If $y^{\frac{1}{m}}+y^{\frac{-1}{m}}=2 x, x^1 1$, then $\left(x^2-1\right)\left(\frac{d y}{d x}\right)^2$ is equal to

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$y^{\frac{1}{m}}+y^{-\frac{1}{m}}=2 x \Rightarrow y^{\frac{2}{m}}+y^{-\frac{2}{m}}+2=4 x^2 \ldots .$. (1) $\begin{aligned} & \frac{1}{m}\left(y^{\frac{1}{m}-1}-y^{-\frac{1}{m}-1}ight) \frac{d x}{d y}=2 \ & \Rightarrow\left(y^{\frac{1}{m}}-y^{-\frac{1}{m}}ight)=\frac{2 m y}{\frac{d y}{d x}} \ & \Rightarrow y^{\frac{2}{m}}+y^{-\frac{2}{m}}-2=\frac{4 m^2 y^2}{\left(\frac{d y}{d x}ight)^2} \ldots \ldots .(2) 	ext { [squaring both sides] }\end{aligned}$ from (1)-(2) $\begin{aligned} & 4=4 x^2-\frac{4 m^2 y^2}{\left(\frac{d y}{d x}ight)^2} \Rightarrow 4\left(\frac{d y}{d x}ight)^2=4 x^2\left(\frac{d y}{d x}ight)^2-4 m^2 y^2 \ & \Rightarrow\left(x^2-1ight)\left(\frac{d y}{d x}ight)^2=m^2 y^2\end{aligned}$

If $y^{\frac{1}{m}}+y^{\frac{-1}{m}}=2 x, x^1 1$, then $\left(x^2-1\right)\left(\frac{d y}{d x}\right)^2$ is…

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