Derive as funções:
f(x)=\frac{\cot g(x)-4^{x}}{\log (x)-2^{x}}

Passo 1

Estamos com
f(x)=\frac{\cot\bigl(g(x)\bigr)-4^{x}}{\log(x)-2^{x}},
que é um quociente u(x)/v(x) onde
u(x)=\cot\bigl(g(x)\bigr)-4^{x}, \quad v(x)=\log(x)-2^{x}.
Para obter f'(x) vamos usar a regra do quociente
f'(x)=\frac{u'(x)\,v(x)-u(x)\,v'(x)}{\bigl(v(x)\bigr)^{2}},
além da regra da cadeia, da derivada de exponencial a^{x} e da derivada de \log(x).

Passo 2

Derivando
u(x)=\cot\bigl(g(x)\bigr)-4^{x},
temos
\frac{d}{dx}\bigl[\cot(g(x))\bigr] =-\csc^{2}\bigl(g(x)\bigr)\,g'(x),
\frac{d}{dx}\bigl[4^{x}\bigr] =4^{x}\ln(4).
Logo
u'(x) =-\csc^{2}\bigl(g(x)\bigr)\,g'(x)\;-\;4^{x}\ln(4).

Passo 3

Derivando
v(x)=\log(x)-2^{x},
obtém-se
\frac{d}{dx}[\log(x)] =\frac{1}{x},
\frac{d}{dx}[2^{x}] =2^{x}\ln(2).
Logo
v'(x) =\frac{1}{x}\;-\;2^{x}\ln(2).

Passo 4

Aplicando a regra do quociente
f'(x) =\frac{u'(x)\,v(x)-u(x)\,v'(x)}{\bigl(v(x)\bigr)^{2}},
substituindo u,u',v,v' achados acima, vem
f'(x) =\frac{\bigl[-\csc^{2}(g(x))\,g'(x)-4^{x}\ln(4)\bigr]\bigl[\log(x)-2^{x}\bigr] -\bigl[\cot(g(x))-4^{x}\bigr]\bigl[\tfrac{1}{x}-2^{x}\ln(2)\bigr]} {\bigl[\log(x)-2^{x}\bigr]^{2}}.

Resposta

f'(x) =\frac{\bigl(-\csc^{2}(g(x))\,g'(x)-4^{x}\ln(4)\bigr)\bigl(\log(x)-2^{x}\bigr) -\bigl(\cot(g(x))-4^{x}\bigr)\bigl(\tfrac{1}{x}-2^{x}\ln(2)\bigr)} {\bigl(\log(x)-2^{x}\bigr)^{2}}.

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