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Derivative of |sinx|

  1. Jan 16, 2012 #1
    1. The problem statement, all variables and given/known data
    Compute the derivatives of the following (where they are differentiable):
    h) |sinx|


    2. Relevant equations
    Chain rule: (f°g)'(c) = f'(g(c))(g'(c))


    3. The attempt at a solution
    Let f=|x| and g=sin x
    (f°g)'(c) = f'(g(x))g'(c) = f'(sin x)(cos x)
    But I don't know what (|x|)' is. It's +1 when x>0 and -1 when x<0 and it's not differentiable at 0, but then there is no x to plug g into, and looking at the graph, I don't think this would be right. Thanks!
     
  2. jcsd
  3. Jan 16, 2012 #2
    Hint:
    [tex]
    \vert x \vert \equiv \left\lbrace\begin{array}{rl}
    x &, \ x \ge 0 \\

    -x &, \ x < 0
    \end{array}\right.
    [/tex]

    Therefore:
    [tex]
    \frac{d}{d x} \vert x \vert = \left\lbrace \begin{array}{rl}
    1 &, \ x > 0 \\

    -1 &, \ x < 0
    \end{array}\right. \equiv \mathrm{sgn}(x)
    [/tex]
    The derivative does not exist at [itex]x = 0[/itex].

    Then, use the chain rule.
     
  4. Jan 16, 2012 #3

    Simon Bridge

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    express the function piecewise to remove the absolute value sign.
    eg |x| = {x: x>0 and -x: x<0 and 0: x=0}

    [edit]Dickfore beat me to it
     
  5. Jan 17, 2012 #4
    So if x>0, [itex](f°g)' = f'(g(x))g'(x) = (1)(cos x) = cos x[/itex]

    If x<0, [itex](f°g)' = f'(g(x))g'(x) = (-1)(cos x) = -cos x[/itex]

    Thanks for the hint!
     
  6. Jan 17, 2012 #5

    SammyS

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    What you have is not correct. sin(x) may be negative when x > 0 .

    Your function is:

    [itex]\vert \sin(x) \vert \equiv \left\lbrace\begin{array}{rl}
    \sin(x) &\text{if } \ \sin(x) \ge 0 \\

    -\sin(x) &\text{if } \ \sin(x) < 0
    \end{array}\right.[/itex]

    So, what matters is the sign of sin(x), not the sign of x itself.
     
    Last edited: Jan 17, 2012
  7. Jan 17, 2012 #6

    ehild

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    You can use the "sign" function sign(a)=a/|a| (http://mathworld.wolfram.com/Sign.html) to express the derivative of |sinx|.

    It is also possible to use the identity

    [itex]|a|=\sqrt{a^2}[/itex]

    and determine the derivative of [itex]\sqrt{sin^2(x)}[/itex].

    ehild
     
    Last edited: Jan 18, 2012
  8. Jan 19, 2012 #7
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