1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Deriv of trig function

  1. Jan 15, 2007 #1
    1. The problem statement, all variables and given/known data

    Find y' of the following

    2. Relevant equations

    y = (sec (2x))/(1 + tan (2x))

    3. The attempt at a solution

    y' = [(1 + tan (2x))(2 sec (2x) tan (2x)) - (sec (2x))(2 (sec (2x))^2)]/[(1 + tan (2x))^2]

    i can not quite figure out how it reduces to:

    y' = [(2 sec (2x))(tan (2x) - 1)]/[(1 + tan (2x))^2] :grumpy:
  2. jcsd
  3. Jan 15, 2007 #2


    User Avatar
    Staff Emeritus
    Science Advisor

    Try using the identity sec2x=1+tan2x on the last term in the numerator. Then it should simplify to the result you state.
    Last edited: Jan 15, 2007
  4. Jan 15, 2007 #3


    User Avatar
    Homework Helper

    You had

    [tex]y^\prime =\frac{(1 + \tan 2x)(2\sec 2x \tan 2x) - (\sec 2x)(2 \sec^{2} 2x)}{(1 + \tan 2x)^2}=\frac{2\sec 2x \left[(1 + \tan 2x)\tan 2x - \sec^{2} 2x\right]}{(1 + \tan 2x)^2}[/tex]
    [tex]=\frac{2\sec 2x \left[\tan 2x + \tan^{2} 2x - (1 + \tan^{2} 2x)\right]}{(1 + \tan 2x)^2}=\frac{2\sec 2x (\tan 2x - 1)}{(1 + \tan 2x)^2}[/tex]
  5. Jan 15, 2007 #4
    thank you cristo and benorin
  6. Jan 15, 2007 #5


    User Avatar
    Homework Helper

Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Similar Discussions: Deriv of trig function