help with differentiation?

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selpak

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Y^x = x^y find dy/dx .. any clues?


PS: its not a homework !
 

Or an interview question? The solution is surprizingly simple, try to find some values that belongs to the solution, could start with tupel (0,0).
 

no, not an interview question :D
my teacher just told me to solve it without having to solve it only by myself.
I`m just gonna get 5+ marks if i got the problem solved

ANYWAY
i can differentiate one of the sides
but i dunno how to differentiate the other, so any clues?
 

I don't see if it could be solved by any general method. However you can easily see that x=y is a solution. You should be able to show, that no other solution exists.
 

this is what i get
dy/dx = ((y/x)-ln(y))÷((x/y)-ln(x))

this is how i did it
y^x = x^y
ln(y^x) = ln(x^y)
x*ln(y) = y*ln(x)
use product rule and differentiate sides using implicit differentiation and solve for dy/dx and you get the answer at the top.

I might be wrong so please check my work.
 

y x^(y-1)-ln y^x

thats what i get using The Microsoft Math software

the left side is clear
but i dunno how it got the right side :S
 

selpak said:
y x^(y-1)-ln y^x

thats what i get using The Microsoft Math software

the left side is clear
but i dunno how it got the right side :S
Click to expand...

what do you mean left and right side? I dont see an equal sign in your reply. please elaborate.
 

before differntiation it was x^y - y^x = 0
or x^y = y^x


plz help... i have no much time :S
 

selpak said:
before differntiation it was x^y - y^x = 0
or x^y = y^x


plz help... i have no much time :S
Click to expand...


I think what I posted earlier is correct, unless you can prove the steps that I took are mathematically incorrect.
 

Hi,
Just for fun, it is a standard text book problem and this is how it is done:

Starting from where vmp has left:
xlny = ylnx
Differentiating both sides,

x*1/y*dy/dx + lny* dx/dx = y*1/x + lnx*dy/dx .

dy/dx(x/y - lnx) = (y/x - lny) ; since dx/dx = 1

So, dy/dx = (y/x - lny ) ÷ (x/y - lnx)
= 1/x(y-xlny) ÷ 1/y(x-ylnx)
= y/x (y-xlny) ÷ (x-ylnx)
and EUREKA ... The answer is same as what vmp has told and it is right too.

Regards,
Laktronics
 

i found answers on yahoo.com
i`m gonna post them tomorrow ..
thanx vmp
 

find dy/dx implies generally a function in x, not in x and y, I think. Thus I wouldn't call the said function a solution. It's rather a rearrangement.

P.S.: I've seen that it could be meaningful to calculate a differential as shown.
 

thanks laktronics for confirming.
 

Laktronics ur the best

thanx ery1
 

I don't know.

The fact that you are looking for dy/dx implies that there is a functional relationship between x and y (i.e. x and y are not just a single value each).

I think what FvM said in the beginning was right.
x^y = y^x implies that y=x

which implies that dy/dx = 1

If you are implying that y=x is not the only function satisfying this relationship, please show another function which which satisfies x^y = y^x ?

Added after 2 hours 23 minutes:

ok, maybe there are other solutions hiding out there somewhere that just don't come to mind easily
 

Attached is a solution of the function that satisfies x^y=y^x using Maple

First just for info I plotted x^y

Then I defined Z = y^x-x^y

Then I substituted y=x^t into z giving
z := (x*t)^x-x^(x*t)

Then I solved for values of t that would make z=0 (t=1 would correspond to y=x).

Maple comes up with the solution:
t = -1/ln(x)*LambertW(-ln(x)/x)

The plot of that functon looks like 1 below e and then drops down above e.
The part below 1 indicates other functions besides y=x.

For example
x=5, y = 5 * (.3529843829051551765517446)
xx^y - y^x = .1e-22

Who'd of thunk it.
 

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