MathematicsMedium63×since 2002Q3693 Let y=loge(1−x21+x2),−1<x<1. Then at x=12, the value of 225(y′−y′′) is equal to \text { Let } y=\log _e\left(\frac{1-x^2}{1+x^2}\right),-1 < x<1 \text {. Then at } x=\frac{1}{2} \text {, the value of } 225\left(y^{\prime}-y^{\prime \prime}\right) \text { is equal to } Let y=loge(1+x21−x2),−1<x<1. Then at x=21, the value of 225(y′−y′′) is equal to A732B736C742D746Check answerSkip