Differentiating Scalar-Vector Products

Introduction

A scalar function can change the size and direction of a vector-valued function by multiplying each of its components by the same scalar. When both functions vary with a parameter, the derivative must account for both changes. The scalar-vector product rule gives a compact way to do this.

The rule resembles the product rule for two scalar functions, but its result is a vector. It applies to a scalar function $f$ and a vector-valued function $\mathbf{r}$; it is not a dot-product or cross-product rule.

Key idea: The derivative of a scalar times a vector has two contributions: one from the changing scalar and one from the changing vector.

Definition and Product Rule

Let $I$ be an open interval, let $f:I\to\mathbb{R}$ be a scalar-valued function, and let $\mathbf{r}:I\to\mathbb{R}^n$ be a vector-valued function. Their scalar-vector product is the function $f\mathbf{r}:I\to\mathbb{R}^n$ defined at each parameter $t$ by multiplying every component of $\mathbf{r}(t)$ by $f(t)$. If $\mathbf{r}(t)=\langle r_1(t),\ldots,r_n(t)\rangle$, then $f(t)\mathbf{r}(t)=\langle f(t)r_1(t),\ldots,f(t)r_n(t)\rangle$.

Theorem: If $f$ and $\mathbf{r}$ are differentiable at $t$, then their scalar-vector product is differentiable at $t$, and $$ \frac{d}{dt}\bigl(f(t)\mathbf{r}(t)\bigr) = f'(t)\mathbf{r}(t)+f(t)\mathbf{r}'(t). $$

Both terms on the right are vectors. In the first, the scalar $f'(t)$ multiplies the vector $\mathbf{r}(t)$. In the second, $f(t)$ multiplies the vector derivative $\mathbf{r}'(t)$. The addition is vector addition.

Why the Rule Works

The proof follows by adding and subtracting a product that separates the change in the scalar from the change in the vector. Write the difference quotient for $f\mathbf{r}$, and in its numerator add and subtract $f(t)\mathbf{r}(t+h)$. For nonzero $h$ sufficiently close to zero, this gives:

\begin{aligned} \frac{f(t+h)\mathbf{r}(t+h)-f(t)\mathbf{r}(t)}{h} &= \frac{f(t+h)\mathbf{r}(t+h)-f(t)\mathbf{r}(t+h)}{h}\\ &\quad+ \frac{f(t)\mathbf{r}(t+h)-f(t)\mathbf{r}(t)}{h}\\ &= \frac{f(t+h)-f(t)}{h}\mathbf{r}(t+h) + f(t)\frac{\mathbf{r}(t+h)-\mathbf{r}(t)}{h}. \end{aligned}

As $h\to0$, differentiability of $\mathbf{r}$ at $t$ implies continuity of $\mathbf{r}$ there, so $\mathbf{r}(t+h)\to\mathbf{r}(t)$. The scalar difference quotient tends to $f'(t)$, and the vector difference quotient tends to $\mathbf{r}'(t)$. Taking the limit of the two terms therefore gives:

f'(t)\mathbf{r}(t)+f(t)\mathbf{r}'(t).

This proves the rule and explains the two contributions: the first term captures the scalar's change while the vector is at its current value, and the second captures the vector's change scaled by the current scalar.

Applying the Rule Componentwise

Because scalar multiplication and addition of vectors act component by component, the theorem agrees with the ordinary scalar product rule in every coordinate. If $\mathbf{r}(t)=\langle r_1(t),\ldots,r_n(t)\rangle$, then:

\frac{d}{dt}(f\mathbf{r}) = \left\langle f'r_1+fr_1',\ldots,f'r_n+fr_n' \right\rangle.

This component form can be useful for checking a calculation. It is usually more efficient, however, to first write the vector rule and then evaluate its two vector terms.

Worked Example: A Polynomial Scalar Factor

Differentiate $f(t)\mathbf{r}(t)$ when $f(t)=t^2+1$ and $\mathbf{r}(t)=\langle t,2t-1,t^2\rangle$.

First calculate the scalar and vector derivatives:

f'(t)=2t, \qquad \mathbf{r}'(t)=\langle 1,2,2t\rangle.

Apply the product rule. The first term is:

f'(t)\mathbf{r}(t) = 2t\langle t,2t-1,t^2\rangle = \langle 2t^2,4t^2-2t,2t^3\rangle.

The second term is:

f(t)\mathbf{r}'(t) = (t^2+1)\langle 1,2,2t\rangle = \langle t^2+1,2t^2+2,2t^3+2t\rangle.

Add the corresponding components:

\frac{d}{dt}(f\mathbf{r}) = \langle 3t^2+1,6t^2-2t+2,4t^3+2t\rangle.

Check by expanding the original product componentwise:

f(t)\mathbf{r}(t) = \langle t^3+t,\;2t^3-t^2+2t-1,\;t^4+t^2\rangle.

Differentiate each component: the derivatives are $3t^2+1$, $6t^2-2t+2$, and $4t^3+2t$, respectively. They match the product-rule result.

Trigonometric Scalar Factors

The scalar factor need not be a polynomial. Trigonometric functions follow the same rule: differentiate the scalar factor, differentiate the vector function, multiply in each term, and add the resulting vectors. The rule requires differentiability of both functions at the parameter being considered.

