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Differentiating trigonometric F'ns with double angles

  1. Mar 9, 2013 #1
    1. The problem statement, all variables and given/known data
    Sin(tan(2x))
    With respect to x


    2. Relevant equations
    Differentiation


    3. The attempt at a solution

    My question is whether I can simply use d/dx (Tan x) = Sec^2(X) to extrapolate that to d/dx(tan 2x) = Sec^2(2x) ?

    Or do I have to convert to sine/cosine and go from there?
     
  2. jcsd
  3. Mar 9, 2013 #2

    MarneMath

    User Avatar
    Education Advisor

    Well,

    You I want to point out that (tan2x)' does not equal to sec^2(2x). (Error in technique).

    However, the generally idea isn ok (assuming you using the chain rule in regards to sin(tan(2x)).
     
  4. Mar 9, 2013 #3
    So this problem can't be solved by using the chain rule thusly:

    d/dx Sin(tan 2x) = cos(tan2x)Sec^2(2x)?

    It turns into:

    Sin(tan 2x) (2)d/dx [sinxcosx / (cos^2(X) - Sin^2(x))]


    quotient rule etc... damn this is gonna be messy. Am I on the right track at least?
     
    Last edited: Mar 10, 2013
  5. Mar 10, 2013 #4

    MarneMath

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    Education Advisor

    "d/dx Sin(tan 2x) = cos(tan2x)Sec^2(2x)?"

    This is essentially correct, but you're making an error when you take the derivative of tan2x. I'm hoping you look at that part just a little harder and figure out what that error exactly is.
     
  6. Mar 10, 2013 #5
    d/dx tan 2x = [sec^2 (2x)] (2)

    ?
     
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