Stationary points of general function When studying the stationary point(s) of the following
$$
Q=\frac{K(x)}{x}
$$
I find 
$$
\frac{dQ}{dx}=\frac{x\frac{dK(x)}{dx}-K(x)}{x^2}=0
$$
and hence
$$
\frac{K(x)}{x}=\frac{dK(x)}{dx}
$$
I'm sure this is elementary, but I'm looking for an intuitive explanation why the stationary point occurs when the original function is equal to the differential of the numerator. Thanks for your help.
 A: A slight recasting gives
$$x\cdot Q(x)=K(x).$$
Differentiating this gives another way of reaching the same conclusion as you without much more illumination. 
Instead let us go softly and induce a change at a stationary point $x_0\rightarrow x_0+\Delta x$. If $Q$ is stationary then $Q_0\rightarrow Q_0+\Delta Q\approx Q_0$. Finally $K_0\rightarrow K_0+\Delta K$.
Now we have
$$\begin{align}(x_0+\Delta x)\cdot Q_0&=K_0+\Delta K
\\\Rightarrow x_0Q_0+\Delta x\cdot Q_0&=K_0+\Delta K
\\ \underset{x_0Q_0=K_0}{\Rightarrow} \Delta K&=Q_0\cdot\Delta x,
\end{align}$$
and so if $Q$ is stationary at $x_0$ then near $x_0$, $\Delta K$ is proportional to $\Delta x$, and we have for points near $x_0$
$$Q(x)=\frac{K(x)}{x}\approx \frac{K_0+Q_0\Delta x}{x_0+\Delta x}=\frac{K_0}{x_0}=Q_0,$$
where we appealed to the lemma
$$\frac{a+x}{b+y}=\frac{a}{b}\Leftrightarrow \frac{a}{b}=\frac{x}{y}.$$
On the other hand (in hindsight), we could say that for $Q$ to be stationary near $x_0$ we need $K(x)$ to grow in proportion to $x$ so $K(x)\approx mx$:
$$Q(x)=\frac{K(x)}{x}\approx \frac{mx}{x}=m\approx \frac{dK}{dx},$$
and for stationarity we need $m=Q_0$.
A: I think a very intuitive, but not rigorous explanation could be that the function (or better the limit of the function for $x\rightarrow 0$ is a (limit) of a difference quotient of the function $K$. 
