How to prove the derivative of $\ln{f(x)}$? I've been trying to demonstrate the derivative formulas for a few functions. I could demonstrate that $\ln [x^{n}]=\frac{n}{x}$. But when I tried to derive $f(x)=\ln[u(x)]$ I couldn't. I just started it and couldn't go foreward. What I did was:
$$
\begin{align} 
f'(x)&=\lim_{h \to 0}\frac{\ln u(x+h)-\ln u(x)}{h}\\&=\lim_{h \to 0}\frac{\ln\frac{u(x+h)}{u(x)}}{h}\\
&=\lim_{h \to 0} \frac{1}{h}\ln\frac{u(x+h)}{u(x)}\\
&=\lim_{h \to 0} \ln\left(\frac{u(x+h)}{u(x)}\right)^{\frac{1}{h}}
\end{align}  $$
What do I do next?
 A: If you really want to use this approach, then write: $$\frac{u(x+h)}{u(x)}=1+\frac{u’(x)h}{u(x)}+\frac{u(x+h)-u(x)-hu’(x)}{u(x)}$$
Show $u(x+h)-u(x)-hu’(x)=o(h).$ This follows from the definition of the derivative.
Then show:

Lemma: If $f$ is defined and positive on some neighborhood of $0$ with $f(h)=1+ah+o(h)$ then $$\lim_{h\to 0} f(h)^{1/h}=e^a.$$

You can prove this first for $a=0,$ and then for general $a$ by showing $$\frac{1+ah+o(h)}{1+ah}=1+o(h).$$
That Lemma means that:
$$\left(\frac{u(x+h)}{u(x)}\right)^{1/h}\to e^{u’(x)/u(x)}$$
But the chain rule is easier.

The Lemma when $a=0.$
Assume $g(h)=f(h)-1=o(h).$
When $g(h)=0,$ $f(h)^{1/h}=1.$
When $g(h)\neq 0,$
$$f(h)^{1/h}=(1+g(h))^{1/h}=\left((1+g(h))^{1/g(h)}\right)^{g(h)/h}$$
But $(1+g(h))^{1/g(h)}\to e$ as $h\to 0,$ and $\frac{g(h)}h\to 0,$ by the definition of $o(h).$ So $f(h)^{1/h}\to 1.$
A: If you know that $\ln'(x)=1/x$, then you can proceed as follows. Suppose $f$ is differentiable at $x$, and $f(x)>0$. Then,
\begin{align}
(\ln\circ f)'(x) &= \lim_{h\to0}\frac{\ln(f(x+h))-\ln(f(x))}{h} \\[5pt]
&= \lim_{h\to0}\frac{\ln(f(x+h))-\ln(f(x))}{f(x+h)-f(x)}\cdot\lim_{h\to0}\frac{f(x+h)-f(x)}{h}
\end{align}
For the first limit, let $u=f(x+h)-f(x)$. Since $f$ is continuous at $x$, as $h\to0$, $u\to0$. Hence,
\begin{align}
(\ln \circ f)'(x) &= \lim_{h\to0}\frac{\ln(f(x)+u)-\ln(f(x))}{u}\cdot\lim_{h\to0}\frac{f(x+h)-f(x)}{h} \\[5pt]
&= \ln'(f(x))\cdot f'(x) \\[5pt]
&= \frac{f'(x)}{f(x)} \, .
\end{align}
Remark: this derivation assumes that there is a $\delta>0$ such that $0<|h|<\delta\implies f(x+h)-f(x)\neq0$. If this is not the case, then note that $f'(x)$ and $(\ln \circ f)'(x)$ must both be equal to zero.
Further remark: this derivation is extremely silly. It just proves the chain rule in the special case of $\ln \circ f$.
