While studying exponential functions, I understood that $$\frac{d}{dx}a^x=(\ln a)a^x.$$ I also learned previously that if $g(x)$ is the inverse of $f(x)$, then the derivative of $g(x)$ and the derivative of $f(x)$ has reciprocal relationship. As stated in my math textbook,
"In general, if $y=g(x)=f^{-1}(x)$, then $f(y)=x$ and $f'(y)=\frac{dx}{dy}$. It follows from [the equation of the derivative of inverse functions] that $$g'(x)=\frac{dy}{dx}=\frac{1}{f'(g(x))}=\frac{1}{f'(y)}=\frac{1}{dx/dy}.$$ This reciprocal relationship is sometimes written as $$\frac{dy}{dx}=\frac{1}{dx/dy}.$$"
Because $log_ax$ is the inverse of $a^x$, it would have made sense to have the derivative of $log_ax$ as $\frac{1}{(\ln a)a^x}$. However, I know that the derivative of $log_ax$ is $\frac{1}{(\ln a)x}$. Could you explain why the derivative of $log_ax$ and that of $a^x$ does not have reciprocal relationship? You do not need to explain the derivative proofs of $log_ax$ and $a^x$. I understand them completely. What I don't understand is why they don't have a reciprocal relationship.
You have $$g'(x) = \frac{1}{f'(g(x))}$$ With $g(x) = \log_{a}(x)$ one obtains
$$\log_{a}'(x) = \frac{1}{(\ln a)a^{\log_{a}(x)}} = \frac{1}{(\ln a)x}$$
Your mistake was that you didn't look at the function at the point $f(x)$ but at the point $x$. Be careful with infinitesimals.