You can use the quotient rule. OR, you can use some trig identities with #cosx# and #sinx#. ...Let's try the quotient rule instead. And maybe some identities along the way. Yeah, we'll need one identity later. You should get:
#-(secx)/(secx + 1)#
Let's see how.
The two derivatives we'll need:
#d/(dx)[tanu] = sec^2u((du)/(dx))#
#d/(dx)[secu] = secutanu((du)/(dx))#
The Quotient Rule:
#d/(dx)[f(x)/g(x)] = [g(x)f'(x) - f(x)g'(x)]/(g(x))^2#
#= [tanx (-secxtanx) - (1-secx)(sec^2x)]/(tan^2x)#
Multiply some things together:
#= [(-secxtan^2x) - (sec^2x-sec^3x)]/(tan^2x)#
Get rid of extra parentheses:
#= [-secxtan^2x - sec^2x + sec^3x]/(tan^2x)#
Move things around a bit (you'll see why soon):
#= [-secxtan^2x + sec^3x - sec^2x]/(tan^2x)#
Factor out #secx#:
#= [secx(-tan^2x + sec^2x - secx)]/(tan^2x)#
Notice the #tan^2x# by itself? An identity is #tan^2x = sec^2x - 1#:
#= [secx(-(sec^2x - 1) + sec^2x - secx)]/(tan^2x)#
Now we can cancel out #sec^2x#:
#= [secx(cancel(-sec^2x) + 1 + cancel(sec^2x) - secx)]/(tan^2x)#
#= [secx(1 - secx)]/(tan^2x)#
Hm... that's interesting. We have #-(secx - 1)#, which is a factor of #sec^2x - 1#... which is in an identity with #tan^2x#:
#= [-secx(secx - 1)]/(tan^2x)#
#= [-secx(secx - 1)]/(sec^2x - 1)#
Nice, now we can cancel more out!
#= [-secxcancel((secx - 1))]/((secx + 1)cancel((secx - 1)))#
And now we're done!
#= color(blue)(-secx/(secx + 1))#
