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Question
If \[e^y = y^x ,\] prove that\[\frac{dy}{dx} = \frac{\left( \log y \right)^2}{\log y - 1}\] ?
Solution
\[\text{ We have }, e^y = y^x \]
Taking log on both sides,
\[\log e^y = \log y^x \]
\[ \Rightarrow y \log e = x \log y \]
\[ \Rightarrow y = x \log y . . . \left( i \right)\]
Differentiating with respect to x,
\[\frac{dy}{dx} = \frac{d}{dx}\left( x \log y \right)\]
\[ \Rightarrow \frac{dy}{dx} = x\frac{dy}{dx}\left( \log y \right) + \log y\frac{d}{dx}\left( x \right) \]
\[ \Rightarrow \frac{dy}{dx} = \frac{x}{y}\frac{dy}{dx} + \log y \]
\[ \Rightarrow \frac{dy}{dx}\left( 1 - \frac{x}{y} \right) = \log y\]
\[ \Rightarrow \frac{dy}{dx}\left( \frac{y - x}{y} \right) = \log y\]
\[ \Rightarrow \frac{dy}{dx} = \frac{y \log \ y}{y - x}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{y \log \ y}{\left( y - \frac{y}{\log \ y} \right)} \left[ \text{ Using equation} \left( i \right) \right]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{y \log y\left( \log y \right)}{y \log \ y - y}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{y \left( \log y \right)^2}{y\left( \log \ y - 1 \right)}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{\left( \log y \right)^2}{\left( \log y - 1 \right)}\]
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