If y = log_2(log_2(x)),then frac{dy}{dx} is e | vedclass

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(A) Given $y = \log_2(\log_2(x))$.Using the change of base formula $\log_a b = \frac{\log_e b}{\log_e a}$,we can write:$y = \frac{\log_e(\log_2 x)}{\l

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$\frac{\log_2 e}{x \log_e x}$

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(A) Given $y = \log_2(\log_2(x))$.Using the change of base formula $\log_a b = \frac{\log_e b}{\log_e a}$,we can write:$y = \frac{\log_e(\log_2 x)}{\log_e 2} = \frac{\log_e(\frac{\log_e x}{\log_e 2})}{\log_e 2}$$y = \frac{\log_e(\log_e x) - \log_e(\log_e 2)}{\log_e 2}$Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{1}{\log_e 2} \cdot \frac{d}{dx} [\log_e(\log_e x) - \log_e(\log_e 2)]$$\frac{dy}{dx} = \frac{1}{\log_e 2} \cdot \frac{1}{\log_e x} \cdot \frac{d}{dx}(\log_e x)$$\frac{dy}{dx} = \frac{1}{\log_e 2} \cdot \frac{1}{\log_e x} \cdot \frac{1}{x}$Since $\frac{1}{\log_e 2} = \log_2 e$,we get:$\frac{dy}{dx} = \frac{\log_2 e}{x \log_e x}$.

If $y = \log_2(\log_2(x))$,then $\frac{dy}{dx}$ is equal to

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