---
title: The Derivative of a Map ℝⁿ → ℝᵐ
module: Functions of Several Variables (Introduction)
moduleNumber: 8
lessonNumber: 1
order: 801
summary: >
  The derivative of a map between Euclidean spaces is the linear transformation
  of vanishing relative error, unique when it exists and represented in
  coordinates by the Jacobian matrix of partial derivatives. Differentiability
  forces continuity through a local Lipschitz bound. Existence of the partial
  derivatives alone does not suffice; continuity of the partials does.
topics: [Functions of Several Variables (Introduction)]
sources:
  - book: Shkoller
    ref: "Ch. 2 — Differentiable mappings of ℝⁿ to ℝᵐ; §2.1 The derivative"
  - book: Shkoller
    ref: "§2.3 Matrix representation (Jacobian); §2.4 Continuity of differentiable mappings; §2.5 Criteria for differentiability"
  - book: Rosenlicht
    ref: "Ch. 9 — Functions of several variables"
draft: false
---

For a function $f : \mathbb{R} \to \mathbb{R}$, the derivative $f'(x_0)$ is a
number: the slope of the tangent line. That reading does not extend to
several variables. A map $f : \mathbb{R}^n \to \mathbb{R}^m$ has no single slope,
and its graph — a surface, or a higher-dimensional object — has no single tangent
direction. What does carry over is the idea behind the slope: near
$x_0$, the function is well approximated by a **linear** one. In one variable that
linear map happens to be multiplication by $f'(x_0)$. In several variables the
derivative _is_ the linear map itself.[^sh-deriv]

## The derivative as best linear approximation

Fix an open set $A \subseteq \mathbb{R}^n$ and a point $x_0 \in A$. We want a
linear transformation $L : \mathbb{R}^n \to \mathbb{R}^m$ such that the affine map
$x \mapsto f(x_0) + L(x - x_0)$ tracks $f$ to first order near $x_0$. "To first
order" is made precise by requiring the leftover error to vanish _faster_ than
the displacement $\lVert x - x_0 \rVert$.

> **Definition (Derivative in several variables).** A map $f : A \to \mathbb{R}^m$
> defined on an open set $A \subseteq \mathbb{R}^n$ is **differentiable** at
> $x_0 \in A$ if there is a linear transformation
> $Df(x_0) : \mathbb{R}^n \to \mathbb{R}^m$, called the **derivative** of $f$ at
> $x_0$, such that
>
> $$
> \lim_{x \to x_0} \frac{\lVert f(x) - f(x_0) - Df(x_0)\,(x - x_0) \rVert}{\lVert x - x_0 \rVert} = 0.
> $$
>
> Here $\lVert \cdot \rVert$ is the Euclidean norm and $Df(x_0)\,(x - x_0)$ is the
> value of the linear map on the vector $x - x_0$. If $f$ is differentiable at
> every point of $A$, it is differentiable on $A$.

The quantity
$r(x) = f(x) - f(x_0) - Df(x_0)\,(x - x_0)$
is the **remainder**: what the linear model misses. Differentiability says the
remainder is $o(\lVert x - x_0 \rVert)$ — negligible compared to how far we have
moved. An equivalent $\varepsilon$–$\delta$ phrasing is often more convenient in
proofs: for every $\varepsilon > 0$ there is a $\delta > 0$ such that
$\lVert x - x_0 \rVert < \delta$ implies

$$
\lVert f(x) - f(x_0) - Df(x_0)\,(x - x_0) \rVert \le \varepsilon \, \lVert x - x_0 \rVert.
$$

