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