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Differentiating x^x is the following according to an online differentiator
x x ^ (x - 1) + ln x x ^ x
But I was wondering how you arrive at this?
igloo myrtilles fourmis
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hey john!
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.
Last edited by Flowers4Carlos (2005-12-30 17:43:51)
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And how is
∫(x^x)dx?
IPBLE: Increasing Performance By Lowering Expectations.
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Before we attack the integral as asked by krassi,
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.
igloo myrtilles fourmis
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johny!!!!!!!!!!!!!!
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!!
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And for the integral - it can't be simplified.
IPBLE: Increasing Performance By Lowering Expectations.
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,I think.
IPBLE: Increasing Performance By Lowering Expectations.
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I couldn't get it to work and my TI-89 Titanium couldn't either and neither could the wolfram Integrator.
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I've tested this long before.
Also Mathematica v5.2 Maple v10 and MatLab v7.1 can't expand it.
Last edited by krassi_holmz (2006-01-01 12:46:29)
IPBLE: Increasing Performance By Lowering Expectations.
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