Worked Example: A Sine Factor

Let $f(t)=\sin t$ and $\mathbf{r}(t)=\langle\cos t,t,1\rangle$. Find the derivative, then evaluate it at $t=\frac{\pi}{2}$.

Differentiate each factor:

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

Use the product rule and simplify each component:

\begin{aligned} \frac{d}{dt}(f\mathbf{r}) &= \cos t\langle\cos t,t,1\rangle + \sin t\langle-\sin t,1,0\rangle\\ &= \langle\cos^2 t,t\cos t,\cos t\rangle + \langle-\sin^2 t,\sin t,0\rangle\\ &= \langle\cos^2 t-\sin^2 t,\;t\cos t+\sin t,\;\cos t\rangle. \end{aligned}

At $t=\frac{\pi}{2}$, $\cos(\frac{\pi}{2})=0$ and $\sin(\frac{\pi}{2})=1$. Thus:

\left.\frac{d}{dt}(f\mathbf{r})\right|_{t=\pi/2} = \langle 0-1,\;0+1,\;0\rangle = \langle-1,1,0\rangle.

For a check, the original product is $\langle\sin t\cos t,t\sin t,\sin t\rangle$. Differentiating these three components gives $\cos^2t-\sin^2t$, $\sin t+t\cos t$, and $\cos t$. At $t=\frac{\pi}{2}$, these are $-1,1,0$, as obtained above.

A Mixed Example at a Specific Parameter

When a problem asks for the derivative at one parameter value, it is still important to include both terms before evaluating. This helps keep track of which values belong to the functions and which belong to their derivatives.

Worked Example: A Linear Factor and a Trigonometric Vector

Let $f(t)=3t-2$ and $\mathbf{r}(t)=\langle\cos t,\sin t,t\rangle$. Find the derivative at $t=0$.

The derivatives are:

f'(t)=3, \qquad \mathbf{r}'(t)=\langle-\sin t,\cos t,1\rangle.

At $t=0$, the function and derivative values are:

f(0)=-2, \qquad f'(0)=3, \qquad \mathbf{r}(0)=\langle1,0,0\rangle, \qquad \mathbf{r}'(0)=\langle0,1,1\rangle.

Substitute into both terms of the rule:

\begin{aligned} \left.\frac{d}{dt}(f\mathbf{r})\right|_{t=0} &=f'(0)\mathbf{r}(0)+f(0)\mathbf{r}'(0)\\ &=3\langle1,0,0\rangle-2\langle0,1,1\rangle\\ &=\langle3,0,0\rangle+\langle0,-2,-2\rangle\\ &=\langle3,-2,-2\rangle. \end{aligned}

Check directly. The product is $\langle(3t-2)\cos t,(3t-2)\sin t,3t^2-2t\rangle$. The derivatives of these components at zero are $3$, $-2$, and $-2$, respectively. Therefore the direct calculation confirms $\langle3,-2,-2\rangle$.

Common Mistake: Omitting One Term

A frequent error is to differentiate only the scalar factor and multiply by the unchanged vector. That accounts for the scalar's change but ignores the vector's change. For example, take $f(t)=t^2$ and $\mathbf{r}(t)=\langle t,1\rangle$. Keeping only $f'(t)\mathbf{r}(t)$ gives the incorrect result:

f'(t)\mathbf{r}(t) = 2t\langle t,1\rangle = \langle2t^2,2t\rangle.

The missing term is $f(t)\mathbf{r}'(t)$. Since $\mathbf{r}'(t)=\langle1,0\rangle$, the correct calculation is:

\begin{aligned} \frac{d}{dt}(f\mathbf{r}) &= \langle2t^2,2t\rangle+t^2\langle1,0\rangle\\ &= \langle3t^2,2t\rangle. \end{aligned}

At $t=1$, the incorrect answer is $\langle2,2\rangle$, whereas the correct answer is $\langle3,2\rangle$. Indeed, the original product is $\langle t^3,t^2\rangle$, whose derivative is $\langle3t^2,2t\rangle$. The correction is to include both terms every time: $f'\mathbf{r}$ and $f\mathbf{r}'$.

When to Use the Rule

Use this rule when a vector-valued function is written as a scalar function multiplying a vector-valued function. The scalar multiplies every coordinate; the two factors must be differentiable at the parameter in question. If either factor is not differentiable there, this theorem does not by itself guarantee that the product is differentiable there.

Keep the operations in the rule distinct. First multiply the scalar derivative by the original vector, then multiply the original scalar by the vector derivative, and finally add the two vectors. This method handles polynomial and trigonometric factors in exactly the same way; only the individual derivatives change.

Check Your Understanding

For each question, write the two product-rule terms explicitly before simplifying.

If $f(t)=t^3$ and $\mathbf{r}(t)=\langle t,1\rangle$, what is $\frac{d}{dt}(f\mathbf{r})$?
If $f(t)=\cos t$ and $\mathbf{r}(t)=\langle t,\sin t\rangle$, what are the two terms in the derivative?
For differentiable $f:I\to\mathbb{R}$ and $\mathbf{r}:I\to\mathbb{R}^n$, state the scalar-vector product rule and its differentiability condition.
In the product rule, which term accounts for the change in $\mathbf{r}$?
For $f(t)=t$ and $\mathbf{r}(t)=\langle t^2,1\rangle$, evaluate the derivative of $f\mathbf{r}$ at $t=1$, and check it by differentiating the product components.