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

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Question

(A) Given $y = \sin^{-1} \left( \frac{2x}{1 + x^2} ight)$.Substitute $x = 	an 	heta$,which implies $	heta = 	an^{-1} x$.Since $0 < x < 1$,we hav

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

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(A) Given $y = \sin^{-1} \left( \frac{2x}{1 + x^2} ight)$.Substitute $x = 	an 	heta$,which implies $	heta = 	an^{-1} x$.Since $0 Using the trigonometric identity $\sin 2	heta = \frac{2 	an 	heta}{1 + 	an^2 	heta}$,we get:$y = \sin^{-1} (\sin 2	heta) = 2	heta$.Substituting back $	heta = 	an^{-1} x$,we have $y = 2 	an^{-1} x$.Differentiating with respect to $x$:$\frac{dy}{dx} = 2 \cdot \frac{1}{1 + x^2} = \frac{2}{1 + x^2}$.

If $y = \sin^{-1} \left( \frac{2x}{1 + x^2} \right)$,where $0 < x < 1$ and $0 < y < \frac{\pi}{2}$,then $\frac{dy}{dx} = $

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