$$
% caption: The affine map $f(x_0)+Df(x_0)(x-x_0)$ is the best linear fit at $x_0$;
% the remainder (vertical gap) shrinks faster than the horizontal displacement.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  % axes
  \draw[black, ->] (-0.3,0) -- (6.4,0) node[right] {$x$};
  \draw[black, ->] (0,-0.3) -- (0,4.0) node[above] {$y$};
  % curve y = f(x): a gentle parabola-like arc
  \draw[thick, black] plot[domain=0.4:5.8, samples=60] (\x, {0.9 + 0.28*(\x-1.2)*(\x-1.2)});
  \node[black] at (5.4,3.7) {$y = f(x)$};
  % point x0
  \def\xz{2.4}
  \pgfmathsetmacro\yz{0.9 + 0.28*(\xz-1.2)*(\xz-1.2)}
  \pgfmathsetmacro\slope{0.56*(\xz-1.2)}
  % tangent (affine approximation) line
  \draw[thick, acc] (0.9, {\yz + \slope*(0.9-\xz)}) -- (5.6, {\yz + \slope*(5.6-\xz)});
  \node[acc, anchor=west] at (4.5, {\yz + \slope*(4.5-\xz) - 0.15}) {tangent line};
  % base points
  \def\xa{4.3}
  \pgfmathsetmacro\yfa{0.9 + 0.28*(\xa-1.2)*(\xa-1.2)}
  \pgfmathsetmacro\yla{\yz + \slope*(\xa-\xz)}
  \fill[acc] (\xz,\yz) circle (2pt);
  \node[acc, anchor=south east] at (\xz,\yz) {$x_0$};
  % remainder gap
  \draw[black, dashed] (\xa,0) -- (\xa,\yfa);
  \draw[->, black, thick] (\xa,\yla) -- (\xa,\yfa);
  \node[black, anchor=west] at (\xa, {0.5*(\yla+\yfa)}) {remainder};
  \draw[black, dashed] (\xz,0) -- (\xz,\yz);
  \draw[<->, black] (\xz,-0.25) -- (\xa,-0.25);
  \node[black, anchor=north] at ({0.5*(\xz+\xa)},-0.25) {displacement};
\end{tikzpicture}
$$

Two immediate sanity checks fix the notation.

- **A linear map is its own derivative.** If $f = L$ is already linear, then
  $f(x) - f(x_0) - L(x - x_0) = 0$ identically, so $DL(x) = L$ for every $x$. The
  best linear approximation to a linear map is itself.
- **A constant map has zero derivative.** If $f$ is constant, every difference
  quotient vanishes, so $Df = 0$.

For $f : \mathbb{R} \to \mathbb{R}$ with $f(x) = x^3$, the derivative in this
sense is the linear map $h \mapsto Df(x)\,h = 3x^2 h$: multiplication by the
familiar number $3x^2$. The scalar derivative is recovered as the sole entry of a
$1 \times 1$ matrix, and everything below reduces to
[the one-variable derivative](/real-analysis/differentiation/the-derivative) when
$n = m = 1$.

## Uniqueness of the derivative

Nothing in the definition names a specific $L$, only demands that one exist. In
fact there is at most one.

> **Theorem (Uniqueness).** If $f : A \to \mathbb{R}^m$ is differentiable at
> $x_0 \in A$, the linear transformation $Df(x_0)$ satisfying the definition is
> uniquely determined by $f$.

> **Proof.** Suppose $L_1$ and $L_2$ both work. Fix a unit vector $e$ and set
> $x = x_0 + \lambda e$ for small $\lambda \neq 0$, so $\lVert x - x_0 \rVert = |\lambda|$.
> By linearity,
>
> $$
> \lVert L_1 e - L_2 e \rVert = \frac{\lVert L_1(x - x_0) - L_2(x - x_0) \rVert}{\lVert x - x_0 \rVert}
> \le \frac{\lVert f(x) - f(x_0) - L_1(x - x_0) \rVert}{\lVert x - x_0 \rVert}
> + \frac{\lVert f(x) - f(x_0) - L_2(x - x_0) \rVert}{\lVert x - x_0 \rVert}.
> $$
>
> Both terms on the right tend to $0$ as $\lambda \to 0$, so $L_1 e = L_2 e$. Since
> $e$ was an arbitrary unit vector and any nonzero $y$ is a positive multiple of a
> unit vector, $L_1 = L_2$. $\blacksquare$

When $f : \mathbb{R}^2 \to \mathbb{R}$, the graph is a surface in $\mathbb{R}^3$
and $Df(x_0)$ encodes a unique tangent plane at each point of differentiability.
Uniqueness is the statement that a differentiable surface has one tangent plane,
not several.

## The Jacobian matrix

The derivative is a coordinate-free object, but once a basis is chosen it becomes
a matrix whose entries are the **partial derivatives**.

> **Definition (Partial derivative).** For $f = (f_1, \dots, f_m)$ with
> components $f_j : A \subseteq \mathbb{R}^n \to \mathbb{R}$, the partial
> derivative of $f_j$ with respect to $x_i$ at $x = (x_1, \dots, x_n)$ is
>
> $$
> \frac{\partial f_j}{\partial x_i}(x) = \lim_{h \to 0} \frac{f_j(x_1, \dots, x_i + h, \dots, x_n) - f_j(x_1, \dots, x_i, \dots, x_n)}{h},
> $$
>
> whenever the limit exists: the ordinary derivative of $f_j$ in the
> $x_i$-direction, holding the other coordinates fixed.

> **Theorem (Matrix of the derivative).** If $f : A \subseteq \mathbb{R}^n \to
> \mathbb{R}^m$ is differentiable at $x_0$, then all partial derivatives
> $\partial f_j / \partial x_i$ exist there, and the matrix of $Df(x_0)$ in the
> standard bases is the $m \times n$ **Jacobian matrix**
>
> $$
> Df(x_0) =
> \begin{pmatrix}
> \dfrac{\partial f_1}{\partial x_1} & \cdots & \dfrac{\partial f_1}{\partial x_n} \\[2mm]
> \vdots & & \vdots \\[1mm]
> \dfrac{\partial f_m}{\partial x_1} & \cdots & \dfrac{\partial f_m}{\partial x_n}
> \end{pmatrix},
> $$
>
> each partial evaluated at $x_0$.

> **Proof.** Let $a_{ji}$ be the $(j,i)$ entry, the $j$-th component of $Df(x_0)\,e_i$
> where $e_i$ is the $i$-th standard basis vector. Take $x = x_0 + h e_i$ in the
> definition. The relative error going to $0$ forces its $j$-th component to $0$,
> which reads
>
> $$
> \frac{|f_j(\dots, x_i + h, \dots) - f_j(\dots, x_i, \dots) - h\,a_{ji}|}{|h|} \to 0
> \quad\text{as } h \to 0.
> $$
>
> That is the statement that $\partial f_j / \partial x_i$ exists and
> equals $a_{ji}$. $\blacksquare$

The matrix records how each output coordinate responds to each input coordinate.
Row $j$ collects the sensitivities of the single output $f_j$; column $i$ collects
the responses of all outputs to the single input $x_i$.

$$
% caption: The Jacobian: row $j$ is the response of output $f_j$; column $i$ is
% the response to input $x_i$. Entry $(j,i)$ is the partial $D_i f_j$.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  % matrix grid 3x3
  \foreach \r in {0,1,2,3} \draw[black] (0,\r) -- (3.6,\r);
  \foreach \c in {0,1.2,2.4,3.6} \draw[black] (\c,0) -- (\c,3);
  % entries D_i f_j  (row j from top, col i)
  \node at (0.6,2.5) {$D_1 f_1$};
  \node at (1.8,2.5) {$D_2 f_1$};
  \node at (3.0,2.5) {$D_3 f_1$};
  \node at (0.6,1.5) {$D_1 f_2$};
  \node at (1.8,1.5) {$D_2 f_2$};
  \node at (3.0,1.5) {$D_3 f_2$};
  \node at (0.6,0.5) {$D_1 f_3$};
  \node at (1.8,0.5) {$D_2 f_3$};
  \node at (3.0,0.5) {$D_3 f_3$};
  % row highlight
  \draw[acc, thick] (0,1.95) rectangle (3.6,3.02);
  \node[acc, anchor=west, align=left] at (4.0,2.5) {row: output $f_1$};
  % column highlight
  \draw[black, thick] (0.02,0) rectangle (1.18,3);
  \node[black, anchor=west, align=left] at (4.0,0.5) {column: input $x_1$};
  \node[black, anchor=south] at (1.8,3.15) {inputs $x_1$, $x_2$, $x_3$};
\end{tikzpicture}
$$

Two special cases deserve names.

- **Scalar fields.** When $m = 1$, the Jacobian is a single row, the **gradient**
  $\nabla f = Df = \left( \partial f / \partial x_1, \dots, \partial f / \partial x_n \right)$,
  and $Df(x)\,w = \sum_i (\partial f/\partial x_i)\,w_i$ is the inner product of
  the [gradient](/real-analysis/several-variables/gradient-chain-rule) with $w$.
- **Paths.** When $n = 1$, a map $c : \mathbb{R} \to \mathbb{R}^m$ is a
  parameterized curve and $Dc(t)$ is the column vector
  $c'(t) = (c_1'(t), \dots, c_m'(t))$, the **tangent** (velocity) vector to the
  curve.

