Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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.
  2. jcsd
  3. Jan 20, 2010 #2


    User Avatar
    Science Advisor

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


    User Avatar
    Science Advisor
    Homework Helper

    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
    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


    User Avatar
    Science Advisor

    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.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook