Exercício 3.14 São dadas funções envolvendo funçōes trigonométricas. Calcule a derivada de:
f(x)=\frac{\operatorname{cotg}(x)+7^{x}}{\operatorname{cossec}(x)+\log _{10}(x)}, x>0, \operatorname{cossec}(x)+\log _{10}(x) \neq 0

Passo 1

O enunciado pede a derivada de
f(x)=\frac{\operatorname{cotg}(x)+7^{x}}{\cossec(x)+\log_{10}(x)}, \quad x>0,\quad \cossec(x)+\log_{10}(x)\neq0.
O conteúdo envolvido é o cálculo de derivada de quociente, aplicando a regra
\biggl(\frac{u}{v}\biggr)'=\frac{u'v - uv'}{v^2},
além das derivadas de funções trigonométricas \operatorname{cotg}, \cossec, de exponencial 7^x e de logaritmo base 10.

Passo 2

Definimos
u(x)=\operatorname{cotg}(x)+7^{x}, \qquad v(x)=\cossec(x)+\log_{10}(x).
Vamos calcular u'(x) e v'(x) separadamente para depois aplicar a regra do quociente.

Passo 3

Derivada de u(x):
a) \displaystyle\frac{d}{dx}\bigl[\operatorname{cotg}(x)\bigr]=-\cossec^{2}(x).
b) \displaystyle\frac{d}{dx}\bigl[7^{x}\bigr]=7^{x}\ln(7).
Logo,
u'(x)=-\cossec^{2}(x)+7^{x}\ln(7).

Passo 4

Derivada de v(x):
a) \displaystyle\frac{d}{dx}\bigl[\cossec(x)\bigr]=-\cossec(x)\operatorname{cotg}(x).
b) \displaystyle\frac{d}{dx}\bigl[\log_{10}(x)\bigr] =\frac{1}{x\ln(10)}.
Logo,
v'(x)=-\cossec(x)\operatorname{cotg}(x)+\frac{1}{x\ln(10)}.

Passo 5

Aplicando a regra do quociente:
f'(x) =\frac{u'(x)\,v(x)-u(x)\,v'(x)}{\bigl[v(x)\bigr]^{2}} =\frac{\bigl(-\cossec^{2}(x)+7^{x}\ln(7)\bigr)\bigl(\cossec(x)+\log_{10}(x)\bigr) -\bigl(\operatorname{cotg}(x)+7^{x}\bigr)\bigl(-\cossec(x)\operatorname{cotg}(x)+\tfrac{1}{x\ln(10)}\bigr)} {\bigl(\cossec(x)+\log_{10}(x)\bigr)^{2}}.

Resposta

f'(x) =\frac{\bigl(-\cossec^{2}(x)+7^{x}\ln(7)\bigr)\bigl(\cossec(x)+\log_{10}(x)\bigr) -\bigl(\operatorname{cotg}(x)+7^{x}\bigr)\bigl(-\cossec(x)\operatorname{cotg}(x)+\tfrac{1}{x\ln(10)}\bigr)} {\bigl(\cossec(x)+\log_{10}(x)\bigr)^{2}}.

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