Question

If $e^x+e^y=e^{x+y}$, prove that $\frac{dy}{dx}+e^{y-x}=0$

Answer 1

$\mathrm{We}\mathrm{have},e^x+e^y=e^{x+y}$ ...(1)
Differentiating both sides using chain rule,
$$\begin{gathered} \frac{d}{dx}\left(e^x\right)+\frac{d}{dx}\left(e^y\right)=\frac{d}{dx}\left(e^{x+y}\right) \\ \Rightarrow e^x+e^y\frac{dy}{dx}=e^{x+y}\frac{d}{dx}\left(x+y\right) \\ \Rightarrow e^x+e^y\frac{dy}{dx}=e^{x+y}\left[1+\frac{dy}{dx}\right] \\ \Rightarrow e^y\frac{dy}{dx}-e^{x+y}\frac{dy}{dx}=e^{x+y}-e^x \\ \Rightarrow \frac{dy}{dx}=\frac{e^{x+y}-e^x}{e^y-e^{x+y}} \\ \Rightarrow \frac{dy}{dx}=\frac{e^{x+y}-e^{x+y}+e^y}{e^{x+y}-e^x-e^{x+y}}\left[\mathrm{Using}\mathrm{eqn}.\left(1\right)\right] \\ \Rightarrow \frac{dy}{dx}=\frac{e^y}{-e^x} \\ \Rightarrow \frac{dy}{dx}=-e^{y-x} \\ \Rightarrow \frac{dy}{dx}+e^{y-x}=0 \end{gathered}$$