Question

$y = \frac{x}{ {x}^{2} + 1 }$
найдите y'(0,5)
пожалуйста..​

Answer 1

$y=\frac{x}{x^2+1}\\\\y'=\frac{1(x^1+1)-2x^2}{(x^2+1)^2}=\frac{1-x^2}{(x^2+1)^2}\\\\y'(0.5)=\frac{1-0.5^2}{(0.5^2+1)^2}=\frac{0.75}{1.25^2}=\frac{3}{4}:(\frac{5}{4})^2=\frac{3}{4}*\frac{16}{25}=\frac{12}{25}$

ОТВЕТ: 12/25=0.48