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## #1 2005-12-30 14:43:02

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

### D(x^x)

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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## #2 2005-12-30 17:42:52

Flowers4Carlos
Member
Registered: 2005-08-25
Posts: 106

### Re: D(x^x)

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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## #3 2005-12-30 23:32:04

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

### Re: D(x^x)

And how is
∫(x^x)dx?

IPBLE:  Increasing Performance By Lowering Expectations.

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## #4 2005-12-31 03:04:03

John E. Franklin
Member
Registered: 2005-08-29
Posts: 3,588

### Re: D(x^x)

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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## #5 2006-01-01 10:46:41

Flowers4Carlos
Member
Registered: 2005-08-25
Posts: 106

### Re: D(x^x)

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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## #6 2006-01-01 11:27:20

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

### Re: D(x^x)

And for the integral - it can't be simplified.

IPBLE:  Increasing Performance By Lowering Expectations.

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## #7 2006-01-01 11:29:15

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

### Re: D(x^x)

,I think.

IPBLE:  Increasing Performance By Lowering Expectations.

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## #8 2006-01-01 12:38:10

God
Member
Registered: 2005-08-25
Posts: 59

### Re: D(x^x)

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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## #9 2006-01-01 12:45:01

krassi_holmz
Real Member
Registered: 2005-12-02
Posts: 1,905

### Re: D(x^x)

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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