\[y=lnx\]
then \[\frac{dy}{dx}=\frac{1}{x}\]
.If
\[y=log_k x\]
what is \[\frac{dy}{dx}\]
?\[y=log_k x \rightarrow k^y=x \rightarrow e^{y ln k}=x\]
Now we can differentiate using
\[\frac{d(e^Bx)}{dx}=Be^{Bx}\]
.We obtain
\[\frac{d( e^{y ln k})}{dx}=(lnk)e^{y ln k}\frac{dy}{dx}=1\]
using the chain rule.Hence
\[\frac{dy}{dx}=\frac{1}{(lnk)e^{y ln k}}=\frac{1}{lnk k^y}=\frac{1}{lnk x}==\frac{1}{xlnk}\]