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Differntiate x^y=y^x

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Differntiate x^y=y^x
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Watch It 4 years, 3 months ago

Log y^x=logx^y Xlogy=ylogx Differntiating with respect to x X×(1/y)×(dy/DX) + logy.1=y.(1/x) + logx×(dy/dx) Dy/DX[x/y - log x]= y/x - log y dy\dx[(x-ylogx)/y]=(y-xlogy)/x Agae khud krlena...
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