A derivative is supposed to be a local linear approximation, but differentiability at isolated points is not enough to support the geometric and analytic arguments that calculus wants to make. A function can have a derivative at every point while its derivative jumps, oscillates, or refuses to behave uniformly. Then tangent planes exist pointwise, yet the function may not be well approximated in a stable way as the base point moves. The class of $C^1$ functions is the first place where pointwise first-order calculus becomes a robust theory.

The central question is this: when does a function have first derivatives that vary continuously with the point? That extra continuity turns the derivative from a list of unrelated local data into a continuous field of linear maps. It is the condition behind the chain rule in its clean form, inverse and implicit function theorems, change of variables, tangent vector fields, and the passage from local estimates to estimates on compact sets.

[example: A Differentiable Function Whose Derivative Is Not Continuous] Let $f: \mathbb R \to \mathbb R$ be given by $f(0)=0$ and $f(x)=x^2\sin(1/x)$ for $x \neq 0$. For $x \neq 0$, the product rule and one-variable chain rule give \begin{align*} f'(x)=2x\sin(1/x)+x^2\cos(1/x)(-x^{-2})=2x\sin(1/x)-\cos(1/x). \end{align*} At $0$, the difference quotient is \begin{align*} \frac{f(h)-f(0)}{h}=\frac{h^2\sin(1/h)}{h}=h\sin(1/h) \end{align*} for $h \neq 0$. Since $|\sin(1/h)|\le 1$, we have \begin{align*} |h\sin(1/h)|\le |h| \end{align*} and therefore $h\sin(1/h)\to 0$ as $h\to 0$. Hence $f'(0)=0$. The derivative is not continuous at $0$. Let $x_k=1/(2\pi k)$ and $y_k=1/((2k+1)\pi)$ for $k\ge 1$. Then $x_k\to 0$ and $y_k\to 0$, while \begin{align*} f'(x_k)=2x_k\sin(2\pi k)-\cos(2\pi k)=0-1=-1 \end{align*} and \begin{align*} f'(y_k)=2y_k\sin((2k+1)\pi)-\cos((2k+1)\pi)=0-(-1)=1. \end{align*} Thus $f'(x)$ has no limit as $x\to 0$, so $f'$ is not continuous at $0$. The function is differentiable everywhere, but it is not $C^1$ on any neighbourhood of $0$. [/example]

This example isolates the missing ingredient. Differentiability says that each point has its own tangent line. The $C^1$ condition says that those tangent lines themselves move continuously. That continuity is not cosmetic: it is exactly what lets local linear approximations be controlled uniformly over nearby points.

## Definition

The one-variable example suggests the right definition, but in analysis the useful form is already multidimensional. A map $f: U \to \mathbb R^m$ on an [open set](/page/Open%20Set) $U \subset \mathbb R^n$ should be called $C^1$ when all first partial derivatives exist and form continuous functions. The extra continuity is what lets the directional rates of change assemble into a derivative field rather than a disconnected collection of pointwise data.

[definition: $C^1$ Function] Let $U \subset \mathbb R^n$ be open, and let $f: U \to \mathbb R^m$ be a function. The function $f$ is a $C^1$ function on $U$ if each component $f_i: U \to \mathbb R$ has first partial derivatives $\partial_{x_j} f_i: U \to \mathbb R$ for every $1 \le i \le m$ and $1 \le j \le n$, and each function $\partial_{x_j} f_i$ is continuous on $U$. [/definition]

For scalar-valued functions this says that the gradient exists and is continuous. For vector-valued functions it says that every entry of the first derivative data varies continuously with the point. The next definition is needed because later estimates and convergence statements refer to the entire collection of such maps, not just to the property of one function.

[definition: $C^1(U; \mathbb R^m)$] Let $U \subset \mathbb R^n$ be open. The space $C^1(U; \mathbb R^m)$ is the set of all $C^1$ functions $f: U \to \mathbb R^m$. [/definition]

When $m=1$, it is common to write $C^1(U)$ for $C^1(U;\mathbb R)$. If the codomain matters, it should be included explicitly.

