Question

Помогите с производной поэтапно
Задана функция f(x)=(x^2)/(0,5^1-2x)
Найдите f'(1)

Answer 1

$f(x)=\frac{x^2}{0.5^{1-2x}}=\frac{x^2}{(0.5^{2x-1})^{-1}}=x^2*0.5^{2x-1}\\\\ f'(x)=[x^2*0.5^{2x-1}]'=[x^2]'*0.5^{2x-1}+x^2*[0.5^{2x-1}]'=\\\\ =2x*0.5^{2x-1}+x^2*0.5^{2x-1}*ln(0.5)*[2x-1]'=\\\\ =2x*0.5^{2x-1}+x^2*0.5^{2x-1}*ln(0.5)*[2]=\\\\ =2x*0.5^{2x-1}*[1+ln(2^{-1})*x]=\\\\ =2x*0.5^{2x-1}*[1-ln(2})*x].\\\\ f'(1)=2*1*0.5^{2*1-1}*[1-ln(2})*1]=\\\\ =2*0.5*[1-ln(2)]=1-ln(2)$