Simple explanation of the differentiation of $\ln(f(x))$ Could somebody explain why the derivative of $\ln[f(x)]$ = $f'(x)/f(x)$ .
Why is it not simply $1/f(x)$ as is the case for the derivative of $\ln(x)$ being $1/x$? 
 A: We note first
$$
x = f(f^{-1}(x)) \Rightarrow \\
1 = f'(f^{-1}(x)) (f^{-1})'(x) \Rightarrow \\
(f^{-1})'(x) = 1 / f'(f^{-1}(x))
$$
so using $f(x) = e^x$ we get
$$
(\ln x)' = 1/e^{\ln(x)} = 1/x
$$
For
$$
(\ln(f(x))' = (\ln)'(f(x)) f'(x) = (1/f(x)) f'(x)
$$
we used the chain rule.
A: The derivative of $\log(f(x))$ with respect to $x$ is given by the chain rule.  Let $y=f(x)$.  
Noting that the derivative of $\log(y)$ with respect to $y$ is $\frac1y$, we can write
$$\begin{align}
\frac{d}{dx}\log(f(x))&=\left(\left.\frac{d\log(y)}{dx}\right)\right|_{y=f(x)}\\\\
&=\left(\left.\frac{d\log(y)}{dy}\frac{dy}{dx}\right)\right|_{y=f(x)}\\\\
&=\left.\left(\frac1y \frac{dy}{dx}\right)\right|_{y=f(x)}\\\\
&=\frac{1}{f(x)}\frac{df(x)}{dx}\\\\
&=\frac{f'(x)}{f(x)}
\end{align}$$
A: what we wish to find is 
$$\lim_{\Delta x\to 0} \frac{\ln f(x+\Delta x)-\ln f(x)}{\Delta x}$$
This can be found easily by using the chain rule (e.g. $\frac{\Delta y}{\Delta x} = \frac{\Delta y}{\Delta u}\cdot\frac{\Delta u}{\Delta x}$).
What $1/f(x)$ represents is
$$\lim_{\Delta f(x)\to 0} \frac{\ln (f(x)+\Delta f(x))-\ln f(x)}{\Delta f(x)}$$
