Find the derivative: frac{d}{dx}(log_7(log_7 | vedclass

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(C) We need to find the derivative of $y = \log_7(\log_7 x)$.Using the change of base formula,$\log_a b = \frac{\log_e b}{\log_e a}$,we can write:$y =

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$\frac{\log_7 e}{x \log_e x}$

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(C) We need to find the derivative of $y = \log_7(\log_7 x)$.Using the change of base formula,$\log_a b = \frac{\log_e b}{\log_e a}$,we can write:$y = \frac{\log_e(\log_7 x)}{\log_e 7} = \frac{1}{\log_e 7} \cdot \log_e(\log_7 x)$.Now,differentiate with respect to $x$:$\frac{dy}{dx} = \frac{1}{\log_e 7} \cdot \frac{d}{dx}(\log_e(\log_7 x))$.Using the chain rule,$\frac{d}{dx}(\log_e u) = \frac{1}{u} \cdot \frac{du}{dx}$,where $u = \log_7 x = \frac{\log_e x}{\log_e 7}$:$\frac{dy}{dx} = \frac{1}{\log_e 7} \cdot \frac{1}{\log_7 x} \cdot \frac{d}{dx}(\log_7 x)$.Since $\frac{d}{dx}(\log_7 x) = \frac{d}{dx}(\frac{\log_e x}{\log_e 7}) = \frac{1}{x \log_e 7}$:$\frac{dy}{dx} = \frac{1}{\log_e 7} \cdot \frac{1}{\log_7 x} \cdot \frac{1}{x \log_e 7} = \frac{1}{x \cdot \log_7 x \cdot (\log_e 7)^2}$.Alternatively,using $\log_7 x = \frac{\log_e x}{\log_e 7}$ and $\log_7 e = \frac{1}{\log_e 7}$:$\frac{dy}{dx} = \frac{1}{\log_e 7} \cdot \frac{1}{\log_7 x} \cdot \frac{\log_7 e}{x} = \frac{\log_7 e}{x \log_e x}$.Thus,the correct option is $C$.

Find the derivative: $\frac{d}{dx}(\log_7(\log_7 x))$

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