Assuming $g(x):R^n\rightarrow R^m$, where $x\in R^n$, does the following equality hold?
$\dfrac{dg(x)^T}{dx}=\dfrac{dg(x)}{dx}=\dfrac{dg(x)^T}{dx^T}=\dfrac{dg(x)}{dx^T}$
I am confused about the dimension. I think they are all equal. Please help me with this. Thanks in advance:)
Given a differentiable function $g:\>{\mathbb R}^n\to{\mathbb R}^m$ at each point $x$ we have its differential (also called Jacobian map, or derivative) $$dg(x):\quad T_x\to T_{g(x)},\qquad X\mapsto Y:=dg(x).X\ ,$$ which is implicitly defined by $$g(x+X)-g(x)=dg(x).X+o\bigl(|X|\bigr)\qquad (X\to0)\ .$$ This differential is here independently of any conventions about matrices.
The usual convention is that the vectors $x$, $X$, $y$, $Y$ are written as column vectors. The matrix of $dg(x)$ then appears as an $(m\times n)$-matrix, often denoted by $J_g(x)$, and one has $$[Y]=\bigl[dg(x)\bigr] [X]$$ (of course the parentheses $[\ ]$ are usually omitted). If for some reason one needs $Y$ as a row vector the rules of matrix algebra at once give $$Y^\top= X^\top\ J_g(x)^\top\ .$$ It seems that the matrix $\bigl[dg(x)\bigr]=J_g(x)$ is denoted by ${dg\over dx}$ in some circles. But this ${dg\over dx}$ is an "icon" and as such indivisible. In particular it is not the quotient of two matrices, nor the limit of such quotients. It follows that three of the four expressions printed in the question make no sense. If for some reason you need the transpose of ${dg\over dx}$ write $\bigl({dg\over dx}\bigr)^\top$.
Answering your comment:
It goes the other way round: Mathematicians have codified the notion of $dg(x)$ after three hundred years of handwaving multivariate calculus. Now it is up to the physicists to web this and similar notions into their tales instead of mindlessly throwing indices and summation conventions around. – To answer your question: All I can say is $dg(x)$ is a linear map as explained above, and sometimes I write $[dg(x)]$ to indicate clearly that I'm talking about the matrix of this map.