Is this proof that the derivative of $\ln(x)$ is $1/x$ correct? \begin{align}
\frac{\mathrm{d}}{\mathrm{d} x}\ln x &= \lim_{h\to0} \frac{\ln(x + h) - \ln(x)}{h} \\
&= \lim_{h\to0} \frac{\ln(\frac{x + h}{x})}{h} \\
&= \lim_{h\to0} \frac{\ln(1 + \frac{h}{x})}{h} \\
\end{align}
\begin{align}
&= \lim_{h\to0}{\ln(1 + \frac{h}{x})}^\frac{1}{h} \\
\end{align}
Now this is the part I'm asking about, because I see most people set the reciprocal of my limit, but this one seems to work:
\begin{align}
n=\frac{x}{h},  h=\frac{x}{n}, \frac{1}{h} = \frac{n}{1}*\frac{1}{x}
\end{align}
So when h tends towards 0, n tends towards infinity.
\begin{align}
&= \lim_{n\to\infty}{\ln\biggl((1 +\frac{1}{n})}^n\biggl)^\frac{1}{x} \\
\end{align}
Here we have
\begin{align}
{\ln\lim_{n\to\infty}\biggl((1 +\frac{1}{n})}^n\biggl)^\frac{1}{x} \\
\end{align}
\begin{align}
=\frac1x\ln(e) = \frac1x
\end{align}
This is my first post, I was just curious. Please don't berate me if it's a silly question.
 A: yeah your answer looks correct , I'd probably do it this way tho ;
$\lim_{h\to0}\frac1h\ln\big(1+\frac hx) \\= \lim_{h\to0}\frac xh \frac1x\ln\big(1+\frac hx) \\= \frac1x\lim_{h\to0}\ln\big(1+\frac 1{\frac xh})^\frac xh\\ = \frac1x\ln(e) \\= \frac1x$
A: I think this is OK. You should acknowledge that at one point you are using the fact that the function raising something to a fixed power is continuous:
$$
\lim (\text{something}^a)=  (\lim (\text{something}) )^a .
$$
Moreover, the usual definition of exponentiation for an arbitrary exponent $a$ depends on the natural logarithm:
$$
\text{something}^a = \exp(a (\ln \text{something})) 
$$
so there's a lot of serious analysis (sometimes skipped over the first time through in calculus) in the correct algebraic manipulations you're doing.
That said, this is a strange exercise. Presumably you have defined $\ln$ as the inverse of exponentiation, so that
$$
\exp(\ln(x)) = x .
$$
Then the formula for the derivative of $\ln$ follows from the chain rule.
