Connection between determinant and quotient rule For the function $\dfrac{f(x)}{g(x)}$, we have, $\left(\dfrac{f}{g}\right)' = \dfrac{gf'-fg'} {g^2}$.
We can write the numerator as
$W(g,f) = \left|\begin{matrix} g & f \\ g' & f'\end{matrix}\right|$ which is called Wronskian.
I wonder why determinant appears in the numerator ? Is there any mathematical relationship between derivative of $\dfrac fg$ and determinant that gives us $W(g,f)$ right away ?
 A: With the risk of making a simple thing confusing, here's another point of view.
Two functions $f,g$ on an interval $I$ give a map to the projective line $I \dashrightarrow  \mathbb P^1$ by $x\mapsto [f(x):g(x)]$ (some indeterminacies can occur if both functions vanish at some point).
You can ask: when does this map have a critical point?
If you choose the chart $[y:1]$ on $\mathbb P^1$ you have to look at the vanishing of the derivative of $f/g$.
If you lift the map to $I\to \mathbb R^2$ via $x\mapsto (f(x),g(x))$ then you get a curve in the plane, with tangent vector $(f'(x),g'(x))$. The map to $\mathbb P^1$ has a critical point precisely when the vectors $(f(x),g(x))$ and $(f'(x),g'(x))$ are proportional, so you get the Wronskian.
The book of Ovsienko and Tabachnikov "Projective Differential Geometry, Old and New" contains a wealth of additional information.
A: If $f$ is $g$ times a constant $c$ then the quotient is $c$ and has derivative zero and the two columns of the Wronskian are linearly dependent, the left column equalling the right column times $c$, and thus the Wronskian has determinant $0$. This immediately suggests a relation.
One can make this a calculation-free proof that the determinant of the Wronskian appears in the numerator by using the ring of dual numbers. If $f$ and $g$ are elements of $\mathbb C[x]/(x^2)$ with $g$ invertible, then $\frac{f}{g}$ has derivative $0$ (at $x=0$) if and only if it is a constant, in which case the determinant of the Wronskian vanishes, so the determinant of the Wronskian must divide $\left( \frac{f}{g} \right)'$.
