Question #320055

Find y' if x^y^2=y^x^2


Expert's answer

xy2=yx2y2ln⁡x=x2ln⁡yd(y2ln⁡x)=d(x2ln⁡y)2yln⁡xdy+y2xdx=2xln⁡ydx+x2ydy(2yln⁡x−x2y)dy=(2xln⁡y−y2x)dxdydx=2xln⁡y−y2x2yln⁡x−x2yx^{y^2}=y^{x^2}\\y^2\ln x=x^2\ln y\\d\left( y^2\ln x \right) =d\left( x^2\ln y \right) \\2y\ln xdy+\frac{y^2}{x}dx=2x\ln ydx+\frac{x^2}{y}dy\\\left( 2y\ln x-\frac{x^2}{y} \right) dy=\left( 2x\ln y-\frac{y^2}{x} \right) dx\\\frac{dy}{dx}=\frac{2x\ln y-\frac{y^2}{x}}{2y\ln x-\frac{x^2}{y}}


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