- #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]

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- Thread starter Orion1
- Start date

- #1

- 970

- 3

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

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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

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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...

[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...

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- #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

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

Daniel.

- #9

Hurkyl

Staff Emeritus

Science Advisor

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If you can find the Taylor Series for the function around

(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

Science Advisor

Homework Helper

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- #12

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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

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