TITLE: Why is the Jacobian matrix the transpose of what I would think it'd be/usefully be (total derivative is a synonym) (EDIT: I was a total wally)
QUESTION [5 upvotes]: I'm sorry this isn't a yes/no/am-I-right question but I seriously cannot see why the Jacobian/total derivative matrix is what it is?
I am also using it as LaTeX practice (for maths) hence the barely relevant blocks of math.
(note the terms in the title and this message is what I hope to get clarified, just incase they are wrong)
$f:\mathbb{R}^n\rightarrow\mathbb{R}^m$
and $f(x)=y$
I will use the notation of $x^i$ being the ith member of a vector x, that is:
$x=(x^1,x^2,x^3, ... ,x^k)$ for a vector of order/length k.
Consider also the linear map: (or should I say transform?)
$A:\mathbb{R}^n\rightarrow\mathbb{R}^m$
Consider now $x,u \in \mathbb{R}^n$ (I would use delta x for u, but then my notation could backfire)
The best linear approximation we can have for the ith coordinate of the image is:
$$f(x+u)^i = f(x)^i + \sum^n_{j=1}(\frac{\partial f(x)^i}{\partial x^j}u^j)$$
Which is basically the definition of a partial derivative. Nothing new here.
Clearly now:
$$f(x) + \begin{bmatrix}
\frac{\partial f(x)^1}{\partial x^1} &
\frac{\partial f(x)^1}{\partial x^2} &
...&
\frac{\partial f(x)^1}{\partial x^n} \\
\frac{\partial f(x)^2}{\partial x^1} &
\frac{\partial f(x)^2}{\partial x^2} &
...&
\frac{\partial f(x)^2}{\partial x^n} \\
...&...&...&...\\
\frac{\partial f(x)^m}{\partial x^1} &
\frac{\partial f(x)^m}{\partial x^2} &
...&
\frac{\partial f(x)^m}{\partial x^n}
\end{bmatrix}*\begin{bmatrix}u^1\\u^2\\...\\u^n\end{bmatrix}
\approx f(x+u)$$
Let:
$$\begin{bmatrix}
\frac{\partial f(x)^1}{\partial x^1} &
\frac{\partial f(x)^1}{\partial x^2} &
...&
\frac{\partial f(x)^1}{\partial x^n} \\
\frac{\partial f(x)^2}{\partial x^1} &
\frac{\partial f(x)^2}{\partial x^2} &
...&
\frac{\partial f(x)^2}{\partial x^n} \\
...&...&...&...\\
\frac{\partial f(x)^m}{\partial x^1} &
\frac{\partial f(x)^m}{\partial x^2} &
...&
\frac{\partial f(x)^m}{\partial x^n}
\end{bmatrix}*\begin{bmatrix}u^1\\u^2\\...\\u^n\end{bmatrix}
=\begin{bmatrix}v^1\\v^2\\...\\v^m\end{bmatrix}$$
Then:
$f(x+u)\approx y+v$
That is a pretty compelling case for not transposing it!
If you transpose it you can then use row-vectors and pre-multiply, rather than post, but column-vectors and linear algebra are something that is so well established, I wont restate.
Now the total derivative matrix is listed in my notes (quite correctly IMO) as what I've just put above.
Writing it like I have above looks nice, because now let that matrix be A:
$f(x+u) \approx f(x)+Au$ isn't that lovely!
The "Jacobian of a transformation" however is written as "the determinant of the transpose of this matrix" and I must ask WHY?
Transposing a matrix does not affect it's determinant! I cannot find any reason to transpose it UNLESS there is some notation I don't know about where:
$$\frac{\partial (x,y)}{\partial(u,v)}$$ = some 2x2 matrix that is the transpose of what I claim it ought to be.
But my book says the above denotes a determinant <---NOT a matrix
So why might I want to transpose it? 
Unless there are several things with "Jacobian" in the name maybe, could someone correct me!
Edit:
The books consider $A^T$ as the Jacobian Matrix (?) and "The Jacobian of the transformation" is defined using the determinant of that matrix, which is written as just:
$$\frac{\partial (y^1,y^2,...,y^n)}{\partial(x^1,x^2,..,x^n)}$$
Obviously in this case n=m
I WAS A HUGE DIPSTICK
I'm so sorry guys, this has wasted everyone's time, 'cept maybe John's.....
In the 2x2 case:
$$r_1=x_1-x_0 = (\frac{\partial x}{\partial u},\frac{\partial y}{\partial u})\Delta u$$
$$r_2=x_2-x_0 = (\frac{\partial x}{\partial v},\frac{\partial y}{\partial v})\Delta v$$
You can then find the area as $\lVert r_1 \times r_2\rVert$
And
$$r_1\times r_2 = det\begin{bmatrix}
\frac{\partial x}{\partial u} &
\frac{\partial y}{\partial u}
\\
\frac{\partial x}{\partial v} &
\frac{\partial y}{\partial v}\end{bmatrix} . \Delta u\Delta v\boldsymbol{k}$$
It is talking about the transpose of that!
I am so sorry everyone!

REPLY [4 votes]: The Jacobian matrix tends to show up in two places: As linear approximation of a map, and in change-of-variables of an integral.
Your explanation of the approximation of a map is correct and you have the formula right, as long as you're representing $\mathbb{R}^n$ with column vectors.
However, in a change of variables, we use the transpose of the Jacobian. Here's why. Let's say we're integrating some function $\phi$ over a domain $\Omega$:
$$\int_\Omega \phi \omega$$
where $\omega$ is the volume form on $\Omega$ (it takes $n$ vector fields and returns a function describing the infinitesimal volume spanned by the vector fields at each point ). In coordinates $x^i$ on $\Omega$, $\omega = dx^1\wedge dx^2\wedge\cdots dx^n$.
If we change variables, we precompose $\phi$ with a diffeomorphism $f$ and get
$$\int_{f^{-1}\Omega}(\phi\circ f)\ f^*\omega.$$
Here $f^*\omega$ is the pullback of the volume form $\omega$, which is defined by taking $n$ vector fields, pushing them through $df$, then feeding them to $\omega$. In coordinates, $f^*\omega = (f^*dx^1)\wedge (f^*dx^2)\wedge\cdots\wedge (f^*dx^n)$. 
That is, $f^*\omega$ is the wedge of the pullbacks of all $n$ coordinate one-forms. Now, in coordinates, a one-form can be represented as a row vector, just as a vector field can be represented as a column vector, and the action of one-forms on vector fields is simply matrix multiplication. The pullback of a one-form, say, $\eta$, is defined by $f^*\eta(v) = \eta(df(v))$. In coordinates, first multiply $v$ by $df$, then multiply by the column vector $\eta$:
$$(\ \ \ \eta\ \ \ )\bigg(\ \ \ df\ \ \ \bigg)\bigg(v\bigg).$$
But if we want the coordinates of $\eta$, represent $\eta$ as a vector and use the transpose: $\eta = (df^T\eta^T)^T$.
So $f^*\omega = (df^Tdx^1)^T\wedge\cdots\wedge(df^Tdx^n)^T$. Whenever you have the $n$-fold wedge product of vector fields, take the determinant of the matrix of the vector fields: $f^*\omega = (\det df^T)dx^1\wedge\cdots dx^n$. So that's why the transpose shows up.