> **Worked example (a map $\mathbb{R}^2 \to \mathbb{R}^3$).** For
> $f(x_1, x_2) = (x_1^2,\ x_1^3 x_2,\ x_1^4 x_2^2)$,
>
> $$
> Df(x_1, x_2) =
> \begin{pmatrix}
> 2x_1 & 0 \\
> 3x_1^2 x_2 & x_1^3 \\
> 4x_1^3 x_2^2 & 2x_1^4 x_2
> \end{pmatrix}.
> $$

> **Worked example (a gradient).** For $f : \mathbb{R}^3 \to \mathbb{R}$ with
> $f(x_1, x_2, x_3) = \dfrac{x_1 \sin x_2}{x_3}$,
>
> $$
> \nabla f = \left( \frac{\sin x_2}{x_3},\ \frac{x_1 \cos x_2}{x_3},\ -\frac{x_1 \sin x_2}{x_3^2} \right).
> $$

The two examples read the derivative off the partials. The definition demands
more — that the resulting linear map actually fit $f$ to first order — and the
next example checks that directly.

> **Worked example (a Jacobian verified from the definition).** For
> $f : \mathbb{R}^2 \to \mathbb{R}^2$, $f(x, y) = (x^2 + y^2,\ xy)$, the partials
> give the candidate derivative at $x_0 = (a, b)$,
>
> $$
> Df(a, b) = \begin{pmatrix} 2a & 2b \\ b & a \end{pmatrix}.
> $$
>
> To confirm this is the derivative, evaluate the remainder at $x_0 + h$ with
> $h = (h_1, h_2)$. The first component expands as
> $(a + h_1)^2 + (b + h_2)^2 = (a^2 + b^2) + (2a\,h_1 + 2b\,h_2) + (h_1^2 + h_2^2)$
> and the second as $(a + h_1)(b + h_2) = ab + (b\,h_1 + a\,h_2) + h_1 h_2$.
> Subtracting $f(x_0)$ and the linear term $Df(a,b)\,h$ leaves
>
> $$
> r(h) = f(x_0 + h) - f(x_0) - Df(a,b)\,h = \bigl( h_1^2 + h_2^2,\ h_1 h_2 \bigr).
> $$
>
> Since $h_1^2 + h_2^2 = \lVert h \rVert^2$ and $|h_1 h_2| \le \tfrac{1}{2}\lVert h \rVert^2$,
>
> $$
> \frac{\lVert r(h) \rVert}{\lVert h \rVert} = \frac{\sqrt{\lVert h \rVert^4 + (h_1 h_2)^2}}{\lVert h \rVert} \le \frac{\sqrt{5}}{2}\,\lVert h \rVert \longrightarrow 0,
> $$
>
> so $f$ is differentiable at every $(a, b)$ and its derivative is the Jacobian above.

As a linear map, $Df(x_0)$ sends a
displacement $w$ in the domain to the linear estimate $Df(x_0)\,w$ of how $f$
changes, an arrow-to-arrow rule between the two spaces.

$$
% caption: $Df(x_0)$ maps a displacement $w$ at $x_0$ to the linear estimate of
% the resulting change in $f$; the true change is that estimate plus a remainder.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  % domain blob
  \draw[black] (0,0) ellipse (1.7 and 1.25);
  \node[black, anchor=south] at (0,1.3) {domain in $R^n$};
  \fill[black] (-0.3,-0.1) circle (1.6pt);
  \node[black, anchor=north east] at (-0.3,-0.1) {$x_0$};
  \draw[->, acc, thick] (-0.3,-0.1) -- (0.9,0.55);
  \node[acc, anchor=south west] at (0.6,0.35) {$w$};
  % arrow across
  \draw[->, black, very thick] (2.2,0) -- (3.6,0);
  \node[black, anchor=south] at (2.9,0.05) {$Df(x_0)$};
  % codomain blob
  \begin{scope}[xshift=6cm]
    \draw[black] (0,0) ellipse (1.9 and 1.25);
    \node[black, anchor=south] at (0,1.3) {codomain in $R^m$};
    \fill[black] (-0.5,-0.2) circle (1.6pt);
    \node[black, anchor=north east] at (-0.5,-0.2) {$f(x_0)$};
    \draw[->, acc, thick] (-0.5,-0.2) -- (0.9,0.35);
    \node[acc, anchor=south] at (0.35,0.15) {$Df(x_0)\,w$};
  \end{scope}
\end{tikzpicture}
$$

## Differentiability implies continuity

The one-variable fact that differentiable functions are continuous generalizes,
and the proof yields something stronger: a local Lipschitz bound.

