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))
