Question

ln(e的x次幂+1）求导得?有点歧义 是ln〔（e的x次幂）+1〕

Answer 1

令u=e^x+1 则y=ln u

y'x=y'u乘以u'x

=(ln u)'乘以（e^x+1)'

=1/u乘以e^x

再把u=e^x+1代入得

y'=e^x/(e^x+1)