Finding the nth Derivative of a Fraction

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Discussion Overview

The discussion revolves around finding a general form for the nth derivative of a fraction, specifically inquiring about the application of product and quotient rules in differentiation. The scope includes theoretical aspects of calculus and mathematical reasoning.

Discussion Character

  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant asks if there is a general form for the nth derivative of a fraction.
  • Another participant provides a general form for the product of two functions and queries how to adapt this for the case of a fraction.
  • A follow-up question seeks clarification on how to determine the coefficients of each term in the general form.
  • A later reply states that the coefficients are binomial coefficients and provides a summation notation for the product rule.

Areas of Agreement / Disagreement

Participants are exploring the topic with some agreement on the use of binomial coefficients, but there is no consensus on the general form for the nth derivative of a fraction itself.

Contextual Notes

There are unresolved aspects regarding the adaptation of the product rule to the case of a fraction and the specific application of the coefficients in the general form.

EngWiPy
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Hello,

Is there a general form for the nth derivative for a fraction?

Thanks in advance.
 
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Hello saeddawoud! :smile:

General form for a product:

(fg)(n) = f(n)g + nC1f(n-1)g(1) + … + fg(n)

so the form for f(1/g) is … ? :wink:
 
tiny-tim said:
Hello saeddawoud! :smile:

General form for a product:

(fg)(n) = f(n)g + nC1f(n-1)g(1) + … + fg(n)

so the form for f(1/g) is … ? :wink:

Thank you, it is really helpful, but how to determine the coefficients of each term in general form?

Regards
 
saeddawoud said:
Thank you, it is really helpful, but how to determine the coefficients of each term in general form?

Regards

The coefficients are just the binomial coefficients. That's what the nC1 in Tiny Tim's expression is. Written in summation notation:

[tex](fg)^{(n)} = \sum_{k=0}^{n}\frac{n!}{k!(n-k)!}f^{(n-k)}g^{(k)}[/tex]
 

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