Solve this

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




$e^{\log x^{y}}+e^{\log y^{x}}=e^{\log (x+y)^{x+y}}$

$e^{y \log x}+e^{x \log y}=e^{(x+y) \log (x+y)}$

Differentiating it with respect to $\mathrm{x}$ using chain rule, product rule,

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

$e^{y \log x}\left[y \frac{d}{d x}(\log x)+\log x \frac{d y}{d x}\right]+e^{x \log y}\left[x \frac{d}{d x}(\operatorname{logy})+\log y \frac{d x}{d x}\right]$

$=e^{(x+y) \log (x+y)} \frac{d}{d x}[(x+y) \log (x+y)]$

$\begin{aligned} e^{\log x} y\left[y\left(\frac{1}{x}\right)\right.&\left.+\log x \frac{d y}{d x}\right]+e^{\log y} x\left[\frac{x}{y} \frac{d y}{d x}+\log y(1)\right] \\ &=e^{\log (x+y)^{x+y}}\left[(x+y) \frac{d}{d x} \log (x+y)+\log (x+y) \frac{d}{d x}(x+y)\right] \end{aligned}$

$\begin{aligned} x^{y}\left[\frac{y}{x}+\log x\right.&\left.\frac{d y}{d x}\right]+y^{x}\left[\frac{x d y}{y d x}+\log y\right] \\ &=(x+y)^{x+y}\left[(x+y) \cdot \frac{1}{(x+y)} \cdot \frac{d}{d x} \cdot(x+y)+\log (x+y)\left(1+\frac{d y}{d x}\right)\right] \end{aligned}$

$\begin{aligned} x^{y} \frac{y}{x}+x^{y} \cdot \log x & \frac{d y}{d x}+y^{x} \cdot \frac{x}{y} \cdot \frac{d y}{d x} \\ &+y^{x} \log y=(x+y)^{x+y}\left[1 \times\left(1+\frac{d y}{d x}\right)+\log (x+y)\left(1+\frac{d y}{d x}\right)\right] \end{aligned}$

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

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

$\frac{d y}{d x}\left[x^{y} \log x+x y^{x-1}-(x+y)^{x+y}(1+\log (x+y))\right]$

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

$\frac{d y}{d x}\left[x^{y} \log x+x y^{x-1}-(x+y)^{x+y}(1+\log (x+y))\right]$

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

$\frac{d y}{d x}=\left[\frac{(x+y)^{x+y}(1+\log (x+y))-x^{y-1} \times y-y^{x} \log y}{x^{y} \log x+x y^{x-1}-(x+y)^{x+y}(1+\log (x+y))}\right]$


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