Differentiating Logarithms

If  
\[y=lnx\]
  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  
\[\frac{dy}{dx}=\frac{1}{(lnk)e^{y ln k}}=\frac{1}{lnk k^y}=\frac{1}{lnk x}==\frac{1}{xlnk}\]