Question

$y = \frac{ {45x }^{2} + 41}{5x + 23}$
y'-?
решите пожалуйста
буду благодарен

Answer 1

$y = \frac{ {45x }^{2} + 41}{5x + 23} \\ \\ y'= \frac{(45x^2+41)'*(5x+23)-(5x+23)'(45x^2+41)}{(5x+23)^2} = \\ \\ \frac{90x*(5x+23)-5(45x^2+41)}{(5x+23)^2} = \\ \\ \frac{450x^2+2070x-225x^2-205}{(5x+23)^2} = \\ \\ \frac{225x^2+2070x-205}{(5x+23)^2}$