If y = sin^{-1}left(frac{log x^2}{1+(log x)^2 | vedclass

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(A) Given $y = \sin^{-1}\left(\frac{\log x^2}{1+(\log x)^2}ight) = \sin^{-1}\left(\frac{2 \log x}{1+(\log x)^2}ight)$.Let $\log x = 	an 	heta$,t

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$2$

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(A) Given $y = \sin^{-1}\left(\frac{\log x^2}{1+(\log x)^2}ight) = \sin^{-1}\left(\frac{2 \log x}{1+(\log x)^2}ight)$.Let $\log x = 	an 	heta$,then $y = \sin^{-1}\left(\frac{2 	an 	heta}{1+	an^2 	heta}ight) = \sin^{-1}(\sin 2	heta) = 2	heta$.Substituting back,$y = 2 	an^{-1}(\log x)$.Differentiating with respect to $x$:$\frac{dy}{dx} = 2 	imes \frac{1}{1+(\log x)^2} 	imes \frac{1}{x}$.At $x = 1$,$\log 1 = 0$,so $\left(\frac{dy}{dx}ight)_{x=1} = 2 	imes \frac{1}{1+0^2} 	imes \frac{1}{1} = 2 	imes 1 	imes 1 = 2$.

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

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