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Thread: nth derivative formula

  1. #1
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    nth derivative formula

    find an explicit formula for the $\displaystyle nth$ derivative of $\displaystyle 1/x$

    so, i know $\displaystyle f^{0}(x)=1/x$
    $\displaystyle f^{1}(x)=-1/x^2$
    $\displaystyle f^{2}(x)= 2/x^3$
    $\displaystyle f^{3}(x)=-6/x^4$

    i'm not sure i see the pattern..
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  2. #2
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    $\displaystyle f(x) = \frac{1}{x} = x^{-1}$

    $\displaystyle f(x)' = -1x^{-2}$

    $\displaystyle f(x)'' = (-1)(-2)x^{-3}$

    $\displaystyle f(x)''' = (-1)(-2)(-3)x^{-4}$

    I suggest looking now into that handy little ! operator.
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  3. #3
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    sorry, i'm a little confused.. is there such a thing as (-n)! ? is that even what you're suggesting? you have to forgive me, i'm not very good at this :/
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  4. #4
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    Google the Gamme function.

    Easier way to get that negative, though:

    When n is -4, we have a negative coefficient. -1^(n - 1)

    $\displaystyle (-1)^{n-1}(n - 1)!x^{-n}$

    Edit: Whoops. Technically, this isn't nth derivative. But you don't have to change much. =p

    $\displaystyle (-1)^{n}(n!)x^{-(n + 1)}$

    So, first derivative, we have -1x^-2, right? Formula says, first derivative means n = 1.

    $\displaystyle (-1)^1(1!)x^{-2} = -x^{-2}$

    And n = 2?

    $\displaystyle (-1)^2(2!)x^{-3} = 2x^{-3}$

    Seems to work
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  5. #5
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    okay, so.. i show work by just showing the calculation of the derivatives that you showed earlier?

    thanks for your help!
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  6. #6
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    That should be good enough. But I don't know your instructor.

    $\displaystyle f^1(x) = (-1)x^{-2}$
    $\displaystyle f^2(x) = (-1)(-2)x^{-3}$
    $\displaystyle f^3(x) = (-1)(-2)(-3)x^{-4}$

    Show a few derivatives. Recognize patterns and relationships between numbers and the order of each derivative.

    $\displaystyle (1)(2)(3) = 3!$ occurs @ 3rd-order derivative. So that's how you get $\displaystyle n!$

    $\displaystyle x^{-4}$ @ 3rd order derivative. So that's $\displaystyle -(n + 1)$ or $\displaystyle (1 - n)$

    First order derivative has a negative coefficient, second has positive, third negative, fourth positive, etc. You know that every even gives a positive and every odd gives a negative. Hence $\displaystyle (-1)^n$

    Edit: Another way to show the $\displaystyle (-1)^n$:

    $\displaystyle (-1)(-2)(-3)$ is the same as $\displaystyle (-1)(-1)(-1)(1)(2)(3)$, right? So that's $\displaystyle (-1)^3(1)(2)(3) = (-1)^33!$ right there.
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  7. #7
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    this was a very big help, thanks so much
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  8. #8
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    Once you have "guessed" a formula for the nth derivative of 1/x, it should not be difficult to prove it by induction on n.
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