Differentiation / Chain rule - Splitting Logarithms

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SUMMARY

The discussion focuses on the differentiation of logarithmic functions, specifically using the chain rule and properties of logarithms. The participants analyze the expression for the derivative of the logarithm of the eta function, particularly the term log(η(-1/τ)). The correct application of the chain rule is emphasized, demonstrating that the derivative involves both the logarithmic differentiation and the derivative of the inner function. The final expression confirms the correct differentiation process, yielding a result involving E2(−1/τ).

PREREQUISITES
  • Understanding of logarithmic differentiation
  • Familiarity with the chain rule in calculus
  • Knowledge of the eta function and its properties
  • Basic concepts of complex analysis
NEXT STEPS
  • Study the properties of the eta function, particularly η(τ) and η(−1/τ)
  • Learn advanced techniques in logarithmic differentiation
  • Explore the applications of the chain rule in complex functions
  • Investigate the implications of E2(τ) in modular forms
USEFUL FOR

Students and professionals in mathematics, particularly those studying calculus, complex analysis, and modular forms, will benefit from this discussion.

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Homework Statement



holdmeinurpantaloonies.png

Use the top line to get 1) and 2)

Homework Equations


above

The Attempt at a Solution


So for 2) split the log up using ##log (AB)=log (A) + log (B) ## and this is simple enough

I think I may be doing something stupid with 1) though. I have

##\frac{\partial}{\partial \tau} log (\eta(\frac{-1}{\tau})) = \frac{\partial}{\partial \tau} log (\eta(\tau)) \frac{\partial}{\partial \tau}(\frac{-1}{\tau})= \frac{1}{\tau^2} \frac{ \pi i}{12} E_2(\tau)##
 
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##(f(g(x)))^\prime = f^\prime(g(x))g^\prime(x)##

You applied the rule incorrectly,

##\displaystyle \frac{\partial}{\partial \tau} log (\eta(\frac{-1}{\tau})) = \frac{\partial}{\partial \tau} log (\eta(\color{red}{\frac{-1}{\tau}})) \frac{\partial}{\partial \tau}(\frac{-1}{\tau})= \frac{1}{\tau^2} \frac{ \pi i}{12} E_2(\color{red}{\frac{-1}{\tau}})##
 

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