If f(x) = sin^{-1}left[frac{2^{x+1}}{1+4^x}ri | vedclass

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(A) Given $f(x) = \sin^{-1}\left[\frac{2 \cdot 2^x}{1 + (2^x)^2}ight]$.Let $2^x = 	an 	heta$,then $	heta = 	an^{-1}(2^x)$.Substituting this into

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$\log 2$

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(A) Given $f(x) = \sin^{-1}\left[\frac{2 \cdot 2^x}{1 + (2^x)^2}ight]$.Let $2^x = 	an 	heta$,then $	heta = 	an^{-1}(2^x)$.Substituting this into the function:$f(x) = \sin^{-1}\left[\frac{2 	an 	heta}{1 + 	an^2 	heta}ight]$Using the identity $\sin 2	heta = \frac{2 	an 	heta}{1 + 	an^2 	heta}$,we get:$f(x) = \sin^{-1}(\sin 2	heta) = 2	heta = 2 	an^{-1}(2^x)$.Now,differentiating with respect to $x$:$f'(x) = 2 \cdot \frac{1}{1 + (2^x)^2} \cdot \frac{d}{dx}(2^x)$$f'(x) = \frac{2}{1 + 4^x} \cdot 2^x \log_e 2$.Evaluating at $x = 0$:$f'(0) = \frac{2}{1 + 4^0} \cdot 2^0 \log_e 2 = \frac{2}{1 + 1} \cdot 1 \cdot \log_e 2 = \frac{2}{2} \log_e 2 = \log_e 2$.

If $f(x) = \sin^{-1}\left[\frac{2^{x+1}}{1+4^x}\right]$,then $f'(0) = $

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