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

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(C) 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)}{\lo

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$\frac{1}{(x \log_e x) \log_e 2}$

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(C) 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 $x$:$\frac{dy}{dx} = \frac{1}{\log_e 2} \cdot \frac{d}{dx} [\log_e(\log_e x) - \log_e(\log_e 2)]$.Since $\log_e(\log_e 2)$ is a constant,its derivative is $0$.$\frac{dy}{dx} = \frac{1}{\log_e 2} \cdot \frac{1}{\log_e x} \cdot \frac{d}{dx}(\log_e x)$.$\frac{dy}{dx} = \frac{1}{\log_e 2} \cdot \frac{1}{\log_e x} \cdot \frac{1}{x}$.$\frac{dy}{dx} = \frac{1}{x \log_e x \log_e 2}$.

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

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