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Question
If \[xy \log \left( x + y \right) = 1\] ,Prove that \[\frac{dy}{dx} = - \frac{y \left( x^2 y + x + y \right)}{x \left( x y^2 + x + y \right)}\] ?
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Solution
\[\text{ We have, xy } \log\left( x + y \right) = 1\]
Differentiating it with respect to x,
\[\Rightarrow \frac{d}{dx}\left[ xy \log\left( x + y \right) \right] = \frac{d}{dx}\left( 1 \right)\]
\[ \Rightarrow xy\frac{d}{dx}\log\left( x + y \right) + x \log\left( x + y \right)\frac{d y}{d x} + y \log\left( x + y \right)\frac{d}{dx}\left( x \right) = 0 \left[ \text{ using chain rule and product rule } \right]\]
\[ \Rightarrow xy\left( \frac{1}{x + y} \right)\frac{d}{dx}\left( x + y \right) + x \log\left( x + y \right)\frac{d y}{d x} + y \log\left( x + y \right)\left( 1 \right) = 0\]
\[ \Rightarrow \left( \frac{xy}{x + y} \right) \left( 1 + \frac{d y}{d x} \right) + x \log\left( x + y \right)\frac{d y}{d x} + y \log\left( x + y \right) = 0\]
\[ \Rightarrow \left( \frac{xy}{x + y} \right)\frac{d y}{d x} + \left( \frac{xy}{x + y} \right) + x\left( \frac{1}{xy} \right)\frac{d y}{d x} + y\left( \frac{1}{xy} \right) = 0 \left[ \because xy \log\left( x + y \right) = 1 \right]\]
\[ \Rightarrow \frac{d y}{d x}\left[ \frac{xy}{x + y} + \frac{1}{y} \right] = - \left[ \frac{1}{x} + \frac{xy}{x + y} \right]\]
\[ \Rightarrow \frac{d y}{d x}\left[ \frac{x y^2 + x + y}{\left( x + y \right)y} \right] = - \left[ \frac{x + y + x^2 y}{x\left( x + y \right)} \right]\]
\[ \Rightarrow \frac{d y}{d x} = - \left[ \frac{x + y + x^2 y}{x\left( x + y \right)} \right]\left[ \frac{y\left( x + y \right)}{x y^2 + x + y} \right]\]
\[ \Rightarrow \frac{d y}{d x} = - \frac{y}{x}\left( \frac{x + y + x^2 y}{x + y + x y^2} \right)\]
Hence proved
