Question

帮忙解一道高数题

y=ln(sec x + tan x)
求“y=ln(sec x + tan x)”的导数

Answer 1

y'=[1/(secx+tanx)]*(secx+tanx)'
而（secx+tanx)'=(1/cosx+sinx/cosx)'
=[(1+sinx)/cosx]'
=[sinx(1+sinx)+cosx^2]/cosx^2
=[sinx+(sinx^2+cosx^2)]/cosx^2
=(1+sinx)/cosx^2
所以y'=上式*[1/(secx+tanx)]
=上式*[cosx/(1+sinx)]
=1/cosx

Answer 2

求啥