# Determinants and taylor expansion

1. Sep 17, 2013

### dipole

I'm doing a proof, and near the last step I want to write the expression,

$$\frac{d}{dt} \det{A(t)} = \lim_{\epsilon \to 0} \frac{\det{(A+\epsilon \frac{dA}{dt})} - \det{A}}{\epsilon}$$

which produces the right answer, so I believe that it may be correct. This looks very much like a Taylor expansion, but I'm having trouble justifying it exactly - does anyone know the proper way to Taylor expand a determinant, and if the above expression is actually true?

2. Sep 17, 2013

### voko

What you are really doing is: $\det A(t + \epsilon) = \det (A(t) + B(t, \epsilon))$, where $B(t, \epsilon) = A(t + \epsilon) - A(t)$, which is a matrix. Show that as $\epsilon \to 0$, $B(t, \epsilon) \to A'(t)\epsilon$.

3. Sep 17, 2013

### dipole

Ah yes I see now. The old "add zero" trick. ;)

Thank you!