How is this Differentiation Trick Manipulated in Calculus?

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Discussion Overview

The discussion revolves around a differentiation technique in calculus, specifically how to manipulate expressions involving derivatives and functions of a variable, r. Participants are exploring the transition between two forms of a derivative expression and seeking clarity on the manipulation involved.

Discussion Character

  • Technical explanation
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant expresses confusion about how to manipulate the expression from one form to another, specifically questioning the steps involved in the differentiation process.
  • Another participant provides a potential expansion of the derivative, suggesting that the two expressions are indeed different, which raises questions about the validity of the manipulation.
  • A third participant humorously questions the role of "x" in the expressions, indicating a lack of clarity about the variables involved.
  • A later reply reiterates the need to start over and emphasizes that "x" is a function of r, suggesting that expanding both expressions will yield the same result, but the direct transition remains unclear.

Areas of Agreement / Disagreement

Participants do not appear to reach a consensus on the manipulation of the expressions. There are competing views on whether the two forms are equivalent and how to properly derive one from the other.

Contextual Notes

There are unresolved assumptions regarding the definitions of the variables and the functions involved, as well as the steps necessary for the manipulation of the expressions. The discussion does not clarify the conditions under which the equivalence holds.

Who May Find This Useful

This discussion may be useful for students and practitioners of calculus who are grappling with differentiation techniques and the manipulation of derivative expressions.

Baggio
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I keep seeing this trick everywhere but I don't see how it is done.

how do we go from

[itex]\frac{1}{r^{2}}\frac{d}{dr}(r^{2}\frac{d \rho}{dr} [\latex]<br /> <br /> to<br /> <br /> [itex]\frac{1}{r} \frac{d^{2} \rhor}{dr^2{}}[\latex]<br /> <br /> Ugggh can't get latex to work anyway it's<br /> <br /> (1/r^2)(d/dr(r^2. dx/dr)<br /> <br /> how do we go from that to<br /> <br /> (1/r)(d/dr(r^2.d(xr)/dr))<br /> <br /> I know for sure that they're equal I just don't know how to manipulate it! :(<br /> <br /> Thanks[/itex][/itex]
 
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Use [ tex ] [ /tex ] (without the spaces).

[tex]\frac{1}{r^{2}}\left(2r\frac{d\rho}{dr}+r^{2}\frac{d^{2}\rho}{dr^{2}}\right)<br /> =\frac{2}{r}\frac{d\rho}{dr}+\frac{d^{2}\rho}{dr}[/tex]

and u can see pretty clearly the 2 things are different.

Daniel.
 
Who's "x" and what does he do...?

Daniel.
 
No, let's start again

(1/r^2)(d/dr(r^2. dx/dr)

&

(1/r)(d/dr(r^2.d(xr)/dr))

x is a function of r, if you expand both you get the same result, but how can I go directly from the top eq to the bottom
 

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