Proof of derivative of $\frac{\ln x}{x}$ from limits I have encountered the following proof of derivative from limits:
let $f(x)= \frac{\ln\ x}{x}$
\begin{equation} f'(x) = \lim_{\Delta x \to 0} \frac{\frac{\ln(x+\Delta x)}{x+\Delta x} - \frac{\ln\ x}{x}}{\Delta x} \end{equation}
\begin{equation} f'(x) = \lim_{\Delta x \to 0} \frac{x\ ln(1+\frac{\Delta x}{x}) - \Delta x\ ln\ x}{x\Delta x(x+\Delta x)}
\end{equation}
\begin{equation} f'(x) = lim_{\Delta x \rightarrow 0} \frac{\Delta x - \Delta x\ ln\ x}{x\Delta x(x+\Delta x)}
\end{equation}
\begin{equation} f'(x) = \frac{1-ln\ x}{x^2}
\end{equation}
From the 2nd to the 3rd line, I don't understand how the term $x\ ln(1+\frac{\Delta x}{x})$ was simplified to $\Delta x$ ?
 A: Hint: $$\ln (1+x) \sim x, x \to 0$$
Addition.
Let's consider $\lim_\limits{\Delta x \to 0} \frac{x\ ln(1+\frac{\Delta x}{x}) - \Delta x\ ln\ x}{x\Delta x(x+\Delta x)}$ and divide it in 2 parts:

*

*$\lim_\limits{\Delta x \to 0} \frac{x\ ln(1+\frac{\Delta x}{x}) }{x\Delta x(x+\Delta x)}=
\lim_\limits{ \Delta x \to 0}\left[\frac{xln(1+\frac{\Delta x}{x})}{\Delta x}\cdot \frac{1}{x(x+\Delta x) }\right]=\\
=\lim_\limits{ \Delta x \to 0} \frac{xln(1+\frac{\Delta x}{x})}{\Delta x} \cdot  \lim_\limits{ \Delta x \to 0} \frac{1}{x(x+\Delta x) }= \frac{1}{x^2}$


*$\lim_\limits{\Delta x \to 0} \frac{\Delta x\ ln\ x}{x\Delta x(x+\Delta x)}=\frac{\ln x}{x^2}$
All steps are exact and based on well known theorems.
And separately about limit $\lim_\limits{ \Delta x \to 0} \frac{xln(1+\frac{\Delta x}{x})}{\Delta x} = \lim_\limits{ \Delta x \to 0} \frac{ln(1+\frac{\Delta x}{x})}{\frac{\Delta x}{x}} =\lim\limits_{t\to 0} \frac{\ln (1+t)}{t}=1 $: here is applied a theorem about limit of functions composition i.e. changing variable, which is standard theorem.
Let me say, that it is important to study and remember, that in multiplication it is possible to change sub-expression with its equivalence expression, if all appropriate conditions hold: assume $f \sim \phi$ i.e. $\lim \frac {f}{\phi} = 1$. Then $\lim (f \cdot g)= \lim  \left(\dfrac {f}{\phi} \cdot \phi \cdot g \right) = \lim \frac {f}{\phi} \cdot \lim (\phi \cdot g) = \lim (\phi \cdot g)$ when all written limits exists.
A: That step is wrong. You can't replace a sub-expression by another unless they are equal. Any appeal to the use of standard limits in this manner is a complete disregard for limit laws. If it's coming from a textbook then shame!!
The correct way to handle this is as follows (using $h$ in place of $\Delta x$ to reduce typing effort)
\begin{align}
f'(x) &=\lim_{h\to 0}\frac{x\log(1+(h/x))-h\log x} {xh(x+h)} \notag\\
&=\lim_{h\to 0}\frac{\log(1+(h/x)) }{h/x} \cdot\frac{1}{x(x+h)}-\frac{\log x} {x(x+h)} \notag\\
&=1\cdot\frac{1}{x\cdot x} - \frac{\log x} {x\cdot x} \notag\\
&=\frac{1-\log x} {x^2}\notag
\end{align}
A: Hint   Using the power law of logarithm and definition of e, we can change 2nd line to $\to$
$f'(x) = \lim_{\Delta x \to 0} \frac{x\    ln(1+\frac{\Delta x}{x}) - \Delta x\ ln\ x}{x\Delta x(x+\Delta    x)}$
$=\lim_{\Delta x \to 0}    \frac{\Delta x\ ln(1+\frac{\Delta x}{x})^{\frac{x}{ \Delta    X}}}{x\Delta x(x+\Delta x)}-\lim_{\Delta x \to 0} \frac{ \Delta x\    ln\ x}{x\Delta x(x+\Delta x)}$
$=\frac{\ \lim_{\Delta x \to 0} ln(1+\frac{\Delta x}{x})^{\frac{x}{ \Delta X}}}{\lim_{\Delta x \to 0} x(x+\Delta x)}-\lim_{\Delta x \to 0} \frac{ \ ln\ x}{x(x+\Delta x)} $
$=\frac{ln(\ \lim_{\Delta x \to 0} (1+\frac{\Delta x}{x})^{\frac{x}{ \Delta X}})}{\lim_{\Delta x \to 0} x(x+\Delta x)}-\lim_{\Delta x \to 0} \frac{ \ ln\ x}{x(x+\Delta x)} $ limit property of continuous function
$=\frac{\ln e}{\lim_{\Delta x \to 0} x(x+\Delta x)}-\lim_{\Delta x \to 0} \frac{ \ ln\ x}{x(x+\Delta x)} $
With these intermediate steps and  definition of e ,  we can go from 2nd line to 3rd line .
A: Using $h$ in place of $\Delta x$ 
$\begin{align}
f'(x) &=\lim_{h\to 0}\frac{x\log(1+(h/x))-h\log x} {xh(x+h)} \notag\\
\end{align}$ 
Use series expansion of $\log (1+h/x)$, where $|h/x|\lt 1$, 
$\log(1+h/x)=\frac hx-\frac 12(\frac hx)^2+\frac 13(\frac hx)^3-\cdots$ 
$\begin{align}
f'(x)&=\lim_{h\to 0}\frac{h(1 +\text{higher powers of $h$)}-h\log x}{xh(h+x)}\\
     &= \lim_{h\to 0}\frac{(1+\text{higher powers of $h$})-\log x}{x(h+x)}\\
     &=\frac{1-\log x}{x^2}\end{align}$
