Differentiating Dot Products of Vector Functions

Introduction

A dot product turns two vectors into a scalar. When both vectors depend on a parameter, their dot product can change because either vector changes. Its derivative must account for both sources of change.

The rule resembles the ordinary product rule for scalar functions, but it is worth seeing precisely how it works for vectors. The component definition of the dot product makes the result both natural and straightforward to verify.

Definition and Differentiation Rule

Let $\mathbf{u}(t)$ and $\mathbf{v}(t)$ be vector-valued functions with values in $\mathbb{R}^n$, where $n$ is a positive integer. Write their components as

\mathbf{u}(t)=\langle u_1(t),u_2(t),\dots,u_n(t)\rangle, \qquad \mathbf{v}(t)=\langle v_1(t),v_2(t),\dots,v_n(t)\rangle.

The dot product is the scalar-valued function defined by multiplying corresponding components and adding the results:

\mathbf{u}(t)\cdot\mathbf{v}(t) = \sum_{i=1}^{n}u_i(t)v_i(t).

Theorem: If $\mathbf{u}$ and $\mathbf{v}$ are differentiable at $t$, then their dot product is differentiable at $t$, and $$ \frac{d}{dt}\bigl(\mathbf{u}(t)\cdot\mathbf{v}(t)\bigr) = \mathbf{u}'(t)\cdot\mathbf{v}(t) + \mathbf{u}(t)\cdot\mathbf{v}'(t). $$

The conditions matter: both vector functions must be differentiable at the parameter value in question. Differentiability of a vector-valued function means that each of its component functions is differentiable there. The theorem applies in any finite dimension, provided both vectors have the same number of components so their dot product is defined.

The two terms represent the two ways the dot product can change: the first differentiates the components of $\mathbf{u}$, while the second differentiates the components of $\mathbf{v}$. As with the scalar product rule, neither contribution can generally be left out.

Why the Rule Works

The component definition reduces the claim to the ordinary product rule. Here is a proof directly from the difference quotient. Fix a parameter $t$ at which both functions are differentiable, and take $h\ne 0$ small enough that the functions are defined at $t+h$. Add and subtract $\mathbf{u}(t)\cdot\mathbf{v}(t+h)$ in the numerator:

\begin{aligned} &\frac{\mathbf{u}(t+h)\cdot\mathbf{v}(t+h) -\mathbf{u}(t)\cdot\mathbf{v}(t)}{h}\\ &= \frac{\bigl(\mathbf{u}(t+h)-\mathbf{u}(t)\bigr) \cdot\mathbf{v}(t+h)}{h} + \mathbf{u}(t)\cdot \frac{\mathbf{v}(t+h)-\mathbf{v}(t)}{h}\\ &= \frac{\mathbf{u}(t+h)-\mathbf{u}(t)}{h} \cdot\mathbf{v}(t+h) + \mathbf{u}(t)\cdot \frac{\mathbf{v}(t+h)-\mathbf{v}(t)}{h}. \end{aligned}

As $h\to0$, the difference quotient for $\mathbf{u}$ tends componentwise to $\mathbf{u}'(t)$, and the difference quotient for $\mathbf{v}$ tends componentwise to $\mathbf{v}'(t)$. Also, differentiability of $\mathbf{v}$ implies continuity at $t$, so $\mathbf{v}(t+h)\to\mathbf{v}(t)$. The dot product is a finite sum of products of components, so these limits give

\lim_{h\to0} \frac{\mathbf{u}(t+h)\cdot\mathbf{v}(t+h) -\mathbf{u}(t)\cdot\mathbf{v}(t)}{h} = \mathbf{u}'(t)\cdot\mathbf{v}(t) + \mathbf{u}(t)\cdot\mathbf{v}'(t).

This is the derivative of the dot product, proving the rule. Equivalently, the definition of the dot product expresses it as the finite sum $\sum_{i=1}^n u_i(t)v_i(t)$; applying the scalar product rule to each summand gives the same result.

Using the Rule in Components

In practice, there are two reliable approaches. You can first calculate $\mathbf{u}\cdot\mathbf{v}$ as a scalar function and differentiate it, or differentiate each vector component and then calculate the two dot products in the theorem. The second approach makes the separate contributions visible; the first provides a useful check.

