Let $A\subset\mathbb R^n$ closed and convex. Its support function is defined as $h_A(u)=\sup_{x\in A}\langle x,u\rangle$, for every $u\in \mathbb R^n$.
Being $A$ closed, there must be, for each $u$, some $x_u\in A$ such that $h_A(u)=\langle x_u,u\rangle$.
For example, the support function of a unit circle is $h(n_\theta)=1$, with $h(n_\theta) = \langle x_\theta,n_\theta\rangle$ with $x_\theta=n_\theta$.
In $\mathbb R^2$, we can parametrise $u$ via its polar angle (sticking to vectors with unit norm). Denote such unit vectors with $n_\theta$. Then we write $h_A(n_\theta) = \langle x_\theta,n_\theta\rangle$ for some $x_\theta\in A$ (there might be more than one choice of $x_\theta$, so let's say we use some criterion to fix one). We want to recover the points $x_\theta$ corresponding to each $\theta$, and thus $\partial A$. Assuming $\theta\mapsto h_A(n_\theta)$ is differentiable around $\theta$, we have $$\partial_\theta h(n_\theta) = \langle \partial_\theta x_\theta, n_\theta\rangle + \langle x_\theta,\hat\theta\rangle,$$ where $\partial_\theta n_\theta=\hat\theta$ is the unit versor orthogonal to $n_\theta$.
If $\langle \partial_\theta x_\theta,n_\theta\rangle=0$, one can easily recover $x_\theta$, whose polar coordinates are $h(n_\theta)$ and $\partial_\theta h(n_\theta)$. This is the case whenever $\partial_\theta x_\theta=0$ (as happens for example for polygons), or when $\partial_\theta x_\theta$ is orthogonal to $n_\theta$ (this is the case e.g. for the case of the unit circle, in which $x_\theta=n_\theta$).
Does $\partial_\theta h_A(n_\theta)$ return the polar coordinate of $x_\theta$ also in the general case? In other words, does $\langle \partial_\theta x_\theta,n_\theta\rangle=0$ always hold? And what about the case of higher dimensions? Does this technique generalise straightforwardly to $\mathbb R^n$?
I asked a closely related question here, and used the same technique discussed here in an answer. This question differs from that one in that it uses a different language/formalism, and focuses on the validity of a specific relation (whether $\partial_\theta x_\theta$ is orthogonal to $n_\theta$) to show that the derivatives of the support function return the coordinates of the corresponding point in its support hyperplane.
Another way to derive the convex region corresponding to a given support function is mentioned in this other answer, but the focus is there on using the convex conjugate (which as far as I understand is not as directly applicable in practice as the method discussed here).
Assuming some smoothness, for example, that the convex body has a strictly convex boundary and that every boundary point has a unique normal, then it follows that the support function is differentiable, and its gradient evaluated at a unit vector $u$ is equal to the unique boundary point of the convex body it supports, where $u$ is a normal vector. Therefore the gradient map provides an effective parametrization of the boundary via the unit Euclidean sphere , and you can use it to actually draw the boundary in two and three dimensions, if you have an explicit expression for the support function.
By a convex body I mean a convex, compact with non-empty interior subset.