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Differentiating a trig function

  1. Oct 14, 2007 #1
    1. The problem statement, all variables and given/known data
    Differentiate using the Chain Rule:
    [tex]y=\cos^2(\frac{x^2 + 2}{x^2 - 2})[/tex]

    2. Relevant equations

    3. The attempt at a solution
    [tex]y' = -2\cos(\frac{x^2 + 2}{x^2 - 2})\sin(\frac{x^2 + 2}{x^2 - 2}) [\frac{2x(x^2 - 2) - (x^2 + 2)2x}{(x^2 - 2)^2}][/tex]
    [tex]\mbox{derivative of cos is -sin so I brought the negative to the front}[/tex]
    [tex]y' = -2\cos(\frac{x^2 + 2}{x^2 - 2})\sin(\frac{x^2 + 2}{x^2 - 2}) [\frac{2x(x^2 - 2 - x^2 - 2)}{(x^2 - 2)^2}][/tex]
    [tex]y' = -2\cos(\frac{x^2 + 2}{x^2 - 2})\sin(\frac{x^2 + 2}{x^2 - 2})[\frac{-8x}{(x^2 - 2)^2}][/tex]
    [tex]y' = 16x(x^2 - 2)^{-2} \cos(\frac{x^2 + 2}{x^2 - 2})\sin(\frac{x^2 + 2}{x^2 - 2})[/tex]

    However the worksheet says the answer is:
    [tex] y' = 8x(x^2 - 2)^{-2} \sin(\frac{2x^2 + 4}{x^2 - 2})[/tex]

    What the *flat line* did I do wrong? My answer isn't even close.
    Last edited: Oct 14, 2007
  2. jcsd
  3. Oct 14, 2007 #2
    they simplified using a formula for:

    when you plug in your values for sin(a)cos(b) the first sin cancels out
  4. Oct 14, 2007 #3
    double angle identity

    [tex]2sinxcosx = sin2x[/tex]
  5. Oct 14, 2007 #4
    Bob, we haven't been shown the half angle identities, which is what I assume that formula is used for. I know those formulae, but can't use them because we haven't been shown them formally :p.

    Roco, that is genius! :D
  6. Oct 14, 2007 #5
    if you find that you're differentiating cosine or sine to the 2nd power, i usually differentiate and simplify using double-angle identity

    also, when you have secx to some power, re-write the power and just add tanx at the end.



    makes the writing a lil easier
  7. Oct 14, 2007 #6
    Roco, thanks for the idea on simplifying the 2nd power part.
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