A useful way to remember the definition is that $C^1$ asks for continuity one level higher than the original function. Since continuity of first partial derivatives forces continuity of the function itself on open subsets of Euclidean space, the notation $C^1$ fits into the larger hierarchy $C^0, C^1, C^2, \ldots$.

[quotetheorem:327]

This theorem explains why the definition is stated using partial derivatives. Partial derivatives are coordinate-accessible quantities, while differentiability is the coordinate-free first-order approximation. To use the theorem in computations, we need a name for the matrix that stores those coordinate derivatives.

[definition: Jacobian Matrix] Let $U \subset \mathbb R^n$ be open, let $f \in C^1(U;\mathbb R^m)$, and let $a \in U$. The Jacobian matrix of $f$ at $a$ is the matrix $Jf_a \in \mathbb R^{m \times n}$ with entries \begin{align*} (Jf_a)_{ij}=\partial_{x_j}f_i(a). \end{align*} [/definition]

The total derivative $Df_a$ is the [linear map](/page/Linear%20Map) $h \mapsto Jf_a h$. The Jacobian matrix is the representation of that map in the standard bases; determinants and matrix ranks belong to the matrix representation.

[example: A Basic $C^1$ Map] Let $f: \mathbb R^2 \to \mathbb R^2$ be \begin{align*} f(x_1,x_2)=(f_1(x_1,x_2),f_2(x_1,x_2))=(x_1^2x_2,e^{x_1}+\sin x_2). \end{align*} We compute its first partial derivatives component by component. For the first component, \begin{align*} \partial_{x_1}f_1(x_1,x_2)=\partial_{x_1}(x_1^2x_2)=x_2\partial_{x_1}(x_1^2)=2x_1x_2. \end{align*} Also, \begin{align*} \partial_{x_2}f_1(x_1,x_2)=\partial_{x_2}(x_1^2x_2)=x_1^2\partial_{x_2}(x_2)=x_1^2. \end{align*} For the second component, \begin{align*} \partial_{x_1}f_2(x_1,x_2)=\partial_{x_1}(e^{x_1}+\sin x_2)=e^{x_1}+0=e^{x_1}. \end{align*} Finally, \begin{align*} \partial_{x_2}f_2(x_1,x_2)=\partial_{x_2}(e^{x_1}+\sin x_2)=0+\cos x_2=\cos x_2. \end{align*} The functions $2x_1x_2$, $x_1^2$, $e^{x_1}$, and $\cos x_2$ are continuous on $\mathbb R^2$, so by the definition of $C^1$ we have $f \in C^1(\mathbb R^2;\mathbb R^2)$. At $a=(1,0)$, the Jacobian entries are \begin{align*} \partial_{x_1}f_1(1,0)=2\cdot 1\cdot 0=0,\quad \partial_{x_2}f_1(1,0)=1^2=1,\quad \partial_{x_1}f_2(1,0)=e^1=e,\quad \partial_{x_2}f_2(1,0)=\cos 0=1. \end{align*} Thus \begin{align*} Jf_a=\begin{pmatrix}0&1\end{pmatrix}\text{ in the first row and }\begin{pmatrix}e&1\end{pmatrix}\text{ in the second row}. \end{align*} Equivalently, for $h=(h_1,h_2)\in \mathbb R^2$, \begin{align*} Df_a(h)=Jf_a h=(h_2,eh_1+h_2). \end{align*} This example shows how the coordinate partial derivatives assemble into the linear map that gives the derivative at a point. [/example]

This computation is typical: to verify that a concrete elementary map is $C^1$, compute first partial derivatives and check their continuity. For more delicate functions, especially ones defined piecewise, the same test must include the boundary where formulas meet.

## First-Order Approximation

The reason $C^1$ functions dominate first-order multivariable analysis is that their local linear approximations are stable. At a point $a$, differentiability gives an approximation by $f(a)+Df_a(h)$. Continuity of $Df$ says that the linear part changes little as $a$ changes little. That turns isolated tangent planes into a controlled first-order calculus.

To state this precisely, we need the map that sends a point to its derivative. This object packages all tangent maps into one function, so that the continuity demanded by the $C^1$ condition can be expressed without referring to individual matrix entries.