Determine a derivada da função dada:
f(x)=\operatorname{sen}^{3}(5 x) \cos ^{2}\left(\frac{x}{3}\right);

Passo 1

O enunciado pede a derivada de
f(x) = \sin^{3}(5x)\,\cos^{2}\!\Bigl(\tfrac{x}{3}\Bigr).
Aqui vamos usar a regra do produto e a regra da cadeia.

Passo 2

Defina
g(x)=\sin^{3}(5x) \quad\text{e}\quad h(x)=\cos^{2}\!\Bigl(\tfrac{x}{3}\Bigr).
Pela regra do produto,
f'(x)=g'(x)\,h(x)+g(x)\,h'(x).

Passo 3

Calcule g'(x). Seja u(x)=\sin(5x), então g(x)=u(x)^{3}. Pela cadeia:
u'(x)=\cos(5x)\cdot5 \quad\Longrightarrow\quad g'(x)=3\,u(x)^{2}\,u'(x) =3\,\sin^{2}(5x)\,(5\cos(5x)) =15\,\sin^{2}(5x)\cos(5x).

Passo 4

Calcule h'(x). Seja v(x)=\cos\bigl(\tfrac{x}{3}\bigr), então h(x)=v(x)^{2}. Pela cadeia:
v'(x)=-\sin\!\Bigl(\tfrac{x}{3}\Bigr)\cdot\tfrac{1}{3} \quad\Longrightarrow\quad h'(x)=2\,v(x)\,v'(x) =2\,\cos\!\Bigl(\tfrac{x}{3}\Bigr)\Bigl(-\tfrac{1}{3}\sin\!\bigl(\tfrac{x}{3}\bigr)\Bigr) =-\tfrac{2}{3}\,\sin\!\Bigl(\tfrac{x}{3}\Bigr)\cos\!\Bigl(\tfrac{x}{3}\Bigr).

Passo 5

Substituindo em f'(x)=g'(x)\,h(x)+g(x)\,h'(x), temos
f'(x) =15\,\sin^{2}(5x)\cos(5x)\,\cos^{2}\!\Bigl(\tfrac{x}{3}\Bigr) +\sin^{3}(5x)\,\Bigl(-\tfrac{2}{3}\sin\!\bigl(\tfrac{x}{3}\bigr)\cos\!\bigl(\tfrac{x}{3}\bigr)\Bigr).
Reorganizando,
f'(x) =15\,\sin^{2}(5x)\cos(5x)\cos^{2}\!\Bigl(\tfrac{x}{3}\Bigr) -\tfrac{2}{3}\,\sin^{3}(5x)\,\sin\!\Bigl(\tfrac{x}{3}\Bigr)\cos\!\Bigl(\tfrac{x}{3}\Bigr).

Resposta

f'(x) =15\,\sin^{2}(5x)\cos(5x)\cos^{2}\!\Bigl(\tfrac{x}{3}\Bigr) -\frac{2}{3}\,\sin^{3}(5x)\,\sin\!\Bigl(\tfrac{x}{3}\Bigr)\cos\!\Bigl(\tfrac{x}{3}\Bigr).

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