If y = sin^{-1}left(frac{2^{x+1}}{1+4^x}right | vedclass

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

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$2$

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(A) Given $y = \sin^{-1}\left(\frac{2 \cdot 2^x}{1 + (2^x)^2}ight)$.Let $2^x = 	an 	heta$,then $	heta = 	an^{-1}(2^x)$.$y = \sin^{-1}\left(\frac{2 	an 	heta}{1 + 	an^2 	heta}ight) = \sin^{-1}(\sin 2	heta) = 2	heta = 2 	an^{-1}(2^x)$.Differentiating with respect to $x$:$\frac{dy}{dx} = 2 \cdot \frac{1}{1 + (2^x)^2} \cdot \frac{d}{dx}(2^x) = \frac{2}{1 + 4^x} \cdot 2^x \log 2 = \frac{2^{x+1} \log 2}{1 + 4^x}$.Comparing this with $\frac{dy}{dx} = \frac{2^{x+1} \log 2}{f(x)}$,we get $f(x) = 1 + 4^x$.Therefore,$f(0) = 1 + 4^0 = 1 + 1 = 2$.

If $y = \sin^{-1}\left(\frac{2^{x+1}}{1+4^x}\right)$ and $\frac{dy}{dx} = \frac{2^{x+1} \log 2}{f(x)}$,then $f(0) = $ . . . . . .

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