Quotient rule with determinants?

I’ve always found the quotient rule for differentiation hard to remember:
$$\dod{}{x}\frac{f}{g} = \frac{f’g – g’f}{g^2}$$

I was looking on the Wikipedia page for the finite difference operator and noticed that they use the determinant in describing the quotient rule for the difference operator. The determinant can also be used to describe the quotient rule for the derivative:
$$\dod{}{x}\frac{f}{g} = \frac{1}{g^2}\begin{vmatrix}f’ & g’ \\f & g\end{vmatrix}$$
I wonder why this is? Is it just a coincidence?

Update: It occurred to me that it can also be written this way:
$$\dod{}{x}\frac{f}{g} = \det\left(\frac{1}{g}\begin{bmatrix}f’ & g’ \\f & g\end{bmatrix}\right)$$

Since the determinant represents the area of a parallelogram, the derivative at $x_0$ is the area of parallelogram formed by $ \langle f'(x_0),\ g'(x_0)\rangle $ and $ \langle f(x_0),\ g(x_0)\rangle $ where $ g(x_0) $ is scaled to 1. Still, this interpretation is kind of forced. What does it mean?

March 25, 2012, 11:34pm by Casey
Categories: Math | Tags: | Permalink | 2 comments

Comments (2)

  1. I’ve noticed this before! It is actually quite easy to see why this is so. It is exactly the same as the quotient rule though. I do, however, know a shortcut for large quotient rule problems so that you can do them IN YOUR HEAD! Let me know if you’d like details. Lee Hedgepeth, The University of Montevallo

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