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Derivitave of a Modulus

  1. Jul 30, 2005 #1

    Zurtex

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    Feeling a little bit more confident about my calculus skills I was hoping to check if this is correct. Let’s say you have:

    [tex]g(|f(x)|)[/tex]

    And you want to take the derivative with respect to x, well using the chain rule you get:

    [tex]\bigl( g(|f(x)|) \bigr)' = (|f(x)|)' g'(|f(x)|)[/tex]

    Looking more closly at [itex](|f(x)|)'[/itex]. Defining it as:

    [tex]\left( +\sqrt{(f(x))^2} \right)'[/tex]

    Then using the chain rule we get:

    [tex](|f(x)|)' = 2f(x) f'(x) \: \frac{1}{2} \: \frac{1}{ +\sqrt{(f(x))^2} }[/tex]

    Simplifying:

    [tex](|f(x)|)' = f'(x) \frac{f(x)}{|f(x)|}[/tex]

    Substituting back in the original equation and sorting out the problem of when f(x) = 0:

    [tex]\bigl( g(|f(x)|) \bigr)' = \begin{cases}
    f'(x) \frac{f(x)}{|f(x)|}g'(|f(x)|) & \text{if $ f(x) \neq 0$} \\ \\
    \lim_{f(x) \rightarrow 0} \left(f'(x) \frac{f(x)}{|f(x)|}g'(|f(x)|) \right) & \text{if $ f(x) = 0$ and $ \lim_{f(x) \uparrow 0} \left( f'(x) \frac{f(x)}{|f(x)|}g'(|f(x)|) \right) = \lim_{f(x) \downarrow 0} \left( f'(x) \frac{f(x)}{|f(x)|}g'(|f(x)|) \right)$} \\ \\
    \text{Undefinied} & \text{if $ f(x) = 0$ and $ \lim_{f(x) \uparrow 0} \left( f'(x) \frac{f(x)}{|f(x)|}g'(|f(x)|) \right) \neq \lim_{f(x) \downarrow 0} \left( f'(x) \frac{f(x)}{|f(x)|}g'(|f(x)|) \right)$}
    \end{cases}
    [/tex]

    Correct? I know it all seems a bit over the top (especially when you look at the tex for it :wink:) but I like things to be well defined and in a form like this where I understand it better.

    Edit: I made a mistake, it should be correct now :smile:
     
    Last edited: Jul 30, 2005
  2. jcsd
  3. Jul 30, 2005 #2

    matt grime

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    it is over the top. it is easier to use mod of Y is Y if Y is positive and -Y if Y is negative, the derivative not defined if Y=0, though it is conceivalbe that g(modY) is differentiable when Y=0.
     
  4. Jul 30, 2005 #3

    Zurtex

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    I'm not trying to find an easier method, but rather checking that my method is correct.

    Although I do not claim this method to be more useful. I think it does yeld a more accurate result over some functions than the method you use, for example if you want to find the derivative with respect of x:

    [tex]\left| x^n \right| \quad n \in \mathbb{N} \backslash \negmedspace \{ 1 \}[/tex]

    But really more than anything I just want to check it is correct.

    Edit: I just realised a big mistake in it, but I've now edited the first post accordingly.
     
    Last edited: Jul 30, 2005
  5. Jul 30, 2005 #4

    matt grime

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    but they are easily seen to be equivalent, indeed your f/modf is simply +/-1 depending on whether f is positive or negative, ie you are just not using the words 'when something is positive/negative' and making it neater, so yes your method is correct in spirit though i haven't checked the detail
     
  6. Jul 30, 2005 #5

    Zurtex

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    Neither do I write:

    The derivative of the positive square root of a squared function of x, instead I represent it like:

    [tex]\left( +\sqrt{(f(x))^2} \right)'[/tex]

    It just fits easier in my head if I have an algebraic notation for something rather than a word notation (probably something to do with my dyslexia but whatever). Anyway, I'm glad the spirit is right, but I think I lack the correct notation with the limits. It is meant to be x approaching from smaller values (values coming from negative infinity) to reach f(x) = 0, but instead I think I've wrote f(x) approaching from smaller values to f(x) = 0, which may not make too much sense for functions like f(x) = x2.
     
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