The Product Rule for Curl

Why Multiplying a Vector Field Changes Its Curl

A scalar factor can change a vector field in two ways. It can amplify or reduce the field at each point, and, when that factor varies in space, it can change how the field rotates from one point to another. The curl of a scalar-weighted field therefore contains both the original curl and a contribution from the variation of the scalar.

The formula for this change is a product rule. Its order matters: the extra term is the gradient of the scalar crossed with the vector field, not the other way around. We will derive the rule from the Cartesian definition of curl and check its use in examples.

Definition and Conditions

Let $U$ be an open subset of $\mathbb{R}^3$, let $f:U\to\mathbb{R}$ be a scalar field, and let $\mathbf{F}:U\to\mathbb{R}^3$ be a vector field. Write the components as

\mathbf{F}(x,y,z)=\langle P(x,y,z),Q(x,y,z),R(x,y,z)\rangle.

If $P,Q,R$ have continuous first partial derivatives, the curl is defined by

\nabla\times\mathbf{F} = \left\langle \frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}, \frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}, \frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y} \right\rangle.

The gradient of a scalar field with continuous first partial derivatives is

\nabla f = \left\langle \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right\rangle.

The product $f\mathbf{F}$ means that the scalar multiplies each component: $f\mathbf{F}=\langle fP,fQ,fR\rangle$. If $f$ and the components of $\mathbf{F}$ have continuous first partial derivatives on $U$, then $f\mathbf{F}$ also has continuous first partial derivatives, so its curl is defined there.

Theorem: For scalar and vector fields with continuous first partial derivatives on an open set $U\subseteq\mathbb{R}^3$, $$ \nabla\times(f\mathbf{F})=(\nabla f)\times\mathbf{F}+f(\nabla\times\mathbf{F}). $$

Here $\times$ is the three-dimensional cross product. The two terms have distinct roles: $f(\nabla\times\mathbf{F})$ scales the original curl, while $(\nabla f)\times\mathbf{F}$ accounts for variation in the scalar factor. If $f$ is constant, then $\nabla f=\mathbf{0}$, and the rule reduces to $\nabla\times(f\mathbf{F})=f(\nabla\times\mathbf{F})$.

Deriving the Rule from Components

The derivation uses the ordinary product rule for partial derivatives. For example, the first component of $\nabla\times(f\mathbf{F})$ is the $y$-partial derivative of $fR$ minus the $z$-partial derivative of $fQ$. Applying the product rule to those two derivatives gives a part involving derivatives of $f$ and a part involving derivatives of $\mathbf{F}$.

Expand the first curl component.
$$ \frac{\partial(fR)}{\partial y}-\frac{\partial(fQ)}{\partial z} = \frac{\partial f}{\partial y}R+f\frac{\partial R}{\partial y} -\frac{\partial f}{\partial z}Q-f\frac{\partial Q}{\partial z}. $$

Group the two types of terms.
$$ \frac{\partial(fR)}{\partial y}-\frac{\partial(fQ)}{\partial z} = \left(\frac{\partial f}{\partial y}R-\frac{\partial f}{\partial z}Q\right) + f\left(\frac{\partial R}{\partial y}-\frac{\partial Q}{\partial z}\right). $$

Identify the corresponding components.
The first parenthesis is the first component of $(\nabla f)\times\mathbf{F}$; the second is $f$ times the first component of $\nabla\times\mathbf{F}$. Expanding the second and third curl components in the same way gives the other two components of the identity.

For completeness, the three components of the cross product are

(\nabla f)\times\mathbf{F} = \left\langle \frac{\partial f}{\partial y}R-\frac{\partial f}{\partial z}Q,\; \frac{\partial f}{\partial z}P-\frac{\partial f}{\partial x}R,\; \frac{\partial f}{\partial x}Q-\frac{\partial f}{\partial y}P \right\rangle.

For the second component, the product rule gives $\frac{\partial(fP)}{\partial z}-\frac{\partial(fR)}{\partial x}=\left(\frac{\partial f}{\partial z}P-\frac{\partial f}{\partial x}R\right)+f\left(\frac{\partial P}{\partial z}-\frac{\partial R}{\partial x}\right)$. For the third, it gives $\frac{\partial(fQ)}{\partial x}-\frac{\partial(fP)}{\partial y}=\left(\frac{\partial f}{\partial x}Q-\frac{\partial f}{\partial y}P\right)+f\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)$. These are exactly the remaining components of the stated identity, proving the rule.

Worked Examples

Worked Example: A Radial Factor Multiplying a Constant Field

Let $r=\sqrt{x^2+y^2+z^2}$, and consider $f=r$ and $\mathbf{F}=\langle 0,0,1\rangle$. The radial function $r$ is differentiable away from the origin, so work on $U=\mathbb{R}^3\setminus\{\mathbf{0}\}$.

First find the gradient and the curl of $\mathbf{F}$:

\nabla r = \left\langle\frac{x}{r},\frac{y}{r},\frac{z}{r}\right\rangle, \qquad \nabla\times\mathbf{F}=\langle 0,0,0\rangle.

Apply the product rule. Since $\mathbf{F}=\langle 0,0,1\rangle$,

\nabla\times(r\mathbf{F}) = \left\langle\frac{x}{r},\frac{y}{r},\frac{z}{r}\right\rangle \times\langle 0,0,1\rangle +r\langle 0,0,0\rangle = \left\langle\frac{y}{r},-\frac{x}{r},0\right\rangle.

