nth derivative of a quotient

There’s a nice formula for the nth derivative of a product. It looks a lot like the binomial theorem.

$(gh)^{(n)} = \sum_{k=0}^n \binom{n}{k} g^{(k)} h^{(n-k)}$

There is also a formula for the nth derivative of a quotient, but it’s more complicated and less known.

We start by writing the quotient rule in an unusual way.

$\left(\frac{g}{h}\right)^{(1)} = \frac{1}{h^2} \left| \begin{array}{cc} h & g \\ h^\prime & g^\prime \\ \end{array} \right|$

Applying the quotient rule twice gives the following.

$\left(\frac{g}{h}\right)^{(2)} = \frac{1}{h^3} \left| \begin{array}{ccc} h & 0 & g \\ h^\prime & h & g^\prime \\ h^{\prime\prime} & 2h^\prime & g^{\prime\prime} \\ \end{array} \right|$

And here’s the general rule in all its glory.

$\left(\frac{g}{h}\right)^{(n)} = \frac{1}{h^{\,n+1}} \left| \begin{array}{cccccc} h & 0 & 0 & \cdots & 0 & g \\[3pt] h^\prime & h & 0 & \cdots & 0 & g^\prime \\[3pt] h^{\prime\prime} & 2h^\prime & h & \cdots & 0 & g^{\prime\prime} \\[3pt] \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\[3pt] h^{(n)} & \binom{n}{1}h^{(n-1)} & \binom{n}{2}h^{(n-2)} & \cdots & \binom{n}{1}h^\prime & g^{(n)} \end{array} \right|$

Source: V. F. Ivanoff. The nth Derivative of a Fractional Function. The American Mathematical Monthly, Vol. 55, No. 8 (Oct., 1948), p. 491

One thought on “nth derivative of a quotient”

ross

26 April 2026 at 00:09

I should really read the paper, but the 2×2 case looks like 1/h^2 multiplying the Determinant of [the matrix multiplication of matrices A and B, where A is a diagonal matrix with g and h on the diagonal, and B is an ‘operator’ matrix with appropriately placed (d/dx) entries.

It’s late and I feel i am missing something interesting here.

One thought on “nth derivative of a quotient”

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