Why a Scalar-Vector Product Needs a Rule
A scalar function can change as its input changes, and a vector-valued function can change as well. When both change, their product reflects both changes. Differentiating only one factor misses part of the variation.
For example, let \(g(t)\) be a real-valued function and let \(\mathbf{r}(t)\) be a vector-valued function. Their scalar-vector product \(g(t)\mathbf{r}(t)\) is formed by multiplying every component of \(\mathbf{r}(t)\) by the same scalar \(g(t)\). The product rule accounts for the change in the scalar and the change in the vector.
The two terms on the right have different roles. The term \(g'(t)\mathbf{r}(t)\) accounts for the scalar changing while the vector is held at its current value. The term \(g(t)\mathbf{r}'(t)\) accounts for the vector changing while the scalar is held at its current value. Both terms are vectors of the same dimension, so they can be added.
Component Form
Suppose \(\mathbf{r}(t)=\langle x(t),y(t)\rangle\). Since scalar multiplication acts on each component,
Applying the ordinary scalar product rule to each component gives
This is precisely the component form of \(g'\mathbf{r}+g\mathbf{r}'\). In three dimensions, if \(\mathbf{r}(t)=\langle x(t),y(t),z(t)\rangle\), the same rule applies to all three components:
All functions and derivatives in these component expressions are evaluated at \(t\). The rule is valid wherever \(g\) and every component of \(\mathbf{r}\) are differentiable.
Why the Product Rule Works
The vector derivative is defined by a limit of difference quotients, computed component by component. The derivation below uses that definition and the ordinary limit laws for scalar and vector functions.
Write the change in the product between \(t\) and \(t+h\), divided by \(h\), for nonzero \(h\).
Insert \(g(t)\mathbf{r}(t+h)\) so the change separates into a scalar change and a vector change.
Use differentiability and continuity to evaluate the two resulting terms.
In equations, the difference quotient is
Add and subtract \(g(t)\mathbf{r}(t+h)\) in the numerator, then factor each difference:
Because \(g\) is differentiable at \(t\), the first scalar difference quotient tends to \(g'(t)\). Because \(\mathbf{r}\) is differentiable at \(t\), its difference quotient tends to \(\mathbf{r}'(t)\). Differentiability of \(\mathbf{r}\) also implies continuity there, so \(\mathbf{r}(t+h)\) tends to \(\mathbf{r}(t)\). Therefore the limit of the first term is \(g'(t)\mathbf{r}(t)\), and the limit of the second is \(g(t)\mathbf{r}'(t)\). Adding these limits proves
The continuity used here follows directly from differentiability: each component’s difference quotient has a finite limit, so the component change is \(h\) times a quantity that stays bounded near \(t\), and therefore tends to zero as \(h\) tends to zero. Thus the vector itself tends to its value at \(t\).
Worked Examples
Worked Example: A Polynomial Scalar Times a Two-Component Vector
Let \(g(t)=t^2+1\) and \(\mathbf{r}(t)=\langle 2t-1,\;t^2-3\rangle\). Find the derivative of \(g(t)\mathbf{r}(t)\), and verify it by expanding the components.
First, differentiate each factor:
Apply the product rule. The first term is
The second term is
Add corresponding components:
To check, expand the original product first:
Differentiate the first component to get \(6t^2-2t+2\), and differentiate the second to get \(4t^3-4t\). These match the product-rule result.
Worked Example: A Trigonometric Scalar and Vector
Let \(g(t)=\sin t\) and \(\mathbf{r}(t)=\langle t,\cos t\rangle\). Find the derivative of their product and evaluate it at \(t=\frac{\pi}{3}\).
The derivatives of the factors are
Use both terms of the product rule:
At \(t=\frac{\pi}{3}\), use \(\cos(\frac{\pi}{3})=\frac12\) and \(\sin(\frac{\pi}{3})=\frac{\sqrt3}{2}\). The first component is
The second component is
Thus the derivative at the specified input is \(\langle\frac{\pi}{6}+\frac{\sqrt3}{2},-\frac12\rangle\). As a check, the original product is \(\langle t\sin t,\sin t\cos t\rangle\). Differentiating these components gives \(\sin t+t\cos t\) and \(\cos^2t-\sin^2t\), respectively, agreeing with the rule and its evaluated value.
Worked Example: A Reciprocal Scalar Evaluated at a Number
For \(t\ne 0\), let \(g(t)=\frac1t\) and \(\mathbf{r}(t)=\langle t^2+1,\;2t-1\rangle\). Find the derivative of their product at \(t=2\).
Differentiate the factors:
The two product-rule terms are
and
Adding and simplifying component by component gives
At \(t=2\), this is \(\langle 1-\frac14,\frac14\rangle=\langle\frac34,\frac14\rangle\). To check, simplify the product first: \(\frac1t\langle t^2+1,2t-1\rangle=\langle t+\frac1t,2-\frac1t\rangle\). Its component derivatives are \(1-\frac1{t^2}\) and \(\frac1{t^2}\), which at \(t=2\) give \(\frac34\) and \(\frac14\), as required.
A Common Mistake: Differentiating Only the Vector
A frequent error is to differentiate \(\mathbf{r}(t)\) but leave out the term involving \(g'(t)\). For a concrete example, take \(g(t)=t^2\) and \(\mathbf{r}(t)=\langle t,1\rangle\). Omitting the scalar’s derivative would give
which at \(t=1\) is \(\langle1,0\rangle\). But the product is actually \(\langle t^3,t^2\rangle\), whose derivative at \(1\) is \(\langle3,2\rangle\). The missing contribution is \(g'(t)\mathbf{r}(t)\): at \(t=1\), it equals \(2\langle1,1\rangle=\langle2,2\rangle\). Adding the other term gives \(\langle2,2\rangle+\langle1,0\rangle=\langle3,2\rangle\), confirming the correct answer.
Using the Rule Reliably
A consistent process reduces sign and component errors. First identify the scalar factor and the vector factor. Then differentiate each separately, keeping the vector’s components in their original order. Form \(g'\mathbf{r}\) and \(g\mathbf{r}'\), and add corresponding components. If the result is complicated, expand the original product component by component and differentiate those scalar expressions as an independent check.
The rule applies to any finite-dimensional vector-valued function as long as the scalar function and vector function are differentiable at the input. It does not require the components to be polynomials; the examples include trigonometric and reciprocal functions. The key is that the product means scalar multiplication of a vector, not multiplication of two scalar quantities.
Check Your Understanding
For each question, keep the scalar and vector factors distinct until you have written both terms of the product rule.
- State the differentiability conditions and formula for differentiating \(g(t)\mathbf{r}(t)\).
- If \(g(t)=t\) and \(\mathbf{r}(t)=\langle t^2,3t\rangle\), find the derivative of their product.
- Explain why differentiability of \(\mathbf{r}\) at \(t\) allows \(\mathbf{r}(t+h)\) to tend to \(\mathbf{r}(t)\) in the derivation.
- For \(g(t)=\cos t\) and \(\mathbf{r}(t)=\langle 1,\sin t\rangle\), write the product-rule derivative in component form.
- What term is omitted in the incorrect calculation that differentiates only the vector factor?
