If y = log_3(log_3 x),then frac{dy}{dx} at x | vedclass

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(C) Given $y = \log_3(\log_3 x)$.Using the change of base formula,$\log_a b = \frac{\log_e b}{\log_e a}$,we can write $y = \frac{\log_e(\log_3 x)}{\lo

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$\frac{1}{3}(\log_e 3)^{-2}$

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(C) Given $y = \log_3(\log_3 x)$.Using the change of base formula,$\log_a b = \frac{\log_e b}{\log_e a}$,we can write $y = \frac{\log_e(\log_3 x)}{\log_e 3} = \frac{\log_e(\frac{\log_e x}{\log_e 3})}{\log_e 3}$.$y = \frac{\log_e(\log_e x) - \log_e(\log_e 3)}{\log_e 3}$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{1}{\log_e 3} \cdot \frac{d}{dx} [\log_e(\log_e x) - \log_e(\log_e 3)]$.$\frac{dy}{dx} = \frac{1}{\log_e 3} \cdot \frac{1}{\log_e x} \cdot \frac{1}{x}$.At $x = 3$:$\frac{dy}{dx} = \frac{1}{\log_e 3} \cdot \frac{1}{\log_e 3} \cdot \frac{1}{3} = \frac{1}{3(\log_e 3)^2} = \frac{1}{3}(\log_e 3)^{-2}$.Since the provided options were incorrect,the correct value is $\frac{1}{3}(\log_e 3)^{-2}$.

If $y = \log_3(\log_3 x)$,then $\frac{dy}{dx}$ at $x = 3$ is $\ldots \ldots$

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