Question

If $e^x+e^y=e^{x+y}$, prove that $\frac{dy}{dx}=-\frac{e^x\left(e^y-1\right)}{e^y\left(e^x-1\right)}$

Answer 1

$\mathrm{We}\mathrm{have},e^x+e^y=e^{x+y}$
Differentiating with respect to x, we get
$$\begin{gathered} \Rightarrow \frac{d}{dx}\left(e^x\right)+\frac{d}{dx}\left(e^y\right)=\frac{d}{dx}\left(e^{x+y}\right) \\ \Rightarrow e^x+e^y\frac{dy}{dx}=e^{x+y}\frac{d}{dx}\left(x+y\right) \\ \Rightarrow e^x+e^y\frac{dy}{dx}=e^{x+y}\left[1+\frac{dy}{dx}\right] \\ \Rightarrow e^y\frac{dy}{dx}-e^{x+y}\frac{dy}{dx}=e^{x+y}-e^x \\ \Rightarrow \frac{dy}{dx}=\left(\frac{e^x\times e^y-e^x}{e^y-e^x\times e^y}\right) \\ \Rightarrow \frac{dy}{dx}=\frac{e^x\left(e^y-1\right)}{e^y\left(1-e^x\right)} \\ \Rightarrow \frac{dy}{dx}=-\frac{e^x\left(e^y-1\right)}{e^y\left(e^x-1\right)} \end{gathered}$$