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Derivative of abs(sin x)

  1. Jan 20, 2010 #1
    i am trying to solve the following, and the derivative of the |sin x| is required. Find the critical points and which ones are stationary points for |sin x|?
    To find the points we must solve [tex] f'(x)=0[/tex]
    Since f'(x)=|cos x|, and solving for this gives [tex] \frac{\pi}{2}, \frac{3\pi}{2} [/tex] etc.
    That is [tex] n\pi + \frac{\pi}{2} [/tex] as the total number of critial points, but using a graphing calculator I see [tex] n\frac{\pi}{2} [/tex] critical points. So I was thinking maybe f'(x)=|cos x| is not correct. I did a sign analysis and only got all +s, suggesting that (according to the calculus of it) f does not have any relative extremum! I am studing this on my own, so please help, Thanks.
     
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  3. Jan 20, 2010 #2

    disregardthat

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    Think about the alternative definition of [tex]|x|[/tex], namely [tex]\sqrt{x^2}[/tex].
     
  4. Jan 20, 2010 #3

    tiny-tim

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    Hi John! :smile:
    Then f'(x) would always be positive, but looking at the graph of f(x), that clearly isn't true.
     
  5. Jan 20, 2010 #4
    f(x)=|sinx| is differentiable at every point in R\{k*pi}. In general: |x|=x if x>0 and -x if x<0. Do the same for |sinx|.
     
  6. Jan 20, 2010 #5
    Hi,
    I was thinking alright of (sin(x))^2, its derivative is 2sinx*cosx and it gives you the correct number of critical points for |sin x|, but when I use the definition of the derivative to find f'(x) I get |cos x|. Is there any way to get [tex] \sqrt{(\sin x )^2} [/tex] from the definition of the derivative? Thanks.
     
  7. Jan 20, 2010 #6

    disregardthat

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    Just differentiate as you would with [tex]\sqrt{f(x)}[/tex]. Don't cancel out, that is; [tex]\sqrt{y^2}[/tex] to [tex]y[/tex], at any point.
     
  8. Jan 20, 2010 #7
    Perform the steps that i suggested, and then apply the def. of the derivative on each interval separately. Because, like i mentoned |sinx|, as you will find out, is not differentiable at any multiple of pi.

    i.e. |sinx|={ sinx if sinx>0 and -sinx if sinx<0}

    so basically, find the values of x for which sinx>o,and the values of x for which sinx<0, at this point you know that |sinx| is differentiable whenever sinx<0 and sinx>0, all you need to do is determine what happens when sinx=0, namely at those points x. and you can use the def. of the derivative in terms of sequences or however you have learned it to show that whenever x=k*pi, |sinx| is not differentiable.
     
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