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Question
If \[y = \frac{x}{x + 2}\] , prove tha \[x\frac{dy}{dx} = \left( 1 - y \right) y\] ?
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Solution
\[\text{We have, y } = \frac{x}{x + 2}\]
Differentiating with respect to x,
\[\frac{d y}{d x} = \frac{d}{dx}\left( \frac{x}{x + 2} \right)\]
\[ \Rightarrow \frac{d y}{d x} = \frac{\left( x + 2 \right)\frac{d}{dx}\left( x \right) - x\frac{d}{dx}\left( x + 2 \right)}{\left( x + 2 \right)^2} \]
\[ \Rightarrow \frac{d y}{d x} = \frac{x + 2 - x}{\left( x + 2 \right)^2}\]
\[ \Rightarrow \frac{d y}{d x} = \frac{x + 2}{\left( x + 2 \right)^2} - \frac{x}{\left( x + 2 \right)^2}\]
\[ \Rightarrow \frac{d y}{d x} = \frac{1}{x + 2} - \frac{x y^2}{x^2} \left[ \because x + 2 = \frac{x}{y} \right]\]
\[ \Rightarrow \frac{d y}{d x} = \frac{y}{x} - \frac{y^2}{x}\]
\[ \Rightarrow \frac{d y}{d x} = \frac{1}{x}y\left( 1 - y \right)\]
\[ \Rightarrow x\frac{d y}{d x} = \left( 1 - y \right)y\]
\[ \text{Hence proved}\]
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