The Derivative of a Map ℝⁿ → ℝᵐ

For a function $f:\mathbb{R}\to \mathbb{R}$, the derivative $f^{\prime }(x_0)$ 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 $f^{\prime }(x_0)$. In several variables the derivative is the linear map itself.¹

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 $\Vert x-x_0\Vert$.

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 $o(\Vert x-x_0\Vert )$ — negligible compared to how far we have moved. An equivalent $\epsilon$–$\delta$ phrasing is often more convenient in proofs: for every $\epsilon >0$ there is a $\delta >0$ such that $\Vert x-x_0\Vert <\delta$ implies

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 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.

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.

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 $x_i$.

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 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 $c^{\prime }(t)=(c_1^{\prime }(t),\dots ,c_m^{\prime }(t))$, the tangent (velocity) vector to the curve.

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.

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.

Differentiability implies continuity

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

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 $f^{\prime }(0)=0$, yet $f^{\prime }(x)=2x\sin (1/x)-\cos (1/x)$ 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.

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.

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.

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 $C^1(A)$.

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;
• differentiating $Df$ again yields the Hessian and Taylor expansion;
• an invertible square $Df(x_0)$ makes $f$ locally invertible, the inverse function theorem.