Prove that $\frac{d(\log(x))}{dx}=\frac{1}{x}$ Usually this is just given as a straight up definition in a calculus course. I am wondering how you prove it?
I tried using the limit definition, $$\lim\limits_{h\rightarrow 0} \dfrac{\log(x+h)-\log(x)}{h}$$
 but this led to no developments.
 A: Set $y = \log_e(x).$ Rearranging, we get $e^{y} = x\,(*).$ Differentiating both sides implicitly with respect to $x$:
$$e^{y}\cdot\frac{dy}{dx} = 1.$$ It follows that $$\frac{dy}{dx} = \frac{1}{e^y}$$ and hence that the derivative is $\frac{1}{x}$ from $(*).$
A: First note that

$$ e= \lim\limits_{z\to 0}\left(1+z\right)^{\frac{1}{z}}$$

So now
$$ \lim\limits_{h\to 0}\frac{\ln(x+h)-\ln (x)}{h} = \lim\limits_{h\to 0}\frac{\ln\left(\frac{x+h}{x}\right)}{h} $$
$$
= \lim\limits_{h\to 0}\frac{1}{h}\ln\left(1+\frac{h}{x}\right) = \lim\limits_{h\to 0} \ln\left(1+\frac{h}{x}\right)^{\frac1h}
$$
Let $z=\frac{h}{x}$, then
$$ \lim\limits_{z\to 0} \ln\left(1+z\right)^{\frac{1}{zx}}= \lim\limits_{z\to 0} \frac{1}{x} \ln\left(1+z\right)^{\frac{1}{z}}
$$
$$= \frac{1}{x} \ln\left(
\lim\limits_{z\to 0}\left(1+z\right)^{\frac{1}{z}}\right)=\frac1x \ln (e)=\frac1x
$$
A: This depends on how you define $\log(x)$.  As you say, this could be taken to be the definition of $\log(x)$:  define $y=\log(x)$ as the unique solution to the first order diff EQ $y'=\frac{1}{x}$ satisfying $y(1)=0$.
You might define $e^x$ as a solution to a diff EQ, and $\log$ as its inverse.  In this case, you can prove the theorem by the standard method of finding derivatives of inverse functions.
What definition of $\log$ do you have in mind?  Only then can we provide you with a computation of its derivative.
A: As I have written here before
(somewhere),
I like using the
functional equation
for logs of
$f(xy) = f(x)+f(y)$.
From this,
$f(xy)-f(x) = f(y)$
or
$f(x)-f(y) = f(x/y)
$.
Getting this into a form
that looks like a derivative,
$f(x+h)-f(x)
=f(1+h/x)
$.
Since $f(1) = 0$
(from $f(1) = f(1)-f(1) = 0$),
$\begin{array}\\
\frac{f(x+h)-f(x)}{h}
&=\frac{f(1+h/x)}{h}\\
&=\frac{f(1+h/x)-f(1)}{h}\\
&=\frac1{x}\frac{f(1+h/x)-f(1)}{h/x}\\
\end{array}
$
If $f$ is differentiable at $1$,
the right-hand side of this,
as $h \to 0$,
is 
$\frac{f'(1)}{x}
$.
Since the right-hand limit exists,
the left-hand limit exists,
so
$f'(x)
=\frac{f'(1)}{x}
$.
The natural log
is the one where
$f'(1) = 1$.
This also works
(for getting the
derivative of a function
from a functional equation)
 for $\exp$
(from $\exp(x)\exp(y) = \exp(x+y)$)
and $\arctan$
( from
$\arctan(x)+\arctan(y)
=\arctan\left(\frac{x+y}{1-xy}\right)
$)
(and probably more functions with
functional equations).
A: $$\begin{align}
f'(x)&=\lim_{h\rightarrow 0} \dfrac{f(x+h)-f(x)}{h}\tag{1}\\
&=\lim_{h\rightarrow 0} \dfrac{\log(x+h)-\log(x)}{h}\tag{2}\\
&=\lim_{h\rightarrow 0} \dfrac{\log\left(\frac{x+h}{x}\right)}{h}\tag{3}\\
&=\lim_{h\rightarrow 0} \dfrac{\log\left(1+\frac{h}{x}\right)}{h}\tag{4}\\
&=\lim_{h\rightarrow 0}\log\left(1+\frac{h}{x}\right)^{1/h}\tag{5}\\
&=\lim_{u\rightarrow 0}\frac1x\log(1+u)^{1/u}\tag{6}\\
&=\frac1x\lim_{u\rightarrow 0}\log(1+u)^{1/u}\tag{7}\\
&=\frac1x\ln e\tag{8}\\
&=\frac1x\tag{9}\\
\end{align}$$

$$\frac{d}{dx}\Big[\log(x)\Big]=\frac{1}{x}$$


$\text{Explanation}$ $6$ substituting $\dfrac hx=u\iff h=ux$
A: Usually, this statement is defined in the following way: $$\int_0^x \frac{1}{t}dt = \ln(x), x \in (0, \infty)$$. 
Nonetheless, suppose you were not working with this as a definition - how could we verify this from, say, what we know about $f(x)=e^x$ and then proceeding to define $\ln(x)$ as that function $f^{-1}(x)$ which is the inverse function of $e^x$?
We know from derivatives of inverses of differentiable functions that $$[f^{-1}]'(y)=\frac{1}{f'(x)}$$ where $y=f(x)$ (draw the tangent lines on $e^x$ and their corresponding reflected points on the inverse function to convince yourself of this geometrically). Therefore (being a bit liberal with our usage of $x$ here), we have that 
$$\frac{d}{dx} \ln(x) = \frac{1}{e^{ln(x)}}=\frac{1}{x}$$ for $x \in (0, \infty)$ (as $f(x)>0$, the above expression of $f'(x)=e^{f^{-1}(x)}=e^{ln(x)} $exists for all such $x$). 
