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Question Number 170415 by Mastermind last updated on 23/May/22

$$\mathrm{Given}{\mathrm{x}}^{\mathrm{y}}={\mathrm{e}}^{\mathrm{x}-\mathrm{y}}.\mathrm{find}\frac{\mathrm{dy}}{\mathrm{dx}}?$$$$$$$$\mathrm{Mastermind}$$

Answered by aleks041103 last updated on 23/May/22

$$x^y=e^{x-y}$$$$e^y{x}^y=e^x$$$${\left(ex\right)}^y=e^x$$$$y={log}_{ex}\left(e^x\right)=\frac{ln\left(e^x\right)}{ln\left(ex\right)}=\frac{x}{1+ln\left(x\right)}=y$$$$\frac{dy}{dx}=\frac{(1+lnx).1-\frac{1}{x}x}{{(1+lnx)}^2}$$$$\frac{dy}{dx}=\frac{lnx}{{(1+lnx)}^2}$$

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