因为函数的导数等于
反函数导数的倒数。arctanx 的反函数是tany=x,所以tany'=(siny/cosy)'=[(siny)'cosy-siny(cosy)']/(cosy)^2=(cos^2y+sin^2y)/cos^2y=1/cos^2y .............tany=siny/cosy=根号下(1-cos^2y)/cosy,,,,,,,,,,两边平方得tan^2y=(1-cos^2y)/cos^2y......因为上面tany=x.........所以cos^2=1/(x^2+1)........所以由上面(tany)'=1/cos^2y的得(tany)'=x^2+1然后再用倒数得(arctany)'=1/(1+x^2))