y'=（根号下（x/1-x))/(2*(1-x)^2)
y=√[x/(1-x)]y'=(1/2)*[x/(1-x)]^(-1/2)*[x/(x-1)]' =1/2*√[(1-x)/x]* [1-x+x]/(1-x)² =1/2*√[(1-x)/x]* 1/(1-x)²
