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