Explanation of ln derivative... So we know that for example :

I'd like to know how do we get to this result based on:

I've been trying to do it for myself but I really have no idea how to get to the result starting.
I'll show you an alternative way of evaluating deriving that $\frac{d}{dx}\ln(x)=\frac{1}{x}$.

We know that if $f$ is differentiable in an interval $I$, with $f'(x) &lt; 0$ or $f'(x) &gt; 0$ for all $x \in I$, then $f$ has an inverse funcion $f^{-1}$which is defined and is differentiable on $f(I)$.

So let $f(x) = e^x$. Then $f$ is defined and differentiable in $\mathbb R$ and $\forall x \in \mathbb R,f'(x) &gt; 0$, so there exists an inverse function defined and differentiable in $f(\mathbb R) = \mathbb R^+$, which is $f^{-1}(x) = \ln(x)$
We know that $f(f^{-1}(x)) = x$. If we differentiate from both sides we get:
$$\begin{align}
&amp;(f^{-1})'(x)f'(f^{-1}(x)) = 1 \iff
\\
\\
&amp;(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}
\end{align}$$
We know that $f'(x) = e^x$ so $f'(f^{-1}(x)) = x$. So we conclude that
$$\begin{align}
&amp;(f^{-1})'(x)=\frac{1}{x} \iff
\\
\\
&amp;\frac{d}{dx}\ln(x)=\frac{1}{x}
\end{align}$$
Using then the chain rule you can get the derivative you want.
This is a common way of finding out derivatives for inverse functions.
hint
If $ f(x) $ has the form
$$f(x)=\ln(g(x))$$
then
$$\frac{f(x+h)-f(x)}{h}=\frac 1h\ln(\frac{g(x+h)}{g(x)})$$
$$\sim \frac 1h(\frac{g(x+h)}{g(x)}-1)$$
$$\sim \frac{g(x+h)-g(x)}{h}\frac{1}{g(x)}$$
thus
$$\lim_{h\to 0}\frac{f(x+h)-f(x)}{h}=\lim_{h\to 0}\frac{g(x+h)-g(x)}{h}\frac{1}{g(x)}$$
or
$$f'(x)=\frac{g'(x)}{g(x)}$$