im x->0,sec^2*x
=lim x->0,1/cosx^2
=1/1^2
=1.
设x=tany tany'=sex^y arctanx'=1/(tany)'=1/sec^y sec^y=1+tan^y=1+x^2 所以(arctanx)'=1/(1+x^2)
cos2X=(cosX)^2-(sinX)^2
=2*(cosX)^2-1
=1-2*(sinX)^2
im x->0,sec^2*x
=lim x->0,1/cosx^2
=1/1^2
=1.
设x=tany tany'=sex^y arctanx'=1/(tany)'=1/sec^y sec^y=1+tan^y=1+x^2 所以(arctanx)'=1/(1+x^2)
cos2X=(cosX)^2-(sinX)^2
=2*(cosX)^2-1
=1-2*(sinX)^2
转载于:https://my.oschina.net/Bettyty/blog/794687