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Homework Help: Differentiation of a trig function using quotient rule

  1. Feb 1, 2010 #1
    1. The problem statement, all variables and given/known data

    Find the derivative of [tex]\frac{sin x}{1 + cos x}[/tex]



    2. Relevant equations

    Quotient rule [tex]\frac{gf' - fg'}{g^{2}}[/tex]



    3. The attempt at a solution

    [tex]\frac{dy}{dx}[/tex] = [tex]\frac{(1 + cos x)(\frac{d}{dx}(sin x)) - sin x(\frac{d}{dx}(1 + cos x)}{(1 + cos x)^{2}}[/tex]

    simplify the derivative so far:

    = [tex]\frac{(1 + cos x)(cos x) - (sin x)(-sin x)}{(1 + cos x)^{2}}[/tex]

    simplify further:

    = [tex]\frac{cos x + cos^{2}x + sin^{2}x}{(1 + cos x)^{2}}[/tex]

    Use angle identity [tex]sin^{2}x[/tex] + [tex]cos^{2}x = 1[/tex]to simplify even further:

    = [tex]\frac{cos x + 1}{(1 + cos)^{2}}[/tex]

    cancel out the common 1 + cos x

    [tex]\frac{dy}{dx}[/tex] = [tex]\frac{1}{1 + cos x}[/tex]

    I was quite confident in my answer, but I was a little teeny bit hesitant, so I used my graphing calculator to double check. When I did so, I found out that I was wrong, the derivative that I calculated (above) is not the actual derivative of the question.

    At which step did I go wrong?

    Thanks so much in advance everyone!
     
    Last edited: Feb 1, 2010
  2. jcsd
  3. Feb 1, 2010 #2

    radou

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    Homework Helper

    Hm, I don't see where the mistake is. If there is one at all. Perhaps you messed up your input in the graphics calculator?
     
  4. Feb 1, 2010 #3

    tiny-tim

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    Science Advisor
    Homework Helper

    Hi stripes! :smile:

    Looks ok to me. :confused:

    As a double-check, using standard trigonometric identities …

    sinx/(1 + cosx) = tan(x/2), so dy/dx = 1/2 sec2(x/2) = 1/(1 + cosx).

    What answer did your calculator give?​
     
  5. Feb 1, 2010 #4
    basically, on my graphing calculator i put y1 as my original function, then y2 as the derivative that I found. I then used the calculator to find dy/dx of y1 for me at various points (maybe 1, 2, and 3), then I used the calculator to find the values of x = 1, 2, and 3 for y2 and the respective values should be the same...but they weren't!

    I'm almost certain i was error free inputting the functions...I did it over and over and over!
     
  6. Feb 1, 2010 #5

    Mark44

    Staff: Mentor

    Might be a dumb question, but was your calculator in radian mode?
     
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