दिया है: \( xy = e^{\left ( x - y \right )} \)
दोनों पक्षों का log लेने पर \[ \Rightarrow \log xy = {\color{Red} \log e} ^{x-y} \] \[ \Rightarrow \log xy = x - y \]
\[ \Rightarrow \log xy = x - y \]
\[ \because \log \left ( {\color{Emerald} m} . {\color{Orange} n} \right ) = {\color{Emerald} \log m} + {\color{Orange} \log n} \] \[ \therefore \log \left ( {\color{Emerald} x} . {\color{Orange} y} \right ) = {\color{Emerald} \log x} + {\color{Orange} \log y} \]
\[ \Rightarrow \log x + \log y = x - y \]
दोनों पक्षों का \( x \) के सापेक्ष अवकलन करने पर \[ \Rightarrow \frac{d}{dx} \left ( {\color{Emerald} \log x} + {\color{Orange} \log y} \right ) = \frac{d}{dx} \left ( {\color{Emerald} x} + {\color{Orange} y} \right ) \]
हम जानते हैं कि : \( \frac{d}{dx} \left ( {\color{Emerald} u} \pm {\color{Orange} v} \right ) = \frac{d}{dx} {\color{Emerald} u} \pm \frac{d}{dx} {\color{Orange} v} \) \[ \Rightarrow \frac{d}{dx} {\color{Emerald} \log x} + \frac{d}{dx} {\color{Orange} \log y} = \frac{d}{dx} {\color{Emerald} x} - \frac{d}{dx} {\color{Orange} y} \]
\[ \Rightarrow \frac{d}{dx} \log x + \frac{d}{dx} \log y = \frac{d}{dx} x - \frac{d}{dx} y \]
\( {\color{Cyan} \frac{d}{dx} \log y } \) को \( {\color{Cyan} \frac{d}{dy} \log y . \frac{dy}{dx}} \) के रूप में लिखें
\[ \Rightarrow {\color{Emerald} \frac{d}{dx} \log x} + {\color{Cyan} \frac{d}{dy} \log y} = {\color{Emerald} \frac{dx}{dx}} - {\color{Cyan} \frac{dy}{dx}} \]
हम जानते हैं कि : \( {\color{Magenta} \frac{d}{dx} \log x = \frac{1}{x}} \) \[ \Rightarrow {\color{Emerald} \frac{1}{x}} + {\color{Cyan} \frac{1}{y}} . \frac{dy}{dx} = 1 - \frac{dy}{dx} \]
\[ \Rightarrow \frac{1}{x} + \frac{1}{y}.\frac{dy}{dx} = 1 - \frac{dy}{dx} \]
\( {\color{Emerald} \frac{dy}{dx}} \) का पक्षान्तर कर बाएं पक्ष में लाएं तथा \( {\color{Orange} \frac{1}{x}} \) का पक्षान्तर कर दाएं पक्ष में लाएं
\[ \Rightarrow \frac{dy}{dx} + \frac{1}{y} . \frac{dy}{dx} = 1 - \frac{1}{x} \]
\( {\color{Emerald} \frac{dy}{dx}} + \frac{1}{y} . {\color{Emerald} \frac{dy}{dx}} \) में \( {\color{Emerald} \frac{dy}{dx}} \) उभयनिष्ठ है।
\[ \Rightarrow \left ( 1 + \frac{1}{y} \right ) \frac{dy}{dx} = 1 - \frac{1}{x} \]
L.H.S. \( = {\color{Orange} 1} + \frac{1}{{\color{Emerald} y}} = \frac{{\color{Orange} y}}{{\color{Emerald} y}} + \frac{1}{{\color{Emerald} y}} = \frac{{\color{Orange} y}+1}{{\color{Emerald} y}} \)
R.H.S. \( ={\color{Orange} 1} - \frac{1}{{\color{Emerald} x}} = \frac{{\color{Orange} x}}{{\color{Emerald} x}} - \frac{1}{{\color{Emerald} y}} = \frac{{\color{Orange} x}-1}{{\color{Emerald} x}} \)
\[ \Rightarrow \left ( \frac{1+y}{y} \right ) \frac{dy}{dx} = \frac{1-x}{x} \]
\( {\color{Emerald} \frac{y-1}{y}} \) का पक्षान्तर करें।
\[ \frac{dy}{dx} = {\color{Orange} \frac{x-1}{x}} \times {\color{Emerald} \frac{y}{y+1}} = \frac{{\color{Emerald} y}{\color{Orange} \left ( x-1 \right )}}{{\color{Orange} x}{\color{Emerald} \left ( y+1 \right )}} \]\[ \Rightarrow \frac{dy}{dx} = \frac{y\left ( x - 1 \right )}{x\left ( y+1 \right )} \]