no, not an interview question
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)÷((x/y)-ln(x))
this is how i did it
y^x = x^y
ln(y^x) = ln(x^y)
x*ln = 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.
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.
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