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Question Number 62759 by naka3546 last updated on 25/Jun/19

$$y=x^y$$$$\frac{dy}{dx}=?$$

Answered by Kunal12588 last updated on 25/Jun/19

$$y=x^y$$$$\Rightarrow lny=ylnx$$$$\Rightarrow \frac{1}{y}\times y^{\prime }=\frac{y}{x}+y^{\prime }lnx$$$$\Rightarrow y^{\prime }(\frac{1}{y}-lnx)=\frac{y}{x}$$$$\Rightarrow y^{\prime }=\frac{y^2}{x(1-ylnx)}=\frac{x^{2y}}{x(1-x^{y}lnx)}=\frac{x^{2y-1}}{1-x^{y}lnx}$$

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