Implicit Differentiation of logarithm 
Differentiate $y=\log_a(x)$ with respect to $x$

I see that $a^y=x$.
My textbook says implicit differentiation gets us \begin{align*}a'(\ln a)\frac{dy}{dx}&=1 \\\implies \frac{dy}{dx}&=\frac{1}{a'\ln a} \\ \frac{dy}{dx}&=\frac{1}{x\ln a}\end{align*}
What I don't understand is why $\frac{d}{dx}[a^y]=a'(\ln a)\cfrac{dy}{dx}$ and why $a'=x$
When I try this using a base of $e$ with the chain rule, I get \begin{align*}\frac{d}{dx}[e^{y\ln a}]&=\frac{d}{dx}[x] \\ &\boxed{u=y\ln a, du=\frac{dy}{dx}\ln a+\frac1ay; \\ f=e^u, df=e^u \\ df/du*du/dx=e^{y\ln a}\frac{dy}{dx}\ln a+\frac1ay} \\ \implies x\frac{dy}{dx}\ln a+\frac1ay&=1 \\ \frac{dy}{dx}&=\frac{1}{\ln a}\biggr(\frac1x-\frac{y}{a}\biggr)\end{align*}
I see here that if I distribute, I get $\cfrac{1}{x\ln a}-\cfrac{y}{a\ln a}$ which implies $y$ must be zero! But I don't know how to show that, either. Can someone fill the gaps I'm missing in my textbooks solution?
UPDATE: I just realized the mistake I made in my differentiation was forgetting that ln (a) is a constant! Once I took out the constant or allowed the constant to be differentiated to $0$ I got the correct answer. 
I will mark the best answer correct soon enough, though, thanks everyone

Option:
$a^y=x$; 
Tale $\log_e$ of both sides: 
$y \log a=\log x$;
Differentiate with respect to $x$:
$y' \log a=\dfrac{1}{x}$;
$y'=\dfrac{1}{x \log a }$;
