# $e^{-xy} + e^{xy} = 2e^{-y}$ - where am I going wrong?

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?

• 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 $f(x)=g(x)$ ? – Maximilian Janisch Jan 27 at 17:47
• Thanks. This explains my mistake! If you post this in the main thread, I will accept it as an answer explaining my mistake. – Srini Jan 27 at 17:53
• You are welcome, I posted it as an answer 😀 – Maximilian Janisch Jan 27 at 17:57

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 $$f'(x)=g'(x)\iff x=0.$$

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

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

• I am not partially differentiating equation in Step 1. I am partially differentiating the orginal equation w.r.t. x – Srini Jan 27 at 17:46

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.

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 $$y$$.