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

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(C) Let $x = 	an 	heta$,where $	heta = 	an^{-1} x$.The expression becomes $y = \sin^{-1} \left( \frac{2 	an 	heta}{1 + 	an^2 	heta} ight) +

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$\frac{4}{1 + x^2}$

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(C) Let $x = 	an 	heta$,where $	heta = 	an^{-1} x$.The expression becomes $y = \sin^{-1} \left( \frac{2 	an 	heta}{1 + 	an^2 	heta} ight) + \sec^{-1} \left( \frac{1 + 	an^2 	heta}{1 - 	an^2 	heta} ight)$.Using trigonometric identities,$\sin 2	heta = \frac{2 	an 	heta}{1 + 	an^2 	heta}$ and $\cos 2	heta = \frac{1 - 	an^2 	heta}{1 + 	an^2 	heta}$.Thus,$y = \sin^{-1} (\sin 2	heta) + \sec^{-1} \left( \frac{1}{\cos 2	heta} ight) = \sin^{-1} (\sin 2	heta) + \sec^{-1} (\sec 2	heta)$.Assuming the principal value branch,$y = 2	heta + 2	heta = 4	heta$.Substituting back $	heta = 	an^{-1} x$,we get $y = 4 	an^{-1} x$.Differentiating with respect to $x$,$\frac{dy}{dx} = 4 \cdot \frac{d}{dx} (	an^{-1} x) = \frac{4}{1 + x^2}$.

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

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