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Topic review (newest first)

2006-01-02 11:45:01

I've tested this long before.
Also Mathematica v5.2 Maple v10 and MatLab v7.1 can't expand it.

2006-01-02 11:38:10

I couldn't get it to work and my TI-89 Titanium couldn't either and neither could the wolfram Integrator.

2006-01-02 10:29:15

,I think.

2006-01-02 10:27:20

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

2006-01-02 09:46:41

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 x or x 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 tongue

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

John E. Franklin
2006-01-01 02:04:03

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.

2005-12-31 22:32:04

And how is

2005-12-31 16:42:52

hey john!

there is a similar problem like this here:

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.

John E. Franklin
2005-12-31 13:43:02

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?

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