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

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