Differentiate each vector.
Find $\mathbf{u}'(t)$ and $\mathbf{v}'(t)$ by differentiating their components.

Form both dot products.
Compute $\mathbf{u}'(t)\cdot\mathbf{v}(t)$ and $\mathbf{u}(t)\cdot\mathbf{v}'(t)$ component by component.

Add and check.
Add the scalar results. When practical, calculate $\mathbf{u}(t)\cdot\mathbf{v}(t)$ first and differentiate it to verify the answer.

Worked Examples

Worked Example: Polynomial Vector Functions

Let $\mathbf{u}(t)=\langle t^2,2t-1,3\rangle$ and $\mathbf{v}(t)=\langle t+1,t^2,2t\rangle$. Find the derivative of their dot product, and check the result.

First differentiate each vector component:

\mathbf{u}'(t)=\langle 2t,2,0\rangle, \qquad \mathbf{v}'(t)=\langle 1,2t,2\rangle.

Now compute the first dot product:

\begin{aligned} \mathbf{u}'(t)\cdot\mathbf{v}(t) &=(2t)(t+1)+(2)(t^2)+(0)(2t)\\ &=2t^2+2t+2t^2\\ &=4t^2+2t. \end{aligned}

Compute the second dot product:

\begin{aligned} \mathbf{u}(t)\cdot\mathbf{v}'(t) &=(t^2)(1)+(2t-1)(2t)+(3)(2)\\ &=t^2+4t^2-2t+6\\ &=5t^2-2t+6. \end{aligned}

Adding the two contributions gives

\frac{d}{dt}\bigl(\mathbf{u}(t)\cdot\mathbf{v}(t)\bigr) =(4t^2+2t)+(5t^2-2t+6) =9t^2+6.

For a check, calculate the dot product before differentiating:

\begin{aligned} \mathbf{u}(t)\cdot\mathbf{v}(t) &=t^2(t+1)+(2t-1)t^2+3(2t)\\ &=t^3+t^2+2t^3-t^2+6t\\ &=3t^3+6t. \end{aligned}

Its derivative is $9t^2+6$, as obtained from the rule. At $t=1$, the two terms from the rule are $4(1)^2+2(1)=6$ and $5(1)^2-2(1)+6=9$, for a total of $15$. The checked derivative gives $9(1)^2+6=15$.

Worked Example: Trigonometric Components

Let $\mathbf{u}(t)=\langle\cos t,\sin t,0\rangle$ and $\mathbf{v}(t)=\langle\sin t,\cos t,t\rangle$. Find the derivative of their dot product.

Differentiate the components:

\mathbf{u}'(t)=\langle-\sin t,\cos t,0\rangle, \qquad \mathbf{v}'(t)=\langle\cos t,-\sin t,1\rangle.

The first dot product is

\begin{aligned} \mathbf{u}'(t)\cdot\mathbf{v}(t) &=(-\sin t)(\sin t)+(\cos t)(\cos t)+0(t)\\ &=-\sin^2 t+\cos^2 t\\ &=\cos(2t). \end{aligned}

The second dot product is

\begin{aligned} \mathbf{u}(t)\cdot\mathbf{v}'(t) &=(\cos t)(\cos t)+(\sin t)(-\sin t)+0(1)\\ &=\cos^2 t-\sin^2 t\\ &=\cos(2t). \end{aligned}

Therefore the derivative is

\frac{d}{dt}\bigl(\mathbf{u}(t)\cdot\mathbf{v}(t)\bigr) =\cos(2t)+\cos(2t)=2\cos(2t).

To check, the original dot product is $\cos t\sin t+\sin t\cos t+0(t)=2\sin t\cos t=\sin(2t)$. Differentiating gives $2\cos(2t)$. At $t=\frac{\pi}{6}$, this equals $2\cos(\frac{\pi}{3})=2\cdot\frac12=1$, confirming the value at that parameter.

