Question

求导 y=(2X-1)/根号(x^2+1)

Answer 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'=2

v'=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+1)]

=(x+2)/[(x^2+1)√(x^2+1)]