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y-x-y-dy-dx-




Question Number 62759 by naka3546 last updated on 25/Jun/19
y  =  x^y   (dy/dx)  =  ?
$${y}\:\:=\:\:{x}^{{y}} \\ $$$$\frac{{dy}}{{dx}}\:\:=\:\:? \\ $$
Answered by Kunal12588 last updated on 25/Jun/19
y=x^y   ⇒ln y = y ln x  ⇒(1/y)×y′=(y/x)+y′ ln x  ⇒y′((1/y)−ln x)=(y/x)  ⇒y′=(y^2 /(x(1−y ln x)))=(x^(2y) /(x(1−x^y  ln x)))=(x^(2y−1) /(1−x^y  ln x))
$${y}={x}^{{y}} \\ $$$$\Rightarrow{ln}\:{y}\:=\:{y}\:{ln}\:{x} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{y}}×{y}'=\frac{{y}}{{x}}+{y}'\:{ln}\:{x} \\ $$$$\Rightarrow{y}'\left(\frac{\mathrm{1}}{{y}}−{ln}\:{x}\right)=\frac{{y}}{{x}} \\ $$$$\Rightarrow{y}'=\frac{{y}^{\mathrm{2}} }{{x}\left(\mathrm{1}−{y}\:{ln}\:{x}\right)}=\frac{{x}^{\mathrm{2}{y}} }{{x}\left(\mathrm{1}−{x}^{{y}} \:{ln}\:{x}\right)}=\frac{{x}^{\mathrm{2}{y}−\mathrm{1}} }{\mathrm{1}−{x}^{{y}} \:{ln}\:{x}} \\ $$

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