This might on first glance seem like a strange question, since it can just be said that $b=e$, hence $\log_b(a)=\ln(a)$, but hear my reasoning first.
So obviously, the "change of base" formula states that $$\log_a(x)=\frac{\log_b(x)}{\log_b(a)}$$ and when $b=e$, $$\log_a(x)=\frac{\ln(x)}{\ln(a)}$$ and then to take the derivative of $y=\log_a(x)$ from here is therefore quite easy: $$\frac{dy}{dx}=\frac{\ln(a)\cdot\frac{d}{dx}(\ln(x))-\ln(x)\cdot\frac{d}{dx}(\ln(a))}{(\ln(a))^2}$$ and since $\ln(a)$ is a constant,$$\frac{dy}{dx}=\frac{\frac{\ln(a)}{x}}{(\ln(a))^2}=\frac{\ln(a)}{x}\cdot\frac{1}{(\ln(a))^2}=\frac{1}{x\ln(a)}$$
now whilst differentiating, I can understand why the natural log is desirable, seeing as at this point, the derivative of $\ln(x)$ is already known, my question however, is: is it necessary to let $b=e$ to easily compute the derivative? The thought is very loose, so I apologise if the question is ill-worded. Any responses are appreciated.