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Derivative of inverse trig functions.

  1. Oct 15, 2011 #1
    1. The problem statement, all variables and given/known data
    Find the derivative:
    y=sec-1(1/2t3)

    2. Relevant equations

    [tex]\frac{\frac{du}{dx}}{|u|\sqrt{u^2-1}}[/tex]

    3. The attempt at a solution

    I have an example to follow, but I dont know how step 1. became step 2.?...or more exactly the last part under the radical? (1-4t^6) instead of just -1 ???

    1.[tex]y'=\frac{\frac{-3}{2t^4}}{|\frac{1}{2t^3}|\sqrt{(\frac{1}{2t^3})^2-1}}[/tex]

    2.[tex]y'=\frac{-3}{(t)\sqrt{\frac{1}{4t^6}(1-4t^6)}}[/tex]
     
  2. jcsd
  3. Oct 15, 2011 #2
    Nevermind... I see it now.
     
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