Differentiating $e^x + e^y = e^{x + y}$

Question

Differentiating with respect to x, $e^x + e^y = e^{x + y}$, could anyone give me a hint? I do not know even how to start, taking the ln of both sides does not solve the problem.

Answer 1

$$ \exp(x)=\exp(x+y)-\exp(y)=\exp(y)(\exp(x)-1) $$

$$ \exp(y)=\frac{\exp(x)}{\exp(x)-1} $$

$$ y(x)=\ln\left[\frac{\exp(x)}{\exp(x)-1}\right]=x-\ln(\exp(x)-1) $$

$$ \frac{\mathrm{d}y}{\mathrm{d}x}=1-\frac{\exp(x)}{\exp(x)-1} $$

$$ \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\exp(x)-1}{\exp(x)-1}-\frac{\exp(x)}{\exp(x)-1} $$

$$ \frac{\mathrm{d}y}{\mathrm{d}x}=\frac{1}{1-\exp(x)} $$

Answer 2

HINT:

As $e^{x+y}\ne0$ divide both sides by $e^{x+y}$

$$e^{-x}+e^{-y}=1\implies-e^{-x}+e^{-y}\left(-\dfrac{dy}{dx}\right)=0$$

Answer 3

You didn't provide enough information, so I am going to assume that you are taking the derivative with respect to $x$, that $y = y(x)$, and that are familiar with the chain rule.

Let $f(x) = e^x$, so $f'(x)= e^x $

$\frac{d}{dx}f(y(x)) = f'(y(x))y'(x) = e^y \frac{dy}{dx} $

$\frac{d}{dx}f(x+y(x)) = f'(x+y(x))(1 + y'(x)) = e^{x+y}(1+ \frac{dy}{dx}) $

So, after differentiating, the equation becomes

$$e^x+e^y \frac{dy}{dx} = e^{x+y}(1+ \frac{dy}{dx}) $$