why does $\frac{d}{dx} log_b(x)$ not = $\frac{lnb}{x}$?

Question

I know that $log_b(x) = \frac{lnx}{lnb}$, and that differentiating $$\frac{d}{dx}(\frac{lnx}{lnb}) = \frac{1}{lnb}\frac{d}{dx}(lnx)=\frac{1}{xlnb}$$, so where is my mistake when I do it this way: $$y=log_b(x); b^y = x$$ , so with implicit differentiation $$\frac{b^y}{lnb} \frac{dy}{dx}=1; \frac{dy}{dx}=\frac{lnb}{b^y}$$ and substitution gives $$\frac{dy}{dx}=\frac{lnb}{b^{log_b(x)}}=\frac{lnb}{x}$$ ??? This is not the same answer.

Answer 1

The derivative of $b^y$ respect to $x$ is $y'b^y\ln b$