Question

If \[xy \log \left( x + y \right) = 1\] , prove that  \[\frac{dy}{dx} = - \frac{y \left( x^2 y + x + y \right)}{x \left( x y^2 + x + y \right)}\] ?

Answer 1

\[\text{ We have}, xy \log\left( x + y \right) = 1 . . . \left( i \right)\]

\[\frac{dy}{dx}\left\{ xy \log\left( x + y \right) \right\} = \frac{d}{dx}\left( 1 \right)\]
\[ \Rightarrow xy\frac{d}{dx}\log\left( x + y \right) + x \log\left( x + y \right)\frac{dy}{dx} + y \log\left( x + y \right)\frac{d}{dx}\left( x \right) = 0\]
\[ \Rightarrow \frac{xy}{\left( x + y \right)}\left( 1 + \frac{dy}{dx} \right) + x\log\left( x + y \right)\frac{dy}{dx} + y \log\left( x + y \right)\left( 1 \right) = 0\]
\[ \Rightarrow \frac{xy}{\left( x + y \right)}\left( 1 + \frac{dy}{dx} \right) + x\log\left( x + y \right)\frac{dy}{dx} + y \log\left( x + y \right) = 0\]
\[ \Rightarrow \frac{xy}{\left( x + y \right)}\frac{dy}{dx} + \frac{xy}{\left( x + y \right)} + x\left( \frac{1}{xy} \right)\frac{dy}{dx} + y\left( \frac{1}{xy} \right) = 0 \left[ \text{ Using equation} \left( i \right) \right]\]
\[ \Rightarrow \frac{dy}{dx}\left[ \frac{xy}{x + y} + \frac{1}{y} \right] = - \left[ \frac{1}{x} + \frac{xy}{x + y} \right]\]
\[ \Rightarrow \frac{dy}{dx}\left[ \frac{x y^2 + x + y}{\left( x + y \right)y} \right] = - \left[ \frac{x + y + x^2 y}{x\left( x + y \right)} \right]\]
\[ \Rightarrow \frac{dy}{dx} = - \frac{y}{x}\left( \frac{x + y + x^2 y}{x + y + x y^2} \right)\]