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Math Help - Numerical Derivative

  1. #1
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    Numerical Derivative

    If you know numerical derivative, that is what I'm finding.
    If not, that's okay. I can translate.


    Find the NDER of [tex]x^(3/5)

    Also x^(2/3) .... but I think learning the top will be enough to grasp the concept, unless there is something else worth mentioning for this.

    Thanks!
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Truthbetold View Post
    If you know numerical derivative, that is what I'm finding.
    If not, that's okay. I can translate.


    Find the NDER of [tex]x^(3/5)

    Also x^(2/3) .... but I think learning the top will be enough to grasp the concept, unless there is something else worth mentioning for this.

    Thanks!
    wat do u mean by numerical derivative?
    if im not wrong, are you looking for the answer \frac{3}{5}x^{\frac{-2}{5}}
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  3. #3
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    if im not wrong, are you looking for the answer
    \
    That looks correct, but I don't know how to do this.

    NUmerical derivative is this:

    \frac {f(a+.001)- f(a - .001)}{.002}

    If you don't get it, just find the derivative.

    I don't know how to deal with (a + 0.001)^(3/5)

    It isn't shortcuts, meaning the power rule and stuff, because they are next lesson.

    Thanks!
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  4. #4
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Truthbetold View Post
    ..
    NUmerical derivative is this:

    \frac {f(a+.001)- f(a - .001)}{.002}

    ..
    I don't know how to deal with (a + 0.001)^(3/5)

    It isn't shortcuts, meaning the power rule and stuff, because they are next lesson.

    Thanks!
    where did \frac {f(a+.001)- f(a - .001)}{.002} come from?
    i just used power rule..
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  5. #5
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    I am taught the power rule next lesson.
    You asked for the numerical derivative, and that is what that is.
    I need to use that to figure out the equation.
    I prefer the power rule also, but I can't use it.

    I need to eventually know this for the test as well.
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  6. #6
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Truthbetold View Post
    I am taught the power rule next lesson.
    You asked for the numerical derivative, and that is what that is.
    I need to use that to figure out the equation.
    I prefer the power rule also, but I can't use it.

    I need to eventually know this for the test as well.
    ok, i just dont get where a, +- 0.001, and 0.002 came from.. sorry..
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Truthbetold View Post
    If you know numerical derivative, that is what I'm finding.
    If not, that's okay. I can translate.


    Find the NDER of [tex]x^(3/5)

    Also x^(2/3) .... but I think learning the top will be enough to grasp the concept, unless there is something else worth mentioning for this.

    Thanks!
    If I am correct your teacher is expecting you to approximate the value of the derivative of a function at a given point.

    So instead of the "traditional" derivative formula we use
    f^{\prime}(x) \approx \frac{f(x + 0.001) - f(x - 0.001)}{0.002}
    where we are approximating the formula
    f^{\prime}(x) \approx \frac{f(x + \Delta h) - f(x - \Delta h)}{2 \Delta h}
    (If the 2 in the denominator looks a little funny remember that the denominator is the size of the interval over which we are applying the secant function.)

    So for f(x) = x^{3/5} we get
    f^{\prime}(x) \approx \frac{(x + 0.001)^{3/5} - (x - 0.001)^{3/5}}{0.002}

    This result only has value if you know the general form for the binomial theorem:
    (x + \Delta h)^n = x^n + n \Delta h x^{n - 1} + ~ ...
    (The rest of the terms don't matter as \Delta h is small.)

    So
    f^{\prime}(x) \approx \frac{(x + 0.001)^{3/5} - (x - 0.001)^{3/5}}{0.002} \approx \frac{x^{3/5} + \frac{3}{5} \cdot 0.001x^{3/5 - 1} - x^{3/5} + \frac{3}{5} \cdot 0.001x^{3/5 - 1}}{0.002}

    = \frac{3}{5} \cdot \frac{2 \cdot 0.001 x^{-2/5}}{2 \cdot 0.001}

    = \frac{3}{5}x^{-2/5}

    (If anyone objects to the concept of using the approximation form of the binomial theorem, which I believe is gotten by using differential Calculus, I agree that this makes my argument rather circular. I simply couldn't find a way to turn the approximate derivative into a function, rather than something that needed to be graphed point by point.)

    -Dan
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