Resposta :
Cheguei a pensar que havia uma nova derivada! [risos].
[tex]y=\frac{5x^2}{5x+5}\\\\y=\frac{5x^2}{5(x+1)}\\\\y=\frac{x^2}{x+1}\\\\y'=\frac{2x\cdot(x+1)-x^2\cdot1}{(x+1)^2}\\\\y'=\frac{2x^2+2x-x^2}{(x+1)^2}\\\\y'=\frac{x^2+2x}{(x+1)^2}\\\\\boxed{y'=\frac{x(x+2)}{(x+1)^2}}[/tex]
[tex]y=\frac{5x^2}{5x+5}\\\\y=\frac{5x^2}{5(x+1)}\\\\y=\frac{x^2}{x+1}\\\\y'=\frac{2x\cdot(x+1)-x^2\cdot1}{(x+1)^2}\\\\y'=\frac{2x^2+2x-x^2}{(x+1)^2}\\\\y'=\frac{x^2+2x}{(x+1)^2}\\\\\boxed{y'=\frac{x(x+2)}{(x+1)^2}}[/tex]