I've looked everywhere!

There has to be a special rule for finding the derivative of (x^x), because (x^x) is not an exponential expression, is it constant? No... can't be....

gaah! Anyone good with math/calc help me! :)

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- 10-02-2002BMJGah! More math!
I've looked everywhere!

There has to be a special rule for finding the derivative of (x^x), because (x^x) is not an exponential expression, is it constant? No... can't be....

gaah! Anyone good with math/calc help me! :) - 10-02-2002BMJ
This isn't homework BTW; I'm working on some calculus algorithms :)

- 10-02-2002ygfperson
the derivative of (x^x) is (ln(x) + 1) * x^x

why? my calculator told me so. :D - 10-02-2002BMJ
That doesn't help... I got the same thing on my calculator; I need to be able to

*get there*, I cannot find**any**forumals, proof, identities... ANYTHING that helps me :mad: grrrr... this bites - 10-02-2002ygfperson
let's start off at the basics

hmm...

the derivative of a*n^r == a*r*n^(r-1)

however, 1 * x^x == 1 * x * x^(x-1) == 1 * x^x

hmm... - 10-02-2002ygfperson
maybe this will help

http://members.lycos.co.uk/kgsfs/xpwrx.html

//edit: here's the definite solution:

http://mathforum.org/library/drmath/view/53419.html - 10-02-2002Captain Penguin
Ahh good 'ol calculus...

had to whip out my stuff from last year, but here you go: (It's called logarithmic differentiation, btw)

y = x^x

take natural log of both sides

ln(y) = x*ln(x) < - this last part is a property of logarithms

now take the derivative of both sides...

(dy/dx)*ln(y) = (dy/dx)(x*ln(x))

because taking the derivative of y gives dy/dx....

(1/y)*(dy/dx) = 1*ln(x) + x*(1/x) < - product rule

(1/y)*(dy/dx) = ln(x) + 1 <- just simplified the right side...

multiply both sides by y...

(dy/dx) = y(ln(x) + 1)

and since we know y = x^x...

(dy/dx) = x^x * (ln(x) + 1)

and voila.

I don't know how that would be programmed easily, but thats how you get there.. - 10-02-2002Captain PenguinQuote:

[edit]

and that drmath explanation.. kinda stinks. He is making it overly complicated by doing e^(log(x^x))... - 10-02-2002BMJ
thanks guys... I figured it out before I checked back here.

I just need to build a math function for the special case of (x^x) because it is neither a power expression nor an exponential expression...

(d/dx)[x^x] = [(ln(x)+1)(x^x)] works great!