> **Theorem (Differentiable ⇒ Lipschitz ⇒ continuous).** Let $A \subseteq
> \mathbb{R}^n$ be open and $f : A \to \mathbb{R}^m$ differentiable on $A$. Then
> $f$ is continuous. In fact, for each $x_0 \in A$ there exist $M > 0$ and
> $\delta_0 > 0$ such that $\lVert x - x_0 \rVert < \delta_0$ implies
>
> $$
> \lVert f(x) - f(x_0) \rVert \le M \, \lVert x - x_0 \rVert.
> $$

> **Proof.** Every linear map $L : \mathbb{R}^n \to \mathbb{R}^m$ satisfies
> $\lVert L x \rVert \le M_0 \lVert x \rVert$ for some constant $M_0$. Apply the
> definition with $\varepsilon = 1$: there is $\delta_0$ with
> $\lVert f(x) - f(x_0) - Df(x_0)(x - x_0) \rVert \le \lVert x - x_0 \rVert$ for
> $\lVert x - x_0 \rVert < \delta_0$. The triangle inequality then gives
>
> $$
> \lVert f(x) - f(x_0) \rVert \le \lVert Df(x_0)(x - x_0) \rVert + \lVert x - x_0 \rVert
> \le (M_0 + 1)\,\lVert x - x_0 \rVert,
> $$
>
> so $M = M_0 + 1$ works, and continuity at $x_0$ follows by taking
> $\delta = \min(\delta_0,\ \varepsilon / M)$.[^sh-cont] $\blacksquare$

The contrapositive is the practical tool: a map that is **discontinuous** at a
point cannot be differentiable there, no matter how many derivatives seem to
exist. That mechanism drives the counterexample below.

**Two cautions from one variable persist.** Continuity does not imply
differentiability: $f(x) = |x|$ is continuous everywhere but has no derivative at
$0$, since the difference quotient $f(x)/x$ equals $+1$ for $x > 0$ and $-1$ for
$x < 0$. And the derivative of a differentiable function need not be continuous:
$f(x) = x^2 \sin(1/x)$ (with $f(0) = 0$) is differentiable at $0$ with $f'(0) = 0$,
yet $f'(x) = 2x \sin(1/x) - \cos(1/x)$ has no limit as $x \to 0$.

## Existence of partials is not enough

The Jacobian theorem runs one way: differentiability produces partial
derivatives. The converse fails, and not on a technicality. A function can have
_every_ partial derivative at a point and still be discontinuous there, hence not
differentiable.

> **Example (partials exist, derivative does not).** Define $f : \mathbb{R}^2 \to
> \mathbb{R}$ by $f(x_1, x_2) = x_1$ when $x_2 = 0$, $f(x_1, x_2) = x_2$ when
> $x_1 = 0$, and $f(x_1, x_2) = 1$ everywhere else. Along the axes $f$ agrees with
> a coordinate, so
> $\partial f/\partial x_1 = \partial f/\partial x_2 = 1$ at the origin. But
> $f \equiv 1$ off the axes while $f(0,0) = 0$, so $f$ is not continuous at the
> origin, and therefore not differentiable there.

The diagnosis is geometric. A partial derivative probes the difference quotient
along **one coordinate axis only**. The two partials at the origin certify good
behavior along the two axes and say nothing about the approach from any other
direction. Differentiability, by contrast, requires a single linear map that
fits _every_ direction of approach at once.

$$
% caption: Partials sample $f$ only along the two axes (where it is tame); off the
% axes the value jumps to a constant, so no single linear map fits every approach.
\begin{tikzpicture}[>=stealth, font=\footnotesize, scale=1.0]
  \definecolor{acc}{HTML}{4A6FA5}
  % domain square
  \draw[black] (-2.2,-2.2) rectangle (2.2,2.2);
  % axes where f is tame
  \draw[acc, very thick] (-2.2,0) -- (2.2,0);
  \draw[acc, very thick] (0,-2.2) -- (0,2.2);
  \node[acc, anchor=south west] at (2.2,0) {axis: $f$ tame};
  \node[acc, anchor=west] at (0.15,-1.75) {axis: $f$ tame};
  % off-axis region label
  \node[black, align=center] at (1.25,1.25) {elsewhere\\ $f = 1$};
  \node[black, align=center] at (-1.25,-1.25) {elsewhere\\ $f = 1$};
  % origin
  \fill[black] (0,0) circle (2pt);
  \node[black, anchor=north east] at (-0.05,-0.05) {origin};
  % diagonal approaches
  \draw[->, black, thick, dashed] (1.9,1.9) -- (0.18,0.18);
  \draw[->, black, thick, dashed] (-1.9,1.9) -- (-0.18,0.18);
  \node[black, anchor=south] at (0,2.35) {approaches that see the jump};
\end{tikzpicture}
$$

## Continuously differentiable maps are differentiable

The counterexample also identifies the missing hypothesis: **continuity** of the
partials. Off the axes the partials never approach the partials at the origin.
Adding continuity restores the converse.

