If y^m + y^{-m} = 2x, then (x^2 - 1) left( fr | vedclass

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(A) Given $y^m + y^{-m} = 2x$. Differentiating both sides with respect to $x$: $m y^{m-1} \frac{dy}{dx} - m y^{-m-1} \frac{dy}{dx} = 2$. $\frac{dy}{dx

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$m^2 y^2$

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(A) Given $y^m + y^{-m} = 2x$. Differentiating both sides with respect to $x$: $m y^{m-1} \frac{dy}{dx} - m y^{-m-1} \frac{dy}{dx} = 2$. $\frac{dy}{dx} (m y^{m-1} - m y^{-m-1}) = 2$. $\frac{dy}{dx} = \frac{2}{m(y^{m-1} - y^{-m-1})} = \frac{2y}{m(y^m - y^{-m})}$. From the given equation, $(y^m - y^{-m})^2 = (y^m + y^{-m})^2 - 4 = (2x)^2 - 4 = 4(x^2 - 1)$. So, $y^m - y^{-m} = 2\sqrt{x^2 - 1}$. Substituting this into the derivative: $\frac{dy}{dx} = \frac{2y}{m(2\sqrt{x^2 - 1})} = \frac{y}{m\sqrt{x^2 - 1}}$. Squaring both sides: $\left( \frac{dy}{dx} \right)^2 = \frac{y^2}{m^2(x^2 - 1)}$. Therefore, $(x^2 - 1) \left( \frac{dy}{dx} \right)^2 = \frac{y^2}{m^2}$.

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

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