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Question
If \[e^x + e^y = e^{x + y}\] , prove that
\[\frac{dy}{dx} + e^{y - x} = 0\] ?
Solution
\[\text{ We have}, e^x + e^y = e^{x + y}\]
Differentiating both sides using chain rule,
\[\frac{d}{dx}\left( e^x \right) + \frac{d}{dx}\left( e^y \right) = \frac{d}{dx}\left( e^{x + y} \right)\]
\[ \Rightarrow e^x + e^y \frac{dy}{dx} = e^{x + y} \frac{d}{dx}\left( x + y \right)\]
\[ \Rightarrow e^x + e^y \frac{dy}{dx} = e^{x + y} \left[ 1 + \frac{dy}{dx} \right]\]
\[ \Rightarrow e^y \frac{dy}{dx} - e^{x + y} \frac{dy}{dx} = e^{x + y} - e^x \]
\[ \Rightarrow \frac{dy}{dx} = \frac{e^{x + y} - e^x}{e^y - e^{x + y}}\]
\[ \Rightarrow \frac{dy}{dx} = \frac{e^{x + y} - e^{x + y} + e^y}{e^{x + y} - e^x - e^{x + y}} \left[ \text{Using eqn } . \left( 1 \right) \right]\]
\[ \Rightarrow \frac{dy}{dx} = \frac{e^y}{- e^x}\]
\[ \Rightarrow \frac{dy}{dx} = - e^{y - x} \]
\[ \Rightarrow \frac{dy}{dx} + e^{y - x} = 0\]
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