Question

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

Answer 1

\[\text{ We have }, x^y \times y^x = 1\] 

Taking log on both sides,

\[\log\left( x^y \times y^x \right) = \log\left( 1 \right)\]

\[ \Rightarrow y\log x + x \log y = \log1 \]

Differentiating with respect to x ,

\[\frac{d}{dx}\left( y \log x \right) + \frac{d}{dx}\left( x \log x \right) = \frac{d}{dx}\left( \log1 \right)\]

\[ \Rightarrow \left[ y\frac{d}{dx}\left( \log x \right) + \log x\frac{dy}{dx} \right] + \left[ x\frac{d}{dx}\left( \log y \right) + \log y\frac{d}{dx}\left( x \right) \right] = 0\]

\[ \Rightarrow \left[ y\left( \frac{1}{x} \right) + \log x\frac{dy}{dx} \right] + \left[ x\left( \frac{1}{y}\frac{dy}{dx} \right) + \log y\left( 1 \right) \right] = 0\]

\[ \Rightarrow \frac{y}{x} + \log x\frac{dy}{dx} + \frac{x}{y}\frac{dy}{dx} + \log y = 0\]

\[ \Rightarrow \frac{dy}{dx}\left( \log x + \frac{x}{y} \right) = - \left[ \log y + \frac{y}{x} \right]\]

\[ \Rightarrow \frac{dy}{dx}\left[ \frac{y \log x + x}{y} \right] = - \left[ \frac{x \log y + y}{x} \right]\]

\[ \Rightarrow \frac{dy}{dx} = - \frac{y}{x}\left[ \frac{x \log y + y}{y \log x + x} \right]\]