Differentiable map

A map is differentiable at $x$ when it differs from an affine map with linear part $D{f}_x$ by $o(\Vert h\Vert )$ at $x+h$.

Functions whose differentials are linearly independent at the point in question. For $F=(F_1,\dots ,F_r)$ this means the derivative has rank $r$, so $F$ is a submersion there. The condition is open. It is distinct from global algebraic independence of functions.

If a differentiable map $F$ takes values in a level set of a differentiable scalar function $H$, the chain rule gives $D{H}_{F(x)}D{F}_x=0$. At a regular point, the image of the derivative lies in the kernel of the nonzero covector $DH$: the tangent space of the level set. For a map into the unit circle this forces rank at most one.

The Frechet derivative of $f$ at $a$ is the unique linear map $D{f}_a$ such that

A map on a subset of a normed vector space is Hadamard differentiable at $s$ if its directional difference quotients converge to a continuous linear map uniformly along all convergent sequences of directions: $t_j\to 0$, $a_j\to a$, and $s+t_j{a}_j$ in the domain imply $(\Phi (s+t_j{a}_j)-\Phi (s))/t_j\to {\dot{\Phi }}_s(a)$. Tangential differentiability restricts the limit directions to a specified subspace. This is the regularity used by the functional delta method.

On bounded continuously differentiable functions with uniformly bounded total variation of a function, the map $T$ is Hadamard differentiable in the supremum norm with derivative ${\dot{T}}_{F,G}(a,b)={\int }_0^{u}b{F}^{\prime }+G(u)a(u)-G(0)a(0)-{\int }_0^u{G}^{\prime }a$. Use weak convergence of bounded-variation integrators for the first term and integration by parts for the remaining terms. No derivatives of the limiting continuous directions are required.

A map is Fréchet differentiable at $x$ if it admits a bounded linear first-order approximation: $f(x+h)=f(x)+Ah+o(\Vert h\Vert )$ for a continuous linear map $A$.

On the open set of invertible square matrices, the inversion map is differentiable and

A function is continuously differentiable when its derivative exists and varies continuously.

An invertible linear map is a linear bijection; its inverse is also linear.

If the line segment from $x$ to $y$ lies in the domain of a differentiable map and $\Vert D{f}_z\Vert \le M$ along it, then

A differentiable map on a connected open subset of Euclidean space whose derivative vanishes everywhere is constant. The mean value inequality makes it constant on every ball in the domain, hence locally constant; connectedness then makes the value global.

If a continuously differentiable map has invertible derivative at a point, it is a local continuously differentiable diffeomorphism there.

If a smooth map has constant rank $r$ near a point, there are local coordinates in which it is

A local diffeomorphism restricts near every point to a diffeomorphism onto an open subset. In particular, every local diffeomorphism is an open map.

The displayed smooth map from ${\mathbb{R}}^2$ to the punctured plane has Jacobian determinant one: in polar coordinates its radius is $e^s$ and its angle is $r{e}^{-2s}$, so the determinant is $e^{2s}{e}^{-2s}=1$. Nevertheless $a(0,2\pi k)=(1,0)$ for every integer $k$. Thus a local volume-preserving map need not be injective or have bounded weighted inverse multiplicity. Compactly supported data distributed over many inverse branches disprove a determinant-only dispersion with a nonlinear velocity map estimate.

An open map sends every open set to an open set.

If $F(x_0,y_0)=0$ and the partial derivative with respect to $y$ is invertible at $(x_0,y_0)$, then the nearby zero set is the graph $y=g(x)$ of a continuously differentiable function.

If a holomorphic function $F(z,w)$ satisfies $F(z_0,w_0)=0$ and $F_w(z_0,w_0)\ne 0$, then near $(z_0,w_0)$ its zero set is the graph $w=g(z)$ of a unique holomorphic function. The analogous statement with $F_z\ne 0$ uses $w$ as the local coordinate.

Implicit differentiation differentiates an identity $F(x,y(x))=0$ and uses the chain rule to solve for derivatives of the implicitly defined function. In one dimension, $y^{\prime }=-F_x/F_y$ when $F_y\ne 0$.