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) 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!
 A: 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.
A: I have always used what you first have (without the transpose) as the Jacobian. It is in every book I have seen. It is also the definition on wikipedia.
https://en.wikipedia.org/wiki/Jacobian_matrix_and_determinant
So basically, you are right, where ever you are reading it is wrong.
A: It is simply because they don't transpose the vector you feed and do transpose the Jacobian, and conversely, those who don't transpose the Jacobian transpose the vector. This is because $$(A\cdot v)^t=v^t\cdot A^t$$
and your book simply uses the right hand side of the equation as opposed to using the left hand side. Note that in all your equalities you are transposing the column vector, which is what made you think something was wrong. You write $A\cdot v=w$ where both $w$ are column vectors, but then write them with the same letter as row vectors. For example, if we agree to write $f=(f_1,f_2,\ldots,f_n)$ as a row vector, the product in this equation $$f(x) + \begin{pmatrix}
\frac{\partial f(x)^1}{\partial x^1} &
\frac{\partial f(x)^1}{\partial x^2} &
...&
\frac{\partial f(x)^1}{\partial x^n} \\
\frac{\partial f(x)^2}{\partial x^1} &
\frac{\partial f(x)^2}{\partial x^2} &
...&
\frac{\partial f(x)^2}{\partial x^n} \\
...&...&...&...\\
\frac{\partial f(x)^m}{\partial x^1} &
\frac{\partial f(x)^m}{\partial x^2} &
...&
\frac{\partial f(x)^m}{\partial x^n}
\end{pmatrix}\cdot \begin{pmatrix}u^1\\u^2\\...\\u^n\end{pmatrix}
\approx f(x+u)$$
is a column vector, so what you write makes no sense. Of course, one can write them as column vectors, as things are just OK. It is simply a matter of convention!
Alternatively, as you suggest, $$f({\bf x})^t+\left({\bf D}f({\bf x})\cdot {\bf u}^t\right)^t\simeq f({\bf x+u})^t $$is correct assuming we're working with row vetors.
And in that case, this is the same as
$$f({\bf x})+{\bf u}\cdot{\bf D}f({\bf x})^t \simeq f({\bf x+u})$$
which is presumably what your book writes, that is, your book works with row vectors all the way and transposes the matrix to make things work.
