What is d(m^m)?

Question

Hi everyone, I was wondering what the derivative of m^m is with respect to m, it seems you could use the product rule and get

d(m) * m^(m-1) + m * d(m^(m-1))
1 * m^(m-1) + m * (d(m) * m^(m-2) + m * d(m^(m-2)))
etc.

which would be just m^m again. But this makes d(m^m) be the same as d(e^m), so that can't be right, can it?

Answer 1

If you let a function y = m^m, you have:

ln y = m ln m

differentiate both sides with respect to m:

1/y * (dy/dm) = m(1/m) + ln m
dy/dm * 1/y = 1+ln m
dy/dm = y(1+ln m)

and therefore dy/dm = m^m (1+ ln m).

and here's a "two wrongs make a right" method by Tiankai Liu (former IMO contestant) You can read about it here: http://www.geocities.com/buniakowski/Math/tips.html

Answer 2

This one is tricky. You cannot use the rules as is.

The key insight is that m^m = e^[m*ln(m)]

Now you can use the chain rule to see that:

d( e^[m*ln(m)]) = (1+ln(m))e^[m*ln(m)] = (1+ln(m))m^m

Answer 3

let T=m^m
taking ln of both sides
ln T=m ln m
dT/T=[1+ln m]dm
dT/dm=T[1+ln m]
d(m^m)/dm=[1+ln m]m^m

Answer 4

d(m^m) = (ln(m) + 1)*m^m

Answer 5

No. It is better to write f(m)=m^m = e^(m ln(m)).
f' = (m ln(m))'e^(m ln(m)) = (1 + ln(m))e^(m ln(m)) = (1 + ln(m)) m^m

Answer 6

Actually, it's m^m(1+Log(m)), so it's not "m^m again!"