If y = cos^{-1} left( frac{1-4^x}{1+4^x} righ | vedclass

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(A) Let $u = 2^x$. Then $4^x = (2^x)^2 = u^2$.The expression becomes $y = \cos^{-1} \left( \frac{1-u^2}{1+u^2} ight)$.Using the substitution $u =

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$\frac{4 \log 2}{5}$

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(A) Let $u = 2^x$. Then $4^x = (2^x)^2 = u^2$.The expression becomes $y = \cos^{-1} \left( \frac{1-u^2}{1+u^2} ight)$.Using the substitution $u = 	an 	heta$, we have $y = \cos^{-1} \left( \frac{1-	an^2 	heta}{1+	an^2 	heta} ight) = \cos^{-1} (\cos 2	heta) = 2	heta = 2 	an^{-1} u = 2 	an^{-1} (2^x)$.Differentiating with respect to $x$:$\frac{dy}{dx} = 2 \cdot \frac{1}{1+(2^x)^2} \cdot \frac{d}{dx}(2^x) = \frac{2}{1+4^x} \cdot 2^x \log 2$.At $x = 1$:$\frac{dy}{dx} = \frac{2 \cdot 2^1 \log 2}{1+4^1} = \frac{4 \log 2}{5}$.

If $y = \cos^{-1} \left( \frac{1-4^x}{1+4^x} \right)$, then find $\frac{dy}{dx}$ at $x = 1$.

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