Solve this


If $x^{x}+y^{x}=1$, find $\frac{d y}{d x}=-\frac{y(y+x \log y)}{x(y \log x+x)}$


Let $x^{x}=u$ and $y^{x}=v$

Taking log on both sides we get,

$x \log x=\log u \ldots \ldots(1)$

$x \log y=\log v \ldots \ldots(2)$

Using $\log a^{b}=b \log a$

Differentiating both sides of equation (1) we get,

$\mathrm{x} \times \frac{1}{\mathrm{x}}+\log \mathrm{x}=\frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}$

$\frac{d u}{d x}=x^{x}(1+\log x)$       ....(3)

Differentiating both sides of equation (2) we get,

$x \times \frac{1}{y} \frac{d y}{d x}+\log y=\frac{1}{v} \frac{d v}{d x}$

$\frac{d v}{d x}=y^{x}\left(\log y+\frac{x}{y} \frac{d y}{d x}\right)$          ....(4)

We know that, from question,


Differentiating both sides we get,

$\frac{d u}{d x}+\frac{d v}{d x}=0$

Putinng the value of eq(4) and eq(5) in equation above we get,

$x^{x}(1+\log x)+y^{x}\left(\log y+\frac{x}{y} \frac{d y}{d x}\right)=0$

$y^{x} \frac{x}{y} \frac{d y}{d x}=\frac{-x^{x}(1+\log x)}{y^{x}(\log y)}$\

$\frac{d y}{d x}=\frac{-x^{x-1}(1+\log x)}{y^{2 x-1} \log y}$

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