If y^{frac{1}{m}}+y^{-frac{1}{m}}=2x,where x | vedclass

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(B) Given $y^{\frac{1}{m}}+y^{-\frac{1}{m}}=2x$. Squaring both sides,we get $(y^{\frac{1}{m}}+y^{-\frac{1}{m}})^2 = (2x)^2$,which implies $y^{\frac{2}

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$m^2 y^2$

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(B) Given $y^{\frac{1}{m}}+y^{-\frac{1}{m}}=2x$. Squaring both sides,we get $(y^{\frac{1}{m}}+y^{-\frac{1}{m}})^2 = (2x)^2$,which implies $y^{\frac{2}{m}}+y^{-\frac{2}{m}}+2 = 4x^2$,so $y^{\frac{2}{m}}+y^{-\frac{2}{m}} = 4x^2-2$ ... $(1)$ Differentiating both sides with respect to $y$,we get $\frac{1}{m} y^{\frac{1}{m}-1} - \frac{1}{m} y^{-\frac{1}{m}-1} = 2 \frac{dx}{dy}$. Multiplying by $my$,we get $y^{\frac{1}{m}} - y^{-\frac{1}{m}} = 2my \frac{dx}{dy} = \frac{2my}{dy/dx}$. Squaring both sides,we get $(y^{\frac{1}{m}}-y^{-\frac{1}{m}})^2 = \frac{4m^2y^2}{(dy/dx)^2}$. This simplifies to $y^{\frac{2}{m}}+y^{-\frac{2}{m}}-2 = \frac{4m^2y^2}{(dy/dx)^2}$ ... $(2)$ Subtracting $(2)$ from $(1)$,we get $(y^{\frac{2}{m}}+y^{-\frac{2}{m}}+2) - (y^{\frac{2}{m}}+y^{-\frac{2}{m}}-2) = 4x^2 - \frac{4m^2y^2}{(dy/dx)^2}$. $4 = 4x^2 - \frac{4m^2y^2}{(dy/dx)^2}$. Dividing by $4$,we get $1 = x^2 - \frac{m^2y^2}{(dy/dx)^2}$. Rearranging the terms,we get $\frac{m^2y^2}{(dy/dx)^2} = x^2-1$. Therefore,$(x^2-1)(\frac{dy}{dx})^2 = m^2y^2$.

If $y^{\frac{1}{m}}+y^{-\frac{1}{m}}=2x$,where $x \neq 1$,then $(x^2-1)(\frac{dy}{dx})^2$ is equal to

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