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Probably really easy, derivative of cotx/sinx

  1. Feb 24, 2010 #1
    1. The problem statement, all variables and given/known data
    Find the derivative of the function.

    f(x) = [tex]\frac{cotx}{sinx}[/tex]


    2. Relevant equations

    [tex]\frac{cotx}{sinx}[/tex] = [tex]\frac{cosx}{sin^{2}x}[/tex]




    3. The attempt at a solution

    I have the solution guide, I think I'm just having a brain fart but this is what I have so far which is the last step before the final answer

    [tex]\frac{-sin^{2}x - 2cos^{2}x}{sin^{3}x}[/tex]

    The next step, which is the answer according to the book is

    [tex]\frac{-1-cos^{2}x}{sin^{3}x}[/tex]

    Am I just not seeing something on my last step? I think they got the -1 from using the pythagorean indentity but then why is the cosx still hanging around? Thanks.
     
    Last edited: Feb 24, 2010
  2. jcsd
  3. Feb 24, 2010 #2

    Borek

    User Avatar

    Staff: Mentor

    You will feel ashamed enough to put a paper bag on the head.

    [tex]-\sin^2x -2\cos^2x = -(\sin^2x+\cos^2x)-\cos^2x[/tex]
     
  4. Feb 24, 2010 #3
    Yep that's most definitely embarrassing! Thanks!
     
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