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3rd Derivative Theorem?

  1. May 21, 2005 #1


    Does a theorem exist for the derivation of a second or third derivative equation, without having to first derive the first and second equation derivatives?

    Example equation:
    [tex]\frac{d^3}{dx^3} \left( \frac{x}{2x - 1} \right)[/tex]

     
  2. jcsd
  3. May 21, 2005 #2
    Take a look at Leibniz Identity.
     
  4. May 21, 2005 #3
    Thats not what he's asking about, and to the OP: I don't believe so.
     
  5. May 21, 2005 #4
    Hmm, yeah - if there is - it will make life much easier :p
     
  6. May 21, 2005 #5
    Well, we can easily make them couldn't we :)?

    [tex]\begin{array}{rcl}\frac{d^3}{dx^3}C &=& \frac{d}{dx}\frac{d}{dx}\frac{d}{dx}C\\
    &=& \frac{d}{dx}\frac{d}{dx}0\\
    &=& \frac{d}{dx}0\\
    &=& 0\end{array}[/tex]

    Thus we get our first theorem:

    [tex]\frac{d^3}{dx^3}C &=& 0[/tex]

    etc etc...
     
    Last edited: May 21, 2005
  7. May 21, 2005 #6

    cronxeh

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    C is a function of x.. you dont get a 0
     
  8. May 21, 2005 #7
    I think he was being sarcastic and claiming C to stand for constant.
     
  9. May 21, 2005 #8

    dextercioby

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    Nope,there's no theorem,but there are functions,"nice" ones,for which u can find,by a mere substitution,the derivative of arbitrary order.

    To give you an example:take "sine".Compute its "n"-th order derivative.

    Daniel.
     
  10. May 21, 2005 #9

    Hurkyl

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    Taylor's theorem. :smile:

    If you can find the Taylor Series for the function around a, then you can read off the value of any derivative you want at a.

    (Say... by using the geometric series formula)
     
  11. May 21, 2005 #10
    Hurkyl's advice is very satisfactory...with the help of taylor series any derivative cn be found..
     
  12. May 21, 2005 #11

    mathwonk

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    actually the example he gave is rather trivial if rewritten as [1/2][ 1 minus something like (2x-1)^(-1)]
     
  13. May 22, 2005 #12
    n factorial...



    My calculus book describes using an [tex]n![/tex] method for obtaining derivatives:

    Example equation:
    [tex]y = \frac{1}{3x^3}[/tex]

    Where:
    [tex]\frac{d^{n}y}{dx^{n}} = \frac{(-1)^n (n+2)!}{(6x^{n+3})}[/tex]

    Then:
    [tex]\boxed{\frac{d^{3}y}{dx^{3}} = - \frac{20}{x^6}}[/tex]

    Unfortunately, the [tex]n![/tex] method originates after obtaining several of the original derivatives first.

    Example equation 2 (post#1):
    [tex]y = \frac{x}{2x - 1}[/tex]

    Where:
    [tex]\frac{d^{n}y}{dx^{n}} = \frac{(-1)^n (2^{n-1}) n!}{(2x-1)^{n+1}}[/tex]

    Then:
    [tex]\boxed{\frac{d^{3}y}{dx^{3}} = - \frac{24}{(2x - 1)^4}}[/tex]

    Which seems to indicate the existence of a missing theorem, therefore, what is the [tex]n![/tex] theorem?

     
    Last edited: May 22, 2005
  14. May 22, 2005 #13
    Theres no theorem here, you're just finding patterns of derivatives and manipulating them. This is taught in Calc I..
     
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