3rd Derivative Theorem?

  • Thread starter Orion1
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  • #1
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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]

 

Answers and Replies

  • #2
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Orion1 said:


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]


Take a look at Leibniz Identity.
 
  • #3
2,210
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Thats not what he's asking about, and to the OP: I don't believe so.
 
  • #4
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Hmm, yeah - if there is - it will make life much easier :p
 
  • #5
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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:
  • #6
cronxeh
Gold Member
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C is a function of x.. you dont get a 0
 
  • #7
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I think he was being sarcastic and claiming C to stand for constant.
 
  • #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.
 
  • #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)
 
  • #10
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Hurkyl's advice is very satisfactory...with the help of taylor series any derivative cn be found..
 
  • #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)]
 
  • #12
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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:
  • #13
2,210
1
Theres no theorem here, you're just finding patterns of derivatives and manipulating them. This is taught in Calc I..
 

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