Question

y＝㏑[﹙1－eⁿ﹚／eⁿ],则求y的导函数.

Answer 1

y=ln[(1-e^n)/e^n]

=ln(1-e^n)-ln(e^n)

=ln(1-e^n)-n

y'=dy/dn

=[ln(1-e^n)-n]'

=ln'(1-e^n)-n'

=(1-e^n)'/(1-e^n)-1

=-(e^n)/(1-e^n)-1

=(-e^n-1+e^n)/(1-e^n)

=1/(e^n-1)