How does the following simplify?

[Rodger]

1. The problem statement

( F(r) $\frac{d}{dr}$ ( $r^{2}$ $\frac{d}{dr}$ ) ) g(r)


2. The attempt at a solution

Does it just melt down to:

( F(r) $r^{2}$ g(r) ) $\frac{d^{2}}{dr^{2}}$

 

[Mark44]

1. The problem statement

( F(r) $\frac{d}{dr}$ ( $r^{2}$ $\frac{d}{dr}$ ) ) g(r)


2. The attempt at a solution

Does it just melt down to:

( F(r) $r^{2}$ g(r) ) $\frac{d^{2}}{dr^{2}}$

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.

 

[haruspex]

( F(r) $\frac{d}{dr}$ ( $r^{2}$ $\frac{d}{dr}$ ) ) g(r)

To add to what Mark44 said, presumably you mean $F(r) \frac{d}{dr}$ ( $r^{2}$ $\frac{d}{dr}g(r)$ ). You can expand the outer d/dr using the product rule.

 

[Rodger]

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) = $r^{2}$

H(r) = $\frac{d}{dr}$ ( k(r) g(r) ) ------> using the product rule

=> $\frac{d}{dr}$ ( F(r) H(r) ) ------> using the product rule...


or this:

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

 

[haruspex]

H(r) = $\frac{d}{dr}$ ( k(r) g(r) ) ------> using the product rule

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

Yes.