Question

# If ${x}^{y}+{y}^{x}={\left(x+y\right)}^{x+y},\mathrm{find}\frac{dy}{dx}$

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Solution

## $\mathrm{We}\mathrm{have},{x}^{y}+{y}^{x}={\left(x+y\right)}^{x+y}\phantom{\rule{0ex}{0ex}}⇒{e}^{\mathrm{log}{x}^{y}}+{e}^{\mathrm{log}{y}^{x}}={e}^{\mathrm{log}{\left(x+y\right)}^{\left(x+y\right)}}\phantom{\rule{0ex}{0ex}}⇒{e}^{y\mathrm{log}x}+{e}^{x\mathrm{log}y}={e}^{\left(x+y\right)\mathrm{log}\left(x+y\right)}$ Differentiating with respect to x using chain rule and product rule, $⇒\frac{d}{dx}\left({e}^{y\mathrm{log}x}\right)+\frac{d}{dx}\left({e}^{x\mathrm{log}y}\right)=\frac{d}{dx}{e}^{\left(x+y\right)\mathrm{log}\left(x+y\right)}\phantom{\rule{0ex}{0ex}}⇒{e}^{y\mathrm{log}x}\left[y\frac{d}{dx}\left(\mathrm{log}x\right)+\mathrm{log}x\frac{dy}{dx}\right]+{e}^{x\mathrm{log}y}\left[x\frac{d}{dx}\mathrm{log}y+\mathrm{log}y\frac{d}{dx}\left(x\right)\right]={e}^{\left(x+y\right)\mathrm{log}\left(x+y\right)}\frac{d}{dx}\left[\left(x+y\right)\mathrm{log}\left(x+y\right)\right]\phantom{\rule{0ex}{0ex}}⇒{e}^{\mathrm{log}{x}^{y}}\left[y\left(\frac{1}{x}\right)+\mathrm{log}x\frac{dy}{dx}\right]+{e}^{\mathrm{log}x}\left[\frac{x}{y}\frac{dy}{dx}+\mathrm{log}y\left(1\right)\right]={e}^{\mathrm{log}{\left(x+y\right)}^{\left(x+y\right)}}\left[\left(x+y\right)\frac{d}{dx}\mathrm{log}\left(x+y\right)+\mathrm{log}\left(x+y\right)\frac{d}{dx}\left(x+y\right)\right]\phantom{\rule{0ex}{0ex}}⇒{x}^{y}\left[\frac{y}{x}+\mathrm{log}x\frac{dy}{dx}\right]+{y}^{x}\left[\frac{x}{y}\frac{dy}{dx}+\mathrm{log}y\right]={\left(x+y\right)}^{\left(x+y\right)}\left[\left(x+y\right)\frac{1}{\left(x+y\right)}\frac{d}{dx}\left(x+y\right)+\mathrm{log}\left(x+y\right)\left(1+\frac{dy}{dx}\right)\right]\phantom{\rule{0ex}{0ex}}⇒{x}^{y}×\frac{y}{x}+{x}^{y}\mathrm{log}x\frac{dy}{dx}+{y}^{x}×\frac{x}{y}\frac{dy}{dx}+{y}^{x}\mathrm{log}y={\left(x+y\right)}^{\left(x+y\right)}\left[1×\left(1+\frac{dy}{dx}\right)+\mathrm{log}\left(x+y\right)\left(1+\frac{dy}{dx}\right)\right]\phantom{\rule{0ex}{0ex}}⇒{x}^{y-1}×y+{x}^{y}\mathrm{log}x\frac{dy}{dx}+{y}^{x-1}×x\frac{dy}{dx}+{y}^{x}\mathrm{log}y={\left(x+y\right)}^{\left(x+y\right)}+{\left(x+y\right)}^{\left(x+y\right)}\frac{dy}{dx}+{\left(x+y\right)}^{\left(x+y\right)}\mathrm{log}\left(x+y\right)+{\left(x+y\right)}^{\left(x+y\right)}\mathrm{log}\left(x+y\right)\frac{dy}{dx}\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}\left[{x}^{y}\mathrm{log}x+x{y}^{x-1}-{\left(x+y\right)}^{\left(x+y\right)}\left\{1+\mathrm{log}\left(x+y\right)\right\}\right]={\left(x+y\right)}^{\left(x+y\right)}\left\{1+\mathrm{log}\left(x+y\right)\right\}-{x}^{y-1}×y-{y}^{x}\mathrm{log}y\phantom{\rule{0ex}{0ex}}⇒\frac{dy}{dx}=\left[\frac{{\left(x+y\right)}^{\left(x+y\right)}\left\{1+\mathrm{log}\left(x+y\right)\right\}-y{x}^{y-1}-{y}^{x}\mathrm{log}y}{{x}^{y}\mathrm{log}x+x{y}^{x-1}-{\left(x+y\right)}^{\left(x+y\right)}\left\{1+\mathrm{log}\left(x+y\right)\right\}}\right]$

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