# Does the Wronskian have anything to do with the product rule in calculus

Does the Wronskian have anything to do with the product rule in calculus. I ask this because i noticed the form looking similar to the product rule. $$W=g(x)f'(x)-g'(x)f(x)$$ where as the product rule is $$(f(x)g(x))' = g(x)f'(x)+g'(x)f(x)$$ they only differ by a plus operation.

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If you assume that $g\neq 0$, then it's $g(x)^2$ times the derivative of the quotient, and when it's $0$ it means that $f$ and $g$ are proportional. – Davide Giraudo Nov 2 '11 at 19:28
If $W = \left( \begin{array}{ccccc} f_1(x) & f_2(x) & f_3(x) & \dots & f_n(x) \\ f_1'(x) & f_2'(x) & f_3'(x)& \dots & f_n'(x) \\ 1 & 1 & 1 & \dots & 1 \\ \vdots & & &\ddots & \\ 1 & 1 & 1 &\dots & 1 \end{array} \right)$, the product rule is determined by the permanent of the matrix $W$. – JavaMan Nov 2 '11 at 19:33
@DavideGiraudo: Careful. A classic example is $f(x)=x^2$ and $g(x)=x|x|$. Then $g'(x) = 2x$ if $x\geq 0$, $g'(x)=-2x$ if $x\lt 0$, $W(f,g)=0$, but $f$ and $g$ are not proportional (they are equal on the nonnegative numbers, and $f=-g$ on the nonpositive ones). – Arturo Magidin Nov 2 '11 at 19:34
@ArturoMagidin yes you're right. So we have to assume that $g$ never vanishes. – Davide Giraudo Nov 2 '11 at 19:40
@DavideGiraudo: There are any number of conditions you can put to ensure the implication"$W(f,g)=0$ implies $f$ and $g$ proportional"; for example, if they are both analytic (due to Peano). See this paper, and Wikipedia for other situations. – Arturo Magidin Nov 2 '11 at 19:45

Not really. The Wronskian is a determinant of functions and their derivatives. In your case you have that, for two functions $f$ and $g$,
$$W(f,g) = \det\begin{pmatrix}f & g \\ f' & g'\end{pmatrix} = fg'-f'g$$
In general, the Wronskian of $n$ different functions is the determininant of the square $n\times n$ matrix where each column is composed of the $f$ and its derivatives up to the $(n-1)$ th derivative (such that every derivative exists in an interval $I$), ordered from lower to higher order from top to bottom. In general, $$W(f,g) \neq W(g,f)$$ as opposed to $(f·g)'$ and the same goes for higher order Wronskians.