Also Mathematica v5.2 Maple v10 and MatLab v7.1 can't expand it.]]>

hehe.. sorry bout that.

think of y as a function in terms of x, that is, y=f(x). can you say [d/dx]f(x) = 1??? no, because f(x) might be x² or x³ or it might not be differentiable. in our problem, we can rewrite it like this:

f(x) = x^x

lnf(x) = xlnx

[d/dx]lnf(x) = [d/dx](xlnx)

f'(x)/f(x) = lnx + x/x

you don't have to replace y=f(x), it just makes the problem look much easier, which is what i should have done in the first place. this is also called "implicit differentiation". go here for more examples and a thorough explanation

http://archives.math.utk.edu/visual.calculus/3/implicit.7/

krassi: oh geee... i tried using integration by parts on that creatuer but i ended up w/ an uglier looking monster. i'm baffled!!

]]>I am most intrigued by these 2 lines.

I wouldn't have figured this part out.

Very nice.

lny = xlnx now differentiate both sides

y'/y = lnx + x/x

I understand the product rule on the right side of equation,

but the y'/y surprised me because I probably would have

just thought of 1/y, but I guess because we are differentiating

with respect to x this happens? Please enlighten me to the

difference with lny going to y'/y, but if lnx was on right side of

equation, we don't write down x'/x, just 1/x. I think I am

missing something key and basic.

∫(x^x)dx?]]>

there is a similar problem like this here:

http://www.mathsisfun.com/forum/viewtopic.php?id=2310

i'll work it out for ya!

y = x^x apply log to both sides

lny= lnx^x

lny = xlnx now differentiate both sides

y'/y = lnx + x/x

y' = y(lnx + 1) remember that y=x^x

y' = (x^x)(lnx + 1)

y' = (x^x)lnx + x^x

y' = (x^x)lnx + x*x^(x-1)

that last part is typical calculator manipulation. they always spit out some funky looking solutions like x*x^(x-1) when it could just be x^x.

]]>x x ^ (x - 1) + ln x x ^ x

But I was wondering how you arrive at this?

]]>