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I am trying to see if there is any $x$ (real or complex) for which this equation can be solved.

$$e^{-xy} + e^{xy} = 2e^{-y}$$

Step 1. Multiplying both sides by y, $$ye^{-xy} + ye^{xy} = 2ye^{-y}$$

Step2. Partially differentiating the original equation (i.e. $e^{-xy} + e^{xy} = 2e^{-y}$ ) w.r.t. x, $$-ye^{-xy} + ye^{xy} = 0$$

Step3. Adding Steps 1 and 2, $$2ye^{xy} = 2ye^{-y}$$ this gives $x=-1$.

Obviously this is not a solution. Where am I going wrong? or is there a complex value of x for which Step 3 holds?

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    $\begingroup$ Why can you partially differentiate with respect to $x$ ? For example $f(x)=x$ and $g(x)=x^3+x+1$ satisfy $f'(x)=g'(x)$ at $x=0$ but this is not a solution to $f(x)=g(x)$ ? $\endgroup$ – Maximilian Janisch Jan 27 at 17:47
  • $\begingroup$ Thanks. This explains my mistake! If you post this in the main thread, I will accept it as an answer explaining my mistake. $\endgroup$ – Srini Jan 27 at 17:53
  • $\begingroup$ You are welcome, I posted it as an answer 😀 $\endgroup$ – Maximilian Janisch Jan 27 at 17:57
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In general the solutions of the equations $f'(x)=g'(x)$ and $f(x)=g(x)$ are different.

For example, if $f(x)=x$ and $g(x)=x^3+x+1$ for real $x$, then $$f(x)=g(x)\iff x=-1$$ but $$f'(x)=g'(x)\iff x=0.$$

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In step 2, when you differentiate with respect to $x$, you will have a factor of $y$.

So, $-y^2e^{-xy}+y^2e^{xy}=0$

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  • $\begingroup$ I am not partially differentiating equation in Step 1. I am partially differentiating the orginal equation w.r.t. x $\endgroup$ – Srini Jan 27 at 17:46
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The derivative of two different constant functions is zero. The two functions are not equal. All equality of derivatives tells you is that the two functions agree up to a constant offset (recall the "${}+C$" from integral calculus).

The partial derivative with respect to $x$ of two different functions depending only on $y$ is zero. All equality of partial derivatives with respect to $x$ tells you is that the two functions agree up to an arbitrary function in $y$. Your right-hand side is not "$0$"; it's "$F(y)$", which isn't nearly as convenient for your solution method.

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The other answers, but your approach in Step 2 has an error for another reason:

Even if you were looking for an $x$ such that $f(x,y)=g(x,y)$ for all $y$, this would hold if you differentiate both sides of this by $x$ and not by $y$.

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