Find dy dx of the function x^y + y^x = 1

Question

Find $\frac{dy}{dx}$ of the function $x^y + y^x = 1$

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Answer

Given: $x^y + y^x = 1$
Let $y = x^y + y^x = 1$
Let $u = x^y$ and $v = y^x$
Then, $u + v = 1$
$\Rightarrow \frac{d u}{d x}+\frac{d v}{d x}=0$
For, $u = x^y $
Taking log on both sides, we get
$\log u = \log x^y$
$\Rightarrow \log u=y \cdot \log (x)$
Now, differentiating both sides with respect to $x$
$\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{u})=\frac{\mathrm{d}}{\mathrm{dx}}[\mathrm{y} \cdot \log (\mathrm{x})]$
$\Rightarrow \frac{1}{\mathrm{u}} \frac{\mathrm{du}}{\mathrm{dx}}=\left\{\mathrm{y} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{x})+\log \mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\mathrm{y})\right\}$
$\Rightarrow \frac{d u}{d x}=u\left[y \cdot \frac{1}{x}+\log x \cdot\left(\frac{d y}{d x}\right)\right]$
$\Rightarrow \frac{\mathrm{du}}{\mathrm{dx}}=\mathrm{x}^{\mathrm{y}}\left[\frac{\mathrm{y}}{\mathrm{x}}+\log \mathrm{x} \cdot\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\right]$
For $v = y^x$
Taking log on both sides, we get
log $v = \log y^x$
$\Rightarrow \log v=x \cdot \log (y)$
Now, differentiate both sides with respect to $x$
$\frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{v})=\frac{\mathrm{d}}{\mathrm{dx}}[\mathrm{x} \cdot \log (\mathrm{y})]$
$\Rightarrow \frac{1}{\mathrm{v}} \frac{\mathrm{dv}}{\mathrm{dx}}=\left\{\mathrm{x} \cdot \frac{\mathrm{d}}{\mathrm{dx}}(\log \mathrm{y})+\log \mathrm{y} \cdot \frac{\mathrm{d}}{\mathrm{dx}} \mathrm{x}\right\}$
$\Rightarrow \frac{d v}{d x}=v\left[x \cdot \frac{1}{y} \cdot \frac{d y}{d x}+\log y \cdot\left(\frac{d x}{d x}\right)\right]$
$\Rightarrow \frac{d v}{d x}=y^{x}\left[\frac{x}{y} \cdot \frac{d y}{d x}+\log y\right]$
because, $\frac{d u}{d x}+\frac{d v}{d x}=0$
So, $\mathrm{x}^{\mathrm{y}}\left[\frac{\mathrm{y}}{\mathrm{x}}+\log \mathrm{x} \cdot\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)\right]+\mathrm{y}^{\mathrm{x}}\left[\frac{\mathrm{x}}{\mathrm{y}} \cdot \frac{\mathrm{dy}}{\mathrm{dx}}+\log \mathrm{y}\right]=0$
$\Rightarrow\left(\mathrm{x}^{\mathrm{y}} \log \mathrm{x}+\mathrm{xy}^{\mathrm{x}-1}\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}}+\left(\mathrm{yx}^{\mathrm{y}-1}+\mathrm{y}^{\mathrm{x}} \log \mathrm{y}\right)=0$
$\Rightarrow\left(\mathrm{x}^{\mathrm{y}} \log \mathrm{x}+\mathrm{xy}^{\mathrm{x}-1}\right) \cdot \frac{\mathrm{dy}}{\mathrm{dx}}=-\left(\mathrm{yx}^{\mathrm{y}-1}+\mathrm{y}^{\mathrm{x}} \log \mathrm{y}\right)$
$\frac{d y}{d x}=-\frac{\left(y x^{y-1}+y^{x} \log y\right)}{\left(x^{y} \log x+x y^{x-1}\right)}$

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