# Reciprocal of n to the -nth power

Split a number into n parts and multiply them.How much parts should you split(to get large product)?
$$(\frac{1}{n})^n=n^{-n}$$
$$n^{-n}\frac{d}{dn}=????$$
I know how to do $$n^c\frac{d}{dn}=cn^{c-1}$$ or $$c^n\frac{d}{dn}=c^n\ln c$$
(yeah ik it has a d/dn===0 point at n=e but i dont want to copy answers)

trial and errored it out:(ln(x)+1)*(-x^-x)

then solve for (ln(x)+1)(-x^-x)=0
you can make the right part (-x) zero:
(ln(x)+1)(0^0)=indeterminiate
whoops
left part (lnx=-1):
(ln(1/e)+1)(1/e^1/e)=(1-1)(some_weird_transcedental)=0
Conclusion:2 roots,x=0 and x=1/e
meaning that you either break the number into 0 parts and explode the calculator or break it making each part close to e

Umm, first, you put this in the wrong category (help with Snap!).

But also, yes, we're happy to have occasional non-Snap!-related threads here, that doesn't mean you should give us a calculus tutorial. There are other places for that.

Thanks.

Oops

(thank u i was just going to change the category)

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