Check by computing directly. The field $r\mathbf{F}$ is $\langle 0,0,r\rangle$, so

\nabla\times\langle 0,0,r\rangle = \left\langle \frac{\partial r}{\partial y}, -\frac{\partial r}{\partial x}, 0 \right\rangle = \left\langle\frac{y}{r},-\frac{x}{r},0\right\rangle.

The direct computation agrees. The formula and the derivatives are valid only away from the origin, where $r$ is differentiable.

Worked Example: A Linear Factor and a Nonconstant Field

Let $f=2x-y+3z$ and $\mathbf{F}=\langle y,z,x\rangle$. The scalar and vector components are differentiable throughout $\mathbb{R}^3$. Compute each part of the product rule:

\nabla f=\langle 2,-1,3\rangle, \qquad \nabla\times\mathbf{F} = \langle 0-1,\;0-1,\;0-1\rangle = \langle-1,-1,-1\rangle.

The cross-product term is

(\nabla f)\times\mathbf{F} = \langle 2,-1,3\rangle\times\langle y,z,x\rangle = \langle -x-3z,\;3y-2x,\;2z+y\rangle.

Therefore, with $f=2x-y+3z$,

\nabla\times(f\mathbf{F}) = \langle -x-3z-f,\;3y-2x-f,\;2z+y-f\rangle.

Check by direct differentiation. The product field is $\langle fy,fz,fx\rangle$, so its curl is

\begin{aligned} \nabla\times\langle fy,fz,fx\rangle &= \left\langle \frac{\partial(fx)}{\partial y}-\frac{\partial(fz)}{\partial z}, \frac{\partial(fy)}{\partial z}-\frac{\partial(fx)}{\partial x}, \frac{\partial(fz)}{\partial x}-\frac{\partial(fy)}{\partial y} \right\rangle\\ &= \langle -x-3z-f,\;3y-2x-f,\;2z+y-f\rangle, \end{aligned}

using $f_x=2,\;f_y=-1,\;f_z=3$. For a numerical check, at $(1,2,-1)$, $f=-3$, and both expressions give $\langle 5,7,3\rangle$.

Worked Example: A Linear Factor with a Curl-Free Field

Let $f=x$ and $\mathbf{F}=\langle 0,z,y\rangle$. First,

\nabla f=\langle 1,0,0\rangle, \qquad \nabla\times\mathbf{F} = \langle 1-1,\;0-0,\;0-0\rangle = \langle 0,0,0\rangle.

Thus the rule gives

\nabla\times(x\mathbf{F}) = \langle 1,0,0\rangle\times\langle 0,z,y\rangle +x\langle 0,0,0\rangle = \langle 0,-y,z\rangle.

To check, $x\mathbf{F}=\langle 0,xz,xy\rangle$, and direct differentiation yields

\nabla\times\langle 0,xz,xy\rangle = \left\langle \frac{\partial(xy)}{\partial y}-\frac{\partial(xz)}{\partial z}, -\frac{\partial(xy)}{\partial x}, \frac{\partial(xz)}{\partial x} \right\rangle = \langle x-x,-y,z\rangle = \langle 0,-y,z\rangle.

Although $\mathbf{F}$ has zero curl, multiplying it by a nonconstant scalar produces a field whose curl need not be zero.

A Sign Pitfall: Reversing the Cross Product

The cross product is not commutative: $\mathbf{a}\times\mathbf{b}=-(\mathbf{b}\times\mathbf{a})$. Reversing its order therefore changes the sign of the extra term. In the radial example above, incorrectly using $\mathbf{F}\times\nabla r$ gives

\langle 0,0,1\rangle\times \left\langle\frac{x}{r},\frac{y}{r},\frac{z}{r}\right\rangle = \left\langle-\frac{y}{r},\frac{x}{r},0\right\rangle.

But direct differentiation of $r\mathbf{F}=\langle 0,0,r\rangle$ gives $\langle y/r,-x/r,0\rangle$, the opposite sign. The correct order is $(\nabla f)\times\mathbf{F}$, as also follows from the component derivation. A useful check is to compare at a point where both cross-product components are nonzero; for example, at $(1,2,0)$ the correct result is $\langle 2/\sqrt{5},-1/\sqrt{5},0\rangle$, not its negative.

Interpreting the Two Contributions

The product rule separates the curl into a contribution from the vector field and a contribution from the scalar’s spatial variation. If $f$ is constant, only the scaled original curl remains. If $\mathbf{F}$ has zero curl, the result is $(\nabla f)\times\mathbf{F}$, which may still be nonzero, as the third example demonstrates. If $\nabla f$ is parallel to $\mathbf{F}$ at a point, their cross product vanishes at that point, so only the scaled original curl contributes there.

This separation is useful whenever a field is naturally expressed as a scalar weight times a vector field. It lets you compute the curl from simpler pieces, and it makes clear that ignoring the derivative of the scalar is valid only when that scalar is constant (or when the cross-product term happens to vanish). The differentiability conditions matter too: the identity applies where the required first partial derivatives exist and are continuous.

Check Your Understanding

Use the definition of curl and the product rule to answer the following.

State the differentiability condition used for the curl product rule.
For $f=x+y$ and $\mathbf{F}=\langle 0,0,1\rangle$, find $(\nabla f)\times\mathbf{F}$.
Why does a vector field with zero curl not necessarily remain curl-free after multiplication by a nonconstant scalar?
What sign change results from replacing $(\nabla f)\times\mathbf{F}$ with $\mathbf{F}\times(\nabla f)$?
For $f$ constant, state the product rule in its simplified form.

What You'll Learn