Calcule as derivadas das funções abaixo:
\begin{aligned} & f(x)=\csc \left(\frac{e^{x}+3 x^{2}}{5 x^{3}}\right) \\\ & f^{\prime}(x)=-\csc \left(\frac{e^{x}+3 x^{2}}{5 x^{3}}\right) \operatorname{cotg}\left(\frac{e^{x}+3 x^{2}}{5 x^{3}}\right)\left(\frac{\left(e^{x}+6 x\right)\left(5 x^{3}\right)-\left(e^{x}+3 x^{2}\right)\left(15 x^{2}\right)}{25 x^{6}}\right) \end{aligned}

Passo 1

O enunciado pede a derivada de
f(x)=\csc\!\bigl(u(x)\bigr) \quad\text{com}\quad u(x)=\frac{e^{x}+3x^{2}}{5x^{3}}.
Vamos usar o Regra da Cadeia e, para u(x), a Regra do Quociente.

Passo 2

Aplicando a derivada de \csc, que é
\frac{d}{dx}\bigl[\csc(u)\bigr] =-\csc(u)\,\cot(u)\,\frac{du}{dx},
ficamos com
f'(x) =-\csc\bigl(u(x)\bigr)\,\cot\bigl(u(x)\bigr)\,\bigl(u'(x)\bigr).

Passo 3

Agora calculamos u'(x) por quociente, onde
N(x)=e^{x}+3x^{2}, \quad D(x)=5x^{3}.
Temos
N'(x)=e^{x}+6x, \quad D'(x)=15x^{2},
e
u'(x) =\frac{N'(x)\,D(x)-N(x)\,D'(x)}{[D(x)]^{2}} =\frac{(e^{x}+6x)\,(5x^{3})-(e^{x}+3x^{2})\,(15x^{2})}{(5x^{3})^{2}}.

Passo 4

Colocando tudo junto em f'(x):
f'(x) =-\csc\!\Bigl(\tfrac{e^{x}+3x^{2}}{5x^{3}}\Bigr)\, \cot\!\Bigl(\tfrac{e^{x}+3x^{2}}{5x^{3}}\Bigr)\, \frac{(e^{x}+6x)\,5x^{3}-(e^{x}+3x^{2})\,15x^{2}}{25\,x^{6}}.

Resposta

f'(x) =-\csc\!\Bigl(\frac{e^{x}+3x^{2}}{5x^{3}}\Bigr)\, \cot\!\Bigl(\frac{e^{x}+3x^{2}}{5x^{3}}\Bigr)\, \frac{(e^{x}+6x)\,5x^{3}-(e^{x}+3x^{2})\,15x^{2}}{25\,x^{6}}.

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