Using implicit differentiation to differentiate log_a (x)

[endeavor]

Could someone please make sure I'm doing this right.
I want to find the derivative of the logarithm to the base a of x, using implicit differentiation.

Let $$y = \log_{a} x$$

$$a^y = x$$

$$\frac{d}{dx} (a^y) = 1$$ (implicit differentiation)

$$\frac{d}{dx} (e^{\ln a})^y = 1$$

$$\frac{d}{dx} (e^{(\ln a)y}) = 1$$

$$e^{(\ln a)y} \frac{d}{dx} ((\ln a)(y)) = 1$$

$$(e^{\ln a})^y \frac{d}{dy} ((\ln a)(y)) \frac{dy}{dx} = 1$$ (did I use the chain rule correctly here?)

$$(a^y)(\ln a) \frac{dy}{dx} = 1$$

$$x \ln a \frac{dy}{dx} = 1$$

$$\frac{dy}{dx} = \frac{1}{x \ln a}$$

 

[HallsofIvy]

Yep, very nicely done!