A rule of logs and derivitaves

[AStaunton]

in my notes I have:

$$\frac{R^{\prime\prime}}{R^{\prime}}=-\frac{1}{r}\implies\frac{d}{dr}(\ln|\frac{dR}{dr}|)=-\frac{d}{dr}ln|r|$$

where R is a function of r.

can someone please explain in as redundant detail as your patience allows, how the first equality implies the second.,..I think it's probably down to rules of logs that I don't quite understand..please feel free to explain steps that you may consider "obvious", as I am a bit of a moron.

 

[tiny-tim]

Hi AStaunton!

Put R' = S …

then S'/S = -1/r

ie dS/S = -dr/r ​

 

[AStaunton]

thanks

I still don't see how that leads to the second equality...would you mind showing a few more intermediate steps?

 

[tiny-tim]

Don't you know ∫ (1/r) dr = ln|r| + constant ?

 

[AStaunton]

I do...it was something else but I see it now. Thanks