I can compute the derivative of $\log_4 x$ using implicit differentiation. I can also compute it by writing it as $\log_4 x = (\ln x)/(\ln 4)$ and using the fact that the derivative of $\ln x$ is $1/x$. But I'm investigating why I can't seem to get it in the following approach, however bad an approach it might be.
Let $y = \log_4 x = \frac{\ln x}{\ln 4}$. Now I raise $e$ to the equation, getting $$\exp(y) = \exp\left(\frac{\ln x}{\ln 4}\right).$$ Using implicit differentiation, I get \begin{align*} \exp(y)y' &= \exp\left(\frac{\ln x}{\ln 4}\right) \left(\frac{\ln x}{\ln 4}\right)' = \exp\left(y\right) y'\\ \exp(y)y' - \exp(y)y' &= 0\\ (\exp(y) - \exp(y))y' &= 0 \end{align*}
In other words, it leads me nowhere. Is it possible at all do this with the conditions imposed in the question? It seems that if the exercise asks me to use a change of base, then I must really use the derivative of $\ln x$ to solve the problem.
let $l=\lim_{h\to0}\frac{\log_4(x+h)-\log_4(x)}{h}$ $$4^l=4^{(\lim_{h\to0}\frac{\log_4(x+h)-\log_4(x)}{h})}$$ $$4^l=\lim_{h\to0}(\frac{x+h}{x})^{\frac{1}{h}}$$ $\lim_{h\to0}(\frac{x+h}{x})^{\frac{1}{h}}$ is the definition of $\exp(\frac{1}{x})$ so $$4^l=\exp(\frac{1}{x})$$ $$l=\log_4({e^{\frac{1}{x}}})$$ then $l=\log_4({e}) \frac{1}{x}$
but this is the same way to proof that the derivative of $\ln(x)$ is$\frac{1}{x}$