设y＝arctanex−lne2xe2x+1，则dydx|x＝1=e−1e2+1e−1e2+1．

Question

设y＝arctanex−ln√((e2x)/(e2x+1))，则(dy)/(dx)|x＝1=

(e−1)/(e2+1)

(e−1)/(e2+1)

．

Answer 1

∵y＝arctanex−ln√((e2x)/(e2x+1))=arctanex−(1)/(2)(lne2x−ln(e2x+1))=arctanex−x+(1)/(2)ln(e2x+1)
∴y′＝(1)/(1+e2x)•ex−1+(e2x)/(e2x+1)=(ex+e2x)/(1+e2x)−1
∴(dy)/(dx)|x＝1＝(e−1)/(e2+1)