x\cos\bigl(x^{2}\bigr) \;+\;x^{2}\,\bigl[\cos\bigl(x^{2}\bigr)\bigr]'.
E \bigl[\cos(x^{2})\bigr]'=-\operatorname{sen}(x^{2})\cdot2x. Logo
v'(x)=2x\cos\bigl(x^{2}\bigr)\;-\;2x^{3}\operatorname{sen}\bigl(x^{2}\bigr).

Passo 4

Usamos a regra do quociente:
f'(x)=\frac{u'(x)\,v(x)-u(x)\,v'(x)}{\bigl[v(x)\bigr]^{2}}.

Passo 5

Substituindo u,u',v,v':
u'(x)\,v(x) =\Bigl(\tfrac13x^{-\tfrac23}\operatorname{sen} x + x^{\tfrac13}\cos x\Bigr)\,x^{2}\cos(x^{2})
= \tfrac13\,x^{\tfrac{4}{3}}\operatorname{sen} x\cos(x^{2}) \;+\;x^{\tfrac{7}{3}}\cos x\cos(x^{2}),
u(x)\,v'(x) = x^{\tfrac13}\operatorname{sen} x\;\bigl[2x\cos(x^{2})-2x^{3}\operatorname{sen}(x^{2})\bigr]
=2x^{\tfrac{4}{3}}\operatorname{sen} x\cos(x^{2}) \;-\;2x^{\tfrac{10}{3}}\operatorname{sen} x\operatorname{sen}(x^{2}).
Então o numerador fica
u'v - u\,v' =\Bigl(\tfrac13-2\Bigr)x^{\tfrac{4}{3}}\operatorname{sen} x\cos(x^{2}) \;+\;x^{\tfrac{7}{3}}\cos x\cos(x^{2}) \;+\;2x^{\tfrac{10}{3}}\operatorname{sen} x\operatorname{sen}(x^{2})
=-\tfrac{5}{3}\,x^{\tfrac{4}{3}}\operatorname{sen} x\cos(x^{2}) \;+\;x^{\tfrac{7}{3}}\cos x\cos(x^{2}) \;+\;2x^{\tfrac{10}{3}}\operatorname{sen} x\operatorname{sen}(x^{2}).

Passo 6

E o denominador é
\bigl[v(x)\bigr]^{2} =\bigl(x^{2}\cos(x^{2})\bigr)^{2} =x^{4}\cos^{2}(x^{2}).
Portanto
f'(x) =\frac{-\tfrac{5}{3}\,x^{\tfrac{4}{3}}\operatorname{sen} x\cos(x^{2}) \;+\;x^{\tfrac{7}{3}}\cos x\cos(x^{2}) \;+\;2x^{\tfrac{10}{3}}\operatorname{sen} x\operatorname{sen}(x^{2})} {x^{4}\cos^{2}(x^{2})}.

Resposta

\boxed{% f'(x) =\frac{-\tfrac{5}{3}\,x^{\tfrac{4}{3}}\operatorname{sen} x\cos(x^{2}) \;+\;x^{\tfrac{7}{3}}\cos x\cos(x^{2}) \;+\;2x^{\tfrac{10}{3}}\operatorname{sen} x\operatorname{sen}(x^{2})} {x^{4}\cos^{2}(x^{2})}\,.}

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