Question

Find $\frac{dy}{dx}$ in each of the following cases:

$e^{x-y}=\log \left(\frac{x}{y}\right)$

Answer 1

$\mathrm{We}\mathrm{have},e^{x-y}=\log \left(\frac{x}{y}\right)$
Differentiate with respect to x,
$$\begin{gathered} \frac{d}{dx}\left(e^{x-y}\right)=\frac{d}{dx}\left\{\log \left(\frac{x}{y}\right)\right\} \\ \Rightarrow e^{\left(x-y\right)}\frac{d}{dx}\left(x-y\right)=\frac{1}{\left({\displaystyle \frac{x}{y}}\right)}\times \frac{d}{dx}\left(\frac{x}{y}\right) \\ \Rightarrow e^{\left(x-y\right)}\left(1-\frac{\mathit{d}y}{\mathit{d}x}\right)=\frac{y}{x}\left[\frac{y{\displaystyle \frac{d}{dx}}\left(x\right)-x{\displaystyle \frac{dy}{dx}}}{y^2}\right] \\ \Rightarrow e^{\left(x-y\right)}-e^{\left(x-y\right)}\frac{\mathit{d}y}{\mathit{d}x}=\frac{1}{xy}\left[y\left(1\right)-x\frac{\mathit{d}y}{\mathit{d}x}\right] \\ \Rightarrow e^{\left(x-y\right)}-e^{\left(x-y\right)}\frac{\mathit{d}y}{\mathit{d}x}=\frac{1}{x}-\frac{1}{y}\frac{\mathit{d}y}{\mathit{d}x} \\ \Rightarrow \frac{1}{y}\frac{\mathit{d}\mathit{y}}{\mathit{d}\mathit{x}}-e^{\left(x-y\right)}\frac{\mathit{d}y}{\mathit{d}x}=\frac{1}{x}-e^{\left(x-y\right)} \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}\left[\frac{1}{y}-\frac{e^{\left(x-y\right)}}{1}\right]=\frac{1}{x}-\frac{e^{\left(x-y\right)}}{1} \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}\left[\frac{1-y{e}^{\left(x-y\right)}}{y}\right]=\frac{1-x{e}^{\left(x-y\right)}}{x} \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=\frac{y}{x}\left[\frac{1-x{e}^{\left(x-y\right)}}{1-y{e}^{\left(x-y\right)}}\right] \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=\frac{-y}{-x}\left[\frac{x{e}^{\left(x-y\right)}-1}{y{e}^{\left(x-y\right)}-1}\right] \\ \Rightarrow \frac{\mathit{d}y}{\mathit{d}x}=\frac{y}{x}\left[\frac{x{e}^{\left(x-y\right)}-1}{y{e}^{\left(x-y\right)}-1}\right] \end{gathered}$$