Why $\nabla f$ do not exactly coincide with $D f$ (it's its transpose)

Is there any reason (historical, or of any other kind) to why $$\nabla f= \begin{bmatrix}\frac{\partial f}{\partial x} \\ \frac{\partial f}{\partial y} \\ \frac{\partial f}{\partial z} \\ \end{bmatrix}$$ for a three variables, scalar-valued function $$f$$ is the transpose of the total derivative of $$f$$ (and is not exactly equal to the total derivative of $$f$$) ?

Also, as an auxiliary: is my following reasoning right, in finding the Hessian of $$f$$ ?

I perform the matrix product between the $$3\times 1$$ nabla operator matrix: $$\nabla = \begin{bmatrix}\frac{\partial}{\partial x} \\ \frac{\partial}{\partial y} \\ \frac{\partial}{\partial z} \end{bmatrix}$$ and the $$1\times 3$$ total derivative matrix.

I came up with this "out of luck", but I have to say that I never know when to use $$\nabla$$ or instead its transpose.

I am also further confused by the fact that $$H$$ is informally described as $$\nabla ^2$$ in its Wikipedia article. Is it supposed to mean $$\nabla$$ applied to $$\nabla f$$ ?

PS: $$f$$ is assumed to be twice differentiable.

• Perhaps someone will correct me, but I think whether these objects are defined as rows or columns is just a matter of taste - depending on what fits neatly onto the page as much as anything else. Commented May 16 at 9:24
• The gradient of a function is typically a vector field, whereas the total derivative of a function should be a $1$-form. The two are dual, hence in coordinates they are transpose to one another. It isn't much deeper than this. Regarding the Hessian, I think you answered your own question already. The Hessian of a function is the matrix of second derivatives. In $3$ dimensions, the first derivative of $f$ is the data of $f_x$, $f_y$, and $f_z$. The second derivative is the data of $f_{xx}$, $f_{xy}$, etc. Commented May 16 at 9:26
• Would this answer help you math.stackexchange.com/a/4022891/319307? Or probably this one math.stackexchange.com/a/4905610/319307? Commented May 16 at 11:48
• @MathArt yes they confirm me that my way of computing the Hessian was right. I'm still wondering why we denote the Hessian as $\nabla ^2$ when it is in fact $\nabla \nabla ^T$. But yes they were helpful. Commented May 16 at 13:34
• Yes, I had been really puzzled by the abuse of publicly acknowledged Laplace operator $\nabla^2$ as Hessian matrix. Commented May 16 at 18:53