The derivative of log_8(log_5 x) with respect | vedclass

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(D) Let $y = \log_8(\log_5 x)$.Using the change of base formula $\log_a b = \frac{\ln b}{\ln a}$, we have $y = \frac{\ln(\log_5 x)}{\ln 8}$.Since $\lo

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$\frac{1}{x \log_e 8 \log_e x}$

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(D) Let $y = \log_8(\log_5 x)$.Using the change of base formula $\log_a b = \frac{\ln b}{\ln a}$, we have $y = \frac{\ln(\log_5 x)}{\ln 8}$.Since $\log_5 x = \frac{\ln x}{\ln 5}$, we have $y = \frac{\ln(\frac{\ln x}{\ln 5})}{\ln 8} = \frac{\ln(\ln x) - \ln(\ln 5)}{\ln 8}$.Differentiating with respect to $x$:$\frac{dy}{dx} = \frac{1}{\ln 8} \cdot \frac{d}{dx} [\ln(\ln x) - \ln(\ln 5)]$.$\frac{dy}{dx} = \frac{1}{\ln 8} \cdot \frac{1}{\ln x} \cdot \frac{d}{dx}(\ln x) = \frac{1}{\ln 8 \cdot \ln x} \cdot \frac{1}{x}$.Thus, $\frac{dy}{dx} = \frac{1}{x \ln x \ln 8}$.

The derivative of $\log_8(\log_5 x)$ with respect to $x$ is:

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