Partial Derivative¶

Definition¶

Let $E \subseteq \mathbb{R}^n$ and consider a function $\boldsymbol{f} \colon E \to \mathbb{R}^k$${\mathbb{R}}^k$. For an interior point $\boldsymbol{x}_0$ of $E$ and an integer $1 \leq j \leq n$$n$, the partial derivative of $\boldsymbol{f}$ with respect to the $x_j$-variable at $\boldsymbol{x} = \boldsymbol{x}_0$${\mathbf{x}}_0$, denoted by $\displaystyle \frac{\partial \boldsymbol{f}}{\partial x_j}(\boldsymbol{x}_0)$, is given by

where $\boldsymbol{e}_j$ represents the unit vector in the positive direction of the $j$-th coordinate, provided the limit exists.

The partial derivative changes only the $j$th coordinate, so it is the Directional Derivative along the coordinate vector $\boldsymbol{e}_j$ when the corresponding two-sided limit exists.

Interpretation¶

The partial derivative measures how the function changes when you nudge only one coordinate while keeping all others fixed. It is an ordinary derivative of the one-variable function $t\mapsto\boldsymbol{f}(\boldsymbol{x}_0+t\boldsymbol{e}_j)$ at $t=0$$0$, as the second line of the Definition shows.

The vertical plane $x_2=b$ cuts the graph of $f$ in the red curve $x_1\mapsto f(x_1,b)$$f(x_1,b)$, which is the graph of a function of $x_1$ alone. Its tangent line at $(a,b)$ has slope $\frac{\partial f}{\partial x_1}(a,b)$. Cutting with the plane $x_1=a$ instead gives a curve in the $x_2$-direction, and its slope at $(a,b)$ is $\frac{\partial f}{\partial x_2}(a,b)$.

In practice, one treats every other variable as a constant and differentiates with the one-variable rules, as in the Example. For a vector-valued $\boldsymbol{f}$, the partial derivative is a vector in $\mathbb{R}^k$, computed component by component. These vectors are the columns of the Jacobian matrix, and for a real-valued $f$ the partial derivatives form the gradient.

A common slip is to expect partial derivatives to describe the function in every direction. They only look along the coordinate axes. The function with $f(x,y)=xy/(x^2+y^2)$ for $(x,y)\neq(0,0)$ and $f(0,0)=0$ vanishes on both axes, so both partial derivatives at the origin are $0$. Yet it is not even continuous there, since $f(t,t)=\tfrac{1}{2}$ for every $t\neq0$$0$. Existence of partial derivatives is therefore weaker than total differentiability; the Key Results give a sufficient condition, continuity of the partial derivatives, that closes the gap.

Key Results¶

• The partial derivative $\displaystyle \frac{\partial \boldsymbol{f}}{\partial x_j}(\boldsymbol{x}_0)$ exists if and only if both $\mathrm{D}_{\boldsymbol{e}_j}\boldsymbol{f}(\boldsymbol{x}_0)$ and $\mathrm{D}_{-\boldsymbol{e}_j}\boldsymbol{f}(\boldsymbol{x}_0)$ exist and are negatives of each other. In such a scenario, we have $\displaystyle \frac{\partial \boldsymbol{f}}{\partial x_j}(\boldsymbol{x}_0)=\mathrm{D}_{\boldsymbol{e}_j}\boldsymbol{f}(\boldsymbol{x}_0).$
• If $\boldsymbol{f}$ is totally differentiable at $\boldsymbol{x}_0\text{,}$ then for every unit vector $\displaystyle \boldsymbol{v} = (v_1,\,\ldots,\,v_n)= \sum_{j=1}^n v_j\boldsymbol{e}_j$ we have $\displaystyle \mathrm{D}_{\boldsymbol{v}}\boldsymbol{f}(\boldsymbol{x}_0) = \sum_{j=1}^n v_j \frac{\partial }{\partial x_j}\boldsymbol{f}(\boldsymbol{x}_0).$ Without total differentiability this can fail even when all partial derivatives exist.
• If all the partial derivatives are both existent within a neighbourhood and continuous at the centre, then $\boldsymbol{f}$ is totally differentiable at this point.

Examples¶

• The partial derivative of $f(x,y) = 3x^2y^2 + 4xy^3 + 7y$ with respect to the first variable is $\dfrac{\partial }{\partial x}f(x,y) = 6xy^2 + 4y^3.$

Prerequisites¶

• Directional Derivative

Related Concepts¶

• Total Derivative as a Linear Map

Sources¶

• Math III: Analysis · Lecture 12

Aliases: partial differentiation Symbols: partial, real numbers, subset, sum

Last updated: 22 October 2026 19:04