Question

F(x)=5ln(3/x)+(0.3x^2+1) найти пройзводства функции

Answer 1

$F'(x)=(5ln(3/x))'+(0.3x^2+1)'=5 \frac{1}{3/x} (3/x)'+(2*0.3x)=$
$=\frac{5x}{3} 3(x^{-1})'+0.6x=-\frac{5x}{3} \frac{3}{x^2} +0.6x=0.6x-5/x$