I am trying to solve the following problem:
If $u(x,y) = x^u + u^y$, find $\frac{\partial u}{\partial x}$ and $\frac{\partial u}{\partial y}$
My first thought was $$\frac{\partial u}{\partial x} = \ln(x)x^u\frac{\partial u}{\partial x} + yu^{y-1}\frac{\partial u}{\partial x} $$
but this simplifies to
$$ \frac{1 - yu^{y-1}}{\ln(x)} = u $$
which doesn't help. Am I on the right track with treating $x^{u(x,y)}$ as an exponential function and $u(x,y)^y$ as a polynomial? Or does this require a totally different approach?
We have, by definition (assuming $x>0$), $x^{u}=e^{u\log x}$, so differentiating with respect to $x$ gives \begin{align} e^{u\log x}\cdot\left(\frac{\partial u}{\partial x}\log x+ \frac{u}{x}\right)= x^u\cdot \left(\frac{\partial u}{\partial x}\log x+ \frac{u}{x}\right). \end{align} You’re missing the second term, which comes from the product rule. So, now the $\frac{\partial u}{\partial x}$ will not cancel from your equation, and you can rearrange to solve for it in terms of $u,x,y$.