Differentiation / Chain rule - Splitting Logarithms

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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}})##
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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