If y = operatorname{Sin}^{-1}left(frac{2x}{1+ | vedclass

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(B) Given $y = \operatorname{Sin}^{-1}\left(\frac{2x}{1+x^2}ight)$.For $|x| > 1$,we use the substitution $x = 	an 	heta$. Since $x=2 > 1$,we use t

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$(-2)^3$

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(B) Given $y = \operatorname{Sin}^{-1}\left(\frac{2x}{1+x^2}ight)$.For $|x| > 1$,we use the substitution $x = 	an 	heta$. Since $x=2 > 1$,we use the identity $\operatorname{Sin}^{-1}\left(\frac{2x}{1+x^2}ight) = \pi - 2	an^{-1}(x)$.Thus,$y = \pi - 2	an^{-1}(x)$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = -\frac{2}{1+x^2}$.Differentiating again,$\frac{d^2y}{dx^2} = -2 \cdot (-1)(1+x^2)^{-2} \cdot (2x) = \frac{4x}{(1+x^2)^2}$.At $x=2$,$k = \left(\frac{d^2y}{dx^2}ight)_{x=2} = \frac{4(2)}{(1+2^2)^2} = \frac{8}{25}$.Therefore,$25k = 25 \cdot \frac{8}{25} = 8$.Comparing with the options,$8 = (-2)^3$.

If $y = \operatorname{Sin}^{-1}\left(\frac{2x}{1+x^2}\right)$ and $\left(\frac{d^2y}{dx^2}\right)_{x=2} = k$,then $25k =$

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