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