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Math Help - derivative of absolute function

  1. #1
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    derivative of absolute function

    I can use Wolfram to derive abs(cos x) and abs(x+2) and abs(ln x) etc. but is there a general derivative for this?
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  2. #2
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    Re: derivative of absolute function

    If f(x) positive, lf(x)l = f(x)
    If f(x) negative, lf(x)l = -f(x)
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  3. #3
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    Re: derivative of absolute function

    The absolute function is differentiable everywhere except x = 0. This is trivially provable.

    $\displaystyle u = |x|,\ and\ x < 0 \implies u(x) = - x \implies u'(x) = \lim_{h \rightarrow 0}\dfrac{u(x + h) - u(x)}{h} = \lim_{h \rightarrow 0}\dfrac{- (x + h) - (- x)}{h} = \lim_{h \rightarrow 0}\dfrac{- x - h + x}{h} = - 1.$

    $\displaystyle u = |x|,\ and\ x > 0 \implies u(x) = x \implies u'(x) = \lim_{h \rightarrow 0}\dfrac{u(x + h) - u(x)}{h} = \lim_{h \rightarrow 0}\dfrac{x + h - x}{h} = + 1.$

    But $\displaystyle \lim_{h \rightarrow 0^+}\dfrac{|0 + h| - |0|}{h} = \dfrac{h}{h} = 1 \ne -1 = \dfrac{-h}{h} = \lim_{h \rightarrow 0^-}\dfrac{|0 + h| - |0|}{h}$. The limit does not exist at x = 0.

    So, by the chain rule, $y(x) = u(v(x)) \implies y'(x) = u'(v) * v'(x)$ unless v(x) = 0.
    Last edited by JeffM; April 23rd 2014 at 05:49 AM. Reason: MORE TYPOS
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  4. #4
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    Re: derivative of absolute function

    There is a good explanation of the general derivative here: [SOLVED] Derivative of function involving absolute values
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    Re: derivative of absolute function

    cos x at x=0 is no problem.
    anywhere f(x) changes sign may be a discontiuity in derivative of absolute value. usually it's obvious by inspection.
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    Re: derivative of absolute function

    Hartlw makes an important clarification to my post. The key point is whether the limit of the Newton quotient exists at a when v(a) = 0, not whether v(a) = 0.
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