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How does the following simplify?

  • Thread starter Rodger
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  • #1
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1. The problem statement

( F(r) [itex]\frac{d}{dr}[/itex] ( [itex]r^{2}[/itex] [itex]\frac{d}{dr}[/itex] ) ) g(r)


2. The attempt at a solution

Does it just melt down to:

( F(r) [itex]r^{2}[/itex] g(r) ) [itex]\frac{d^{2}}{dr^{2}}[/itex]
 

Answers and Replies

  • #2
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1. The problem statement

( F(r) [itex]\frac{d}{dr}[/itex] ( [itex]r^{2}[/itex] [itex]\frac{d}{dr}[/itex] ) ) g(r)


2. The attempt at a solution

Does it just melt down to:

( F(r) [itex]r^{2}[/itex] g(r) ) [itex]\frac{d^{2}}{dr^{2}}[/itex]
It pretty much doesn't make sense at all. Except in the context of operators, which I don't think is applicable here, d/dr and d2/dr2 don't stand on their own.

The derivative operator d/dr should be applied to some function of r, such as d/dr(r2), which simplifies to 2r.
 
  • #3
haruspex
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( F(r) [itex]\frac{d}{dr}[/itex] ( [itex]r^{2}[/itex] [itex]\frac{d}{dr}[/itex] ) ) g(r)
To add to what Mark44 said, presumably you mean [itex]F(r) \frac{d}{dr}[/itex] ( [itex]r^{2}[/itex] [itex]\frac{d}{dr}g(r)[/itex] ). You can expand the outer d/dr using the product rule.
 
  • #4
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sure, I don't trust operators so forgive me but I have to ask if I can presume that the order of working through such an example might then follows something like this...

k(r) = [itex]r^{2}[/itex]

H(r) = [itex]\frac{d}{dr}[/itex] ( k(r) g(r) ) ------> using the product rule

=> [itex]\frac{d}{dr}[/itex] ( F(r) H(r) ) ------> using the product rule...


or this:

=> F(r) x { [itex]\frac{d}{dr}[/itex] [ [itex]r^{2}[/itex] x [itex]\frac{dg(r)}{dr}[/itex] ] }, using the product rule for the parts in the [] brackets
 
  • #5
haruspex
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H(r) = [itex]\frac{d}{dr}[/itex] ( k(r) g(r) ) ------> using the product rule

=> [itex]\frac{d}{dr}[/itex] ( F(r) H(r) ) ------> using the product rule...
I see no way to move the F(r) inside the derivative like that. It certainly is not the product rule.
or this:

=> F(r) x { [itex]\frac{d}{dr}[/itex] [ [itex]r^{2}[/itex] x [itex]\frac{dg(r)}{dr}[/itex] ] }, using the product rule for the parts in the [] brackets
Yes.
 

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