Im trying to solve for

$$\frac{\partial^2 u}{\partial x^2}$$

for the following system:

  1. $ v + \ln(u) =xy$
  2. $u + \ln(v)= x-y $

I have worked out that $$\frac{\partial u}{\partial x}= \frac{u(y-v)}{1-uv}$$

I applied the quotient rule while trying to solve for $\frac{\partial^2 u }{\partial x^2}$ $$\frac{\partial }{\partial x}\left[\frac{u(y-v)}{1-uv}\right]$$ but got huge messy derivative. But when I put it into a calculator the answer was $0$.

$0$ would make sense if $u$ and $v$ were no longer implicitly defined in terms of $x$. But I don't see why they shouldn't be. Is the answer here $0$ by treating $u, v y$ as constants or should I apply the quotient rule and get what seems to be a long messy answer? If so why??


I think when you put it in calculator you did not consider either $u$, $y$ and $v$ as a function of $x$.

  • 1
    $\begingroup$ Yes. There was no room to define the functions.Thats probably why. So you would say my original way was correct? $\endgroup$ – Red Apr 27 '17 at 11:20
  • $\begingroup$ That is I believe. Be careful with the mess and good luck. $\endgroup$ – SaeidAli Apr 27 '17 at 11:21

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