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What is the derivative of the absolute value of cos(x)?

  1. Oct 22, 2005 #1
    What is the derivative of the absolute value of cos(x)?
     
  2. jcsd
  3. Oct 22, 2005 #2

    TD

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    The derivative of cos(x) is -sin(x) and the derivative of |x| is sgn(x), can you now combine them?
     
  4. Oct 22, 2005 #3
    Thanks, but what does sgn stand for? Is the derivative just -sin(x)*Abs(cos(x))'?
     
    Last edited: Oct 22, 2005
  5. Oct 22, 2005 #4

    TD

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    The 'sign' or 'signum' function, which returns 1 or -1, whether the argument in question was positive or negative.

    See Mathworld or Wikipedia.
     
  6. Oct 22, 2005 #5

    benorin

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    [tex]\frac{d}{dx}|\cos(x)|=-\frac{|\cos(x)|}{\cos(x)}\sin(x)[/tex]
    Note that the signum function can be defined by [tex]sgn(x)=\frac{|x|}{x}[/tex] for nonzero x, and is zero when x is zero. The signum function cannot be use in this case as [tex]|\cos(x)|[/tex] is not differentiable at the values of x for which [tex]\cos(x)=0[/tex] as the lefthand and righthand derivative are not equal there (by lefthand or righthand derivates, what is meant is the left or right-handed limit of the difference quotient at a particular value of x).
     
  7. Oct 22, 2005 #6
    Thank you so much. I've never even heard about the signum function before until now. How would I go about taking higher order derivatives of the signum function like the second and third, etc. How does that work?
     
  8. Oct 22, 2005 #7
    Look at its graph. The derivative should be apparent.
     
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