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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

    cristo

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    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

    benorin

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    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

    benorin

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