求导 y=(2X-1)/根号(x^2+1)

y=u/v则y'=(u'v-uv')/v^2这里u=2x-1,v=(x^2+1)^(1/2)所以u'=2v'=1/2*(x^2+1)^(1/2-1)*(x^2+1)'=1/2*(x^2+1)^(-1/2)*2x=x/√(x^2+1)所以y'=[2√(x^2+1)-(2x-1)*x/√(x^2+1)]/(x^2+1)=(2x^2+2-2x^2+x)/[(x^2+1)√(x^2+...