# 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 ?

• At the very least, they both vanish iff $f$ and $g$ are linearly dependent (under some mild conditions on $f$ and $g$). Commented Jun 30, 2021 at 14:39

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 $$\left( \frac{f}{g} \right)'$$.

• This does not answer the question. The fact that the Wronskian is the numerator is obvious and does not need new proofs. The question is what is a connection between $W(f,g)$ and $(f/g)'$ that makes it true. An answer can be something like "$W(f,g)$ and $(f/g)'g^2$ measure the same <<geometric>> object". Commented Jun 30, 2021 at 2:47
• @MarkSapir The question specifically asked why the determinant appears in the numerator. Commented Jun 30, 2021 at 2:49
• @WillSavin: Exactly. And you do not answer that question. You gave new proofs of a trivial fact. Commented Jun 30, 2021 at 3:05

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.