How to Differentiate the Cross Product of Vector Functions

If two vectors depend on a parameter, their cross product also depends on that parameter. To differentiate it, use a product rule much like the one for scalar functions, but preserve the order of the vectors in each cross product.

For differentiable functions $\mathbf{u}(t)$ and $\mathbf{v}(t)$ with values in $\mathbb{R}^3$,

$$\frac{d}{dt}\bigl(\mathbf{u}(t)\times\mathbf{v}(t)\bigr) =\mathbf{u}'(t)\times\mathbf{v}(t)+\mathbf{u}(t)\times\mathbf{v}'(t).$$

The formula has two terms because either vector can change as $t$ changes. It is not generally correct to differentiate only one of the vectors.

Why the product rule works

The cross product is bilinear: it distributes over addition and scalar multiplication in either input. For a small change $h$,

$$\mathbf{u}(t+h)\times\mathbf{v}(t+h) =\bigl(\mathbf{u}(t)+h\mathbf{u}'(t)+o(h)\bigr)\times \bigl(\mathbf{v}(t)+h\mathbf{v}'(t)+o(h)\bigr).$$

After expanding, the terms that are first order in $h$ are $h\,\mathbf{u}'(t)\times\mathbf{v}(t)$ and $h\,\mathbf{u}(t)\times\mathbf{v}'(t)$. Divide the change by $h$ and take the limit as $h$ approaches zero. The terms involving two small changes vanish in that limit, leaving the stated rule.

How to calculate it componentwise

If

$$\mathbf{u}(t)=(u_1(t),u_2(t),u_3(t)),\qquad \mathbf{v}(t)=(v_1(t),v_2(t),v_3(t)),$$

then

$$\mathbf{u}\times\mathbf{v} =\bigl(u_2v_3-u_3v_2,\;u_3v_1-u_1v_3,\;u_1v_2-u_2v_1\bigr).$$

You can differentiate these three components directly, or compute the two cross products in the product rule. Both methods give the same result. The product-rule form is usually shorter and makes clear that both vector functions must be differentiated.

Worked example

Let

$$\mathbf{u}(t)=(t,t^2,1),\qquad \mathbf{v}(t)=(1,t,t^3).$$

First differentiate each vector componentwise:

$$\mathbf{u}'(t)=(1,2t,0),\qquad \mathbf{v}'(t)=(0,1,3t^2).$$

Now calculate the two terms separately:

$$\begin{aligned} \mathbf{u}'\times\mathbf{v} &=(2t^4,-t^3,-t),\\ \mathbf{u}\times\mathbf{v}' &=(3t^4-1,-3t^3,t). \end{aligned}$$

Adding them gives

$$\frac{d}{dt}(\mathbf{u}\times\mathbf{v}) =(5t^4-1,-4t^3,0).$$

As a check, compute the original cross product first:

$$\mathbf{u}\times\mathbf{v}=(t^5-t,1-t^4,0).$$

Differentiating its components gives $(5t^4-1,-4t^3,0)$, confirming the result.

Common mistakes

• Leaving out a term: The derivative is not just $\mathbf{u}'\times\mathbf{v}$ or just $\mathbf{u}\times\mathbf{v}'$. Include both.
• Reversing the order: In general, $\mathbf{a}\times\mathbf{b}=-\mathbf{b}\times\mathbf{a}$. Thus the terms are $\mathbf{u}'\times\mathbf{v}$ and $\mathbf{u}\times\mathbf{v}'$, in that order.
• Treating the cross product as a scalar product: The result is a vector. Differentiate or add its components as vectors, not as a single scalar expression.
• Forgetting the parameter dependence: Even if one vector looks constant in one part of a calculation, check whether its components actually depend on $t$ before setting its derivative to zero.

If the vectors depend on several variables

If $\mathbf{u}$ and $\mathbf{v}$ depend on a parameterized path $\mathbf{r}(s)$, apply the same rule to their compositions. For example,

$$\frac{d}{ds}\bigl(\mathbf{u}(\mathbf{r}(s))\times\mathbf{v}(\mathbf{r}(s))\bigr) =\frac{d\mathbf{u}}{ds}\times\mathbf{v} +\mathbf{u}\times\frac{d\mathbf{v}}{ds},$$

where each derivative on the right includes the chain rule. If the differentiation is with respect to one coordinate, such as $x$, the same product rule applies using partial derivatives: $\partial_x(\mathbf{u}\times\mathbf{v})=(\partial_x\mathbf{u})\times\mathbf{v}+\mathbf{u}\times(\partial_x\mathbf{v})$.