I know determinants tell you the oriented volume of the parallelepiped after the linear transformation, but if you define the derivative as I do below then it seems equivalent, at least in terms of the “slope” of the linear transformation. Or at least I hope someone can answer what exactly it is I’m defining below. It gives the derivative in single variable but doesn’t give best linear approx. in multivariate case. Is it a kind of directional derivative?
The typical definition of a derivative is $$f'(x) \approx \frac{f(x+dx) - f(x)}{dx}$$ where $dx$ is an infinitesimal (appealing to nonstandard analysis here for simplicity).
This is usually motivated as the slope of the tangent line to a point at $f(x)$, but the more general formulation of the derivative is the ratio of oriented intervals:
Let $I_{ab}$ be an oriented interval in the domain of a function $f$, in most cases we will want this to be an infinitesimal interval so an interval of $[a, a+dx]$. More generally an oriented interval is an oriented n-dimensional hypercube/hyper-rectangle/parallelotope, if $f: \mathbb{R} \rightarrow \mathbb{R}$ then $I_{ab}$ is the interval $[a,b]$ where $a,b \in \mathbb{R}$, so it is a 1-dimensional oriented hypercube, such that $ab = -ba$ (suppresed $I$ notation, treating the interval algebraically like a simplicial complex).
The magnitude of the coefficient to the interval is the size of the interval (i.e. the volume of the interval hypercube) and the sign of the coefficient is the orientation, so we start with a positive unit interval $I_{ab}$ and see how the function changes the interval size and orientation. Then the derivative of a function $f: \mathbb{R}\rightarrow \mathbb{R}$ is:
$$ f'(x) \approx \frac{f(I_{ab})}{I_{ab}} $$
For example, if $f(x) = -2x$ then $f(I_{ab}) = 2I_{ba} = -2I_{ab}$ and hence $f'(x) = \frac{-2I_{ab}}{I_{ab}} = -2$. So more generally for a function $f: \mathbb{R}^M \rightarrow \mathbb{R}^N, f' = \frac{f(I^N)}{I^M}$ where $I^N$ refers to an N-dimensional oriented interval.
For a 2x2 matrix, representing a linear map from $g: \mathbb{R}^2 \rightarrow \mathbb{R}^2$, the interval is an oriented 2-dimensional hypercube (a square), with corners denoted $abcd$, read clockwise for a positive orientation, such that $abcd = -adcb$.
Consider the linear map $F: \begin{bmatrix}-2&0\\0&2 \end{bmatrix}$, which simply inverts one orthonormal basis and scales by 2. We know the determinant is -4. Label an arbitrary 2-hypercube with corner points $abcd$ read in clockwise orientation.
This linear map will map this interval such that $F(I_{abcd}) = 4I_{adcb} = -4I_{abcd}$ (suppresing $I$ for easier reading, $F(abcd) = 4adcb = -4abcd$, since the map flips the originally clockwise orientation of the corners to counter-clockwise, hence negative sign. Now we calculate $F' = \frac{-4I_{abcd}}{I_{abcd}} = -4$, the same as the determinant. In the general case, $f(I^N)$ may map an oriented unit hyper-cube to an arbitrary N-parallelotope, but it is still a kind of interval with orientation which gives the sign of the determinant.
I already knew determinants tell you the volume of the parallelepiped after the mapping but in the way I've defined a derivative here, in what sense is the determinant any different than the slope of the derivative of the linear map represented by the matrix?
The determinant seems analogous to the derivative to me. The derivative tells you how much an (infinitesimal) interval is scaled (magnitude of derivative) and if its orientation changes (sign of derivative), which is exactly what the determinant is telling you about a linear map.
Short answer, no. The derivative of the linear map is the map itself.
What you are computing is how the linear map transforms volumes, which of course is the determinant.