Question

Find  \[\frac{dy}{dx}\] in the following case \[e^{x - y} = \log \left( \frac{x}{y} \right)\] ?

 

Answer 1

\[\text{ We have, } e^{x - y} = \log\left( \frac{x}{y} \right)\]

Differentiate with respect to x,

\[\frac{d}{dx}\left( e^{x - y} \right) = \frac{d}{dx}\left\{ \log\left( \frac{x}{y} \right) \right\}\]
\[ \Rightarrow e^\left( x - y \right) \frac{d}{dx}\left( x - y \right) = \frac{1}{\left( \frac{x}{y} \right)} \times \frac{d}{dx}\left( \frac{x}{y} \right) \]
\[ \Rightarrow e^\left( x - y \right) \left( 1 - \frac{d y}{d x} \right) = \frac{y}{x}\left[ \frac{y\frac{d}{dx}\left( x \right) - x\frac{d y}{d x}}{y^2} \right] \]
\[ \Rightarrow e^\left( x - y \right) - e^\left( x - y \right) \frac{d y}{d x} = \frac{1}{xy}\left[ y\left( 1 \right) - x\frac{d y}{d x} \right]\]
\[ \Rightarrow e^\left( x - y \right) - e^\left( x - y \right) \frac{d y}{d x} = \frac{1}{x} - \frac{1}{y}\frac{d y}{d x}\]
\[ \Rightarrow \frac{1}{y}\frac{d y}{d x} - e^\left( x - y \right) \frac{d y}{d x} = \frac{1}{x} - e^\left( x - y \right) \]
\[ \Rightarrow \frac{d y}{d x}\left[ \frac{1}{y} - \frac{e^\left( x - y \right)}{1} \right] = \frac{1}{x} - \frac{e^\left( x - y \right)}{1}\]
\[ \Rightarrow \frac{d y}{d x}\left[ \frac{1 - y e^\left( x - y \right)}{y} \right] = \frac{1 - x e^\left( x - y \right)}{x}\]
\[ \Rightarrow \frac{d y}{d x} = \frac{y}{x}\left[ \frac{1 - x e^\left( x - y \right)}{1 - y e^\left( x - y \right)} \right]\]
\[ \Rightarrow \frac{d y}{d x} = \frac{- y}{- x}\left[ \frac{x e^\left( x - y \right) - 1}{y e^\left( x - y \right) - 1} \right]\]
\[ \Rightarrow \frac{d y}{d x} = \frac{y}{x}\left[ \frac{x e^\left( x - y \right) - 1}{y e^\left( x - y \right) - 1} \right]\]