##y## isn't a constant with respect to ##x##. The original equation defines a function implicitly; the value of any parameter is dependent upon the other two. I've plugged the equation into mathematica, and the ##\frac{\partial y}{\partial x}## and ##\frac{\partial x}{\partial y}## terms which...
\begin{cases}
\frac{\partial u}{\partial x}=\frac{u}{x(1-u\ln{x})}+\frac{u\ln{u}}{1-y}\frac{\partial y}{\partial x}\\
\frac{\partial u}{\partial y}=\frac{u\ln{x}}{1-y}+\frac{u^2}{x(1-u\ln{x})}\frac{\partial x}{\partial y}
\end{cases}
is what I've got so far.
I began doing a bunch of algebra...
Thanks, Ayre!
Seems to boil down to a combination of the ubiquitous chain rule for logarithms of functions and the product rule. So clear now! I should've seen it... hopefully a bit of mathematical intuition returns with study.
Homework Statement
u = x^u + u^y
Find the partial derivatives of ##u## w.r.t. ##x, y##.Homework Equations
Only the one.
The Attempt at a Solution
I've attempted reducing the problem using logs, but the resulting equations seem no more tenable to me. I'm sure there is a nice trick... it...