Question

Найти производную первого порядка
$frac{3+6x}{ sqrt{3-4x+5 x^{2} } } }$

Answer 1

$y= frac{3+6x}{ sqrt{3-4x+5x^2}}\\y'=frac{6cdot sqrt{3-4x+5x^2}-(3+6x)cdot frac{1}{2sqrt{3-4x+5x^2}}cdot (10x-4)}{3-4x+5x^2}=\\=frac{6cdot (3-4x+5x^2)-(3+6x)cdot (5x-2)}{sqrt{(3-4x+5x^2)^3}}= frac{24-27x}{sqrt{(3-4x+5x^2)^3}}$