Worked Example: A Derivative That Vanishes at One Parameter

Let $\mathbf{u}(t)=\langle t,1-t,2\rangle$ and $\mathbf{v}(t)=\langle 2t-1,t+2,-t\rangle$. Compute the derivative of their dot product and evaluate it at $t=2$.

The component derivatives are

\mathbf{u}'(t)=\langle 1,-1,0\rangle, \qquad \mathbf{v}'(t)=\langle 2,1,-1\rangle.

Using the first term in the rule gives

\begin{aligned} \mathbf{u}'(t)\cdot\mathbf{v}(t) &=(1)(2t-1)+(-1)(t+2)+(0)(-t)\\ &=2t-1-t-2\\ &=t-3. \end{aligned}

The second term is

\begin{aligned} \mathbf{u}(t)\cdot\mathbf{v}'(t) &=(t)(2)+(1-t)(1)+(2)(-1)\\ &=2t+1-t-2\\ &=t-1. \end{aligned}

Thus

\frac{d}{dt}\bigl(\mathbf{u}(t)\cdot\mathbf{v}(t)\bigr) =(t-3)+(t-1)=2t-4.

At $t=2$, the first term is $2-3=-1$, and the second is $2-1=1$. Their sum is $0$. To check, expand the original dot product:

\begin{aligned} \mathbf{u}(t)\cdot\mathbf{v}(t) &=t(2t-1)+(1-t)(t+2)+2(-t)\\ &=2t^2-t+t+2-t^2-2t-2t\\ &=t^2-4t+2. \end{aligned}

Differentiating gives $2t-4$, which is $0$ at $t=2$, matching the result from the dot product rule.

Common Mistake: Omitting One Term

A frequent error is to differentiate only the first vector and write $\frac{d}{dt}(\mathbf{u}\cdot\mathbf{v})=\mathbf{u}'\cdot\mathbf{v}$. This misses the change in $\mathbf{v}$. For a concrete counterexample, take $\mathbf{u}(t)=\langle t,1\rangle$ and $\mathbf{v}(t)=\langle 1,t\rangle$.

\mathbf{u}(t)\cdot\mathbf{v}(t)=t+t=2t, \qquad \frac{d}{dt}\bigl(\mathbf{u}(t)\cdot\mathbf{v}(t)\bigr)=2.

The incorrect one-term calculation gives $\mathbf{u}'(t)=\langle1,0\rangle$, so $\mathbf{u}'(t)\cdot\mathbf{v}(t)=1$, not $2$. The missing contribution is $\mathbf{u}(t)\cdot\mathbf{v}'(t)=\langle t,1\rangle\cdot\langle0,1\rangle=1$. The correct calculation is $1+1=2$, in agreement with direct differentiation.

Takeaway: When both factors depend on the parameter, include one term for the derivative of each factor: $$ (\mathbf{u}\cdot\mathbf{v})'=\mathbf{u}'\cdot\mathbf{v}+\mathbf{u}\cdot\mathbf{v}'. $$

Why the Rule Is Useful

The dot product is a scalar measure of how two vectors align, so its derivative describes how that scalar relationship changes as the parameter varies. In applications, the vectors may represent changing directions, coordinates, or other quantities. The rule separates the rate of change into the contribution from each vector, which can make the source of a changing dot product easier to identify.

For computation, a practical safeguard is to check the result in two ways when the expressions are manageable: apply the two-term rule, then form the dot product first and differentiate the resulting scalar function. Agreement between the methods is a useful check on component arithmetic and signs.

Check Your Understanding

Use the definition of the dot product and its derivative rule to answer the following questions.

State the condition on $\mathbf{u}$ and $\mathbf{v}$ needed to apply the dot product derivative rule at a parameter value.
If $\mathbf{u}(t)=\langle t^2,1\rangle$ and $\mathbf{v}(t)=\langle t,3t\rangle$, what two dot products appear in the derivative rule?
For $\mathbf{u}(t)=\langle t,1\rangle$ and $\mathbf{v}(t)=\langle 1,t\rangle$, explain why differentiating only $\mathbf{u}$ gives an incomplete answer.
Why does the component definition of the dot product reduce the proof to scalar product rules?
Describe one direct check you can use after computing a dot product derivative.