Let $y=\log _e\left(\frac{1-x^2}{1+x^2}\right),-1

Question

Let $y=\log _e\left(\frac{1-x^2}{1+x^2}\right),-1<x<1$. Then at $x=\frac{1}{2}$, the value of $225\left(y^{\prime}-y^{\prime \prime}\right)$ is equal to [JEE Main 2024, 29 Jan (Shift 2)]

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Let \(y=\log _e\left(\frac{1-x^2}{1+x^2}\right),-1<x<1\). Then at \(x=\frac{1}{2}\), the value of \(225\left(y^{\p…

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Let \(y=\log _e\left(\frac{1-x^2}{1+x^2}\right),-1&lt;x&lt;1\). Then at \(x=\frac{1}{2}\), the value of \(225\left(y^{\p…

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Let \(y=\log _e\left(\frac{1-x^2}{1+x^2}\right),-1<x<1\). Then at \(x=\frac{1}{2}\), the value of \(225\left(y^{\p…