> **Theorem ($C^1$ ⇒ differentiable).** Let $A \subseteq \mathbb{R}^n$ be open and
> $f = (f_1, \dots, f_m) : A \to \mathbb{R}^m$. If every partial derivative
> $\partial f_j / \partial x_i$ exists and is continuous on $A$, then $f$ is
> differentiable on $A$.

> **Proof (scalar case; the vector case is componentwise).** Write $f_1 = g$ and fix
> $x$. For $y$ near $x$, expand the difference by changing one coordinate at a time,
>
> $$
> g(y) - g(x) = \sum_{i=1}^{n} \bigl[ g(x_1, \dots, x_{i-1}, y_i, \dots, y_n) - g(x_1, \dots, x_i, y_{i+1}, \dots, y_n) \bigr].
> $$
>
> Apply the one-variable
> [mean value theorem](/real-analysis/differentiation/mean-value-theorem) to each
> bracket: there are intermediate points $u_i$ with
>
> $$
> g(y) - g(x) = \sum_{i=1}^{n} \frac{\partial g}{\partial x_i}(\dots, u_i, \dots)\,(y_i - x_i).
> $$
>
> Subtract $Dg(x)(y - x) = \sum_i \frac{\partial g}{\partial x_i}(x)\,(y_i - x_i)$
> and use the triangle inequality:
>
> $$
> \lVert g(y) - g(x) - Dg(x)(y - x) \rVert \le \sum_{i=1}^{n} \left| \frac{\partial g}{\partial x_i}(\dots, u_i, \dots) - \frac{\partial g}{\partial x_i}(x) \right| \lVert y - x \rVert.
> $$
>
> Each $u_i$ lies between $x$ and $y$, so continuity of the partials makes each
> bracketed difference smaller than any prescribed tolerance once $\lVert y - x
> \rVert$ is small. The relative error goes to $0$, which is differentiability.
> $\blacksquare$

This is the practical test. To certify that a concrete map is differentiable,
compute its partials and check they are continuous — almost always immediate for
formulas built from polynomials, exponentials, and trigonometric functions. Maps
whose partials exist and are continuous on $A$ are called **continuously
differentiable**, or of class $C^1(A)$.

| Property | Holds for $f : \mathbb{R}^n \to \mathbb{R}^m$? | Direction |
| --- | --- | --- |
| differentiable $\Rightarrow$ partials exist | yes | Jacobian theorem |
| differentiable $\Rightarrow$ continuous | yes | Lipschitz bound |
| partials exist $\Rightarrow$ differentiable | **no** | axis counterexample |
| partials exist and continuous $\Rightarrow$ differentiable | yes | $C^1$ theorem |
| differentiable $\Rightarrow$ partials continuous | **no** | $x^2\sin(1/x)$ |

The two "no" rows are the content of the theory. Between the strong hypothesis
($C^1$) and the weak conclusion (partials exist) sits differentiability, and the
counterexamples show it is strictly between them.

## Consequences of the derivative

Near $x_0$ the map $f$ is the linear transformation $Df(x_0)$ plus a controlled
error, and the standard operations act on that linear part directly:

- composing maps multiplies their Jacobians, the
  [chain rule](/real-analysis/several-variables/gradient-chain-rule);
- differentiating $Df$ again yields the
  [Hessian and Taylor expansion](/real-analysis/several-variables/higher-derivatives-taylor-extrema);
- an invertible square $Df(x_0)$ makes $f$ locally invertible, the
  [inverse function theorem](/real-analysis/several-variables/inverse-implicit-theorems).

[^sh-deriv]: **Shkoller**, _MAT125B Lecture Notes_, §2.1 — the derivative of $f : \mathbb{R}^n \to \mathbb{R}^m$ as the unique linear map with vanishing relative error (Definitions 2.2–2.4, Theorem 2.5), and §2.3 for the Jacobian matrix (Definition 2.12, Theorem 2.13).
[^sh-cont]: **Shkoller**, _MAT125B Lecture Notes_, §2.4 (Theorem 2.17, local Lipschitz bound) and §2.5 (Theorem 2.22, continuity of partials gives differentiability; Example 2.21, the axis counterexample). See also Rosenlicht, Ch. 9, for the parallel development of the several-variable derivative.