Question

Найдите:
$y= \frac{ x^{6}-16 x^{2} }{ x^{2} -5x+105} ; y'=?$

Answer 1

$y'= \frac{(6x^5-32x)(x^2-5x+105)-(2x-5)*(x^6-16x^2)}{ (x^{2} -5x+105)^2}$

Answer 2

$y=\frac{x^6-16x^2}{x^2-5x+105}\\\\y'=\frac{(x^6-16x^2)'(x^2-5x+105)-(x^2-5x+105)'(x^6-16x^2)}{(x^2-5x+105)^2}=$

$=\frac{(6x^5-32x)(x^2-5x+105)-(2x-5)(x^6-16x^2)}{(x^2-5x+105)^2}$