Evaluating a 'logarithmic' derivative

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SUMMARY

The integral evaluation of the logarithmic derivative \(\frac{\partial \log\rho (r)}{\partial \log r}\) for the function \(\rho (r) = \rho_0 \Big(1+\big(\frac{r}{\alpha}\big)^2\Big)^{-\frac{3 \beta}{2}}\) is a common task in advanced physics. The simplification provided by the user leads to \(-\frac{3\beta}{2}\log \Bigg[\rho_0^{\frac{-2}{3\beta}} \bigg(1+\big(\frac{r}{A}\big)^2\bigg) \Bigg]\). The discussion emphasizes the application of the chain rule for derivatives, specifically \(\frac{\partial}{\partial\log r} = \frac{\partial r}{\partial\log r}\frac{\partial}{\partial r}\), which is crucial for further evaluation.

PREREQUISITES
  • Understanding of logarithmic derivatives
  • Familiarity with chain rule in calculus
  • Knowledge of physics concepts related to density functions
  • Basic proficiency in handling integrals and derivatives
NEXT STEPS
  • Study the application of logarithmic derivatives in physics
  • Learn advanced calculus techniques, focusing on the chain rule
  • Explore the properties of density functions in statistical mechanics
  • Investigate the implications of constants \(\rho_0\), \(\alpha\), and \(\beta\) in physical models
USEFUL FOR

Students in advanced physics courses, researchers in theoretical physics, and anyone interested in the application of calculus in physical models will benefit from this discussion.

ausdreamer
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Homework Statement



I have to evaluate the following integral:

\frac{\partial \log\rho (r)}{\partial \log r}

for

\rho (r) = \rho_0 \Big(1+\big(\frac{r}{\alpha}\big)^2\Big)^\frac{-3 \beta}{2}

where \rho_0,\alpha,\beta are constants and r is a random variable.

Homework Equations



-

The Attempt at a Solution



I've simplified the derivative to

\frac{\partial \log \rho (r)}{\partial \log r} = -\frac{3\beta}{2}\log \Bigg[\rho_0^\frac{-2}{3\beta} \bigg(1+\big(\frac{r}{A}\big)^2\bigg) \Bigg]

but I'm stuck on where to go from here...I've almost finished my physics degree without encountering such a derivative :P
 
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This comes up a lot in certain fields of research-level physics, but you don't see it that much at undergrad level. Still, it's easy to evaluate using the chain rule:
\frac{\partial}{\partial\log r} = \frac{\partial r}{\partial\log r}\frac{\partial}{\partial r} = \biggl(\frac{\partial\log r}{\partial r}\biggr)^{-1}\frac{\partial}{\partial r}
I'll let you take it from there :wink:

EDIT: actually, your simplification doesn't seem quite right... in any case, I'd recommend starting by using the identity I mentioned. See if that gives you what you need, and if not, post your work in detail.
 
Last edited:

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