Why Linearity Matters
Vector-valued functions describe quantities with several components, such as a position with horizontal and vertical coordinates. When two vector functions are added or subtracted, their components combine in the same way. Differentiation respects those combinations: the derivative of a sum is the sum of the derivatives, and a constant multiple can be carried through the derivative.
These facts are useful because a vector function is often easier to differentiate one component at a time. The rules also let us reorganize a complicated vector expression before differentiating. Their justification comes directly from the limit definition of a vector derivative and the corresponding limit laws for real-valued functions.
Definition and Conditions
Let \(\mathbf{r}:I\to\mathbb{R}^n\) be a vector-valued function on an interval \(I\), and let \(t_0\) be an interior point of \(I\). Its derivative at \(t_0\), when the limit exists, is defined by the limit of its difference quotient:
Division by the scalar \(h\) means multiplying every component of the vector by \(1/h\). If \(\mathbf{r}(t)=\langle r_1(t),r_2(t),\ldots,r_n(t)\rangle\), then the derivative exists precisely when each component difference quotient has a limit, and those limits form the derivative vector.
For two vector functions to be added or subtracted, they must take values in the same space, so their corresponding components can be combined. The rules below apply at any point where both functions are differentiable. A constant-multiple rule also requires that the multiplier be a fixed real number, not a function of the parameter.
The Linearity Rules
Suppose \(\mathbf{u}\) and \(\mathbf{v}\) are differentiable vector-valued functions with values in the same \(\mathbb{R}^n\), and \(c\) is a constant real number. Then:
The first two rules are the sum and difference rules for vector differentiation. The third is the constant-multiple rule. Together they imply that any fixed linear combination of differentiable vector functions can be differentiated by applying the same coefficients to their derivatives.
Why the Rules Work
We prove all three rules from the derivative definition. The key fact is that a vector limit exists exactly when the limits of all its components exist, and in that case the vector limit is the vector of those component limits.
For the sum rule, use the definition at a point \(t_0\). For nonzero \(h\) sufficiently close to zero, the difference quotient for the sum can be rearranged as follows:
Because both functions are differentiable at \(t_0\), the two vector difference quotients have limits. Taking the limit and using the sum law for limits in each component gives:
For the difference rule, subtraction is also performed component by component. The difference quotient separates into the first function's quotient minus the second function's quotient, and the difference law for limits applies to every component:
Finally, a constant \(c\) can be factored out of each component limit. Indeed, the difference quotient for \(c\mathbf{u}\) is \(c\) times the difference quotient for \(\mathbf{u}\). Since \(c\) does not depend on \(h\), the constant-multiple limit law gives:
Each step is valid because the relevant component limits exist by differentiability, and the real-number limit laws apply to those components. This proves the three rules at \(t_0\); since the argument applies at every point where both functions are differentiable, it establishes the result throughout their common differentiability domain.
Worked Examples
Worked Example: Differentiate a Sum from Its Difference Quotient
Let \(\mathbf{r}(t)=\langle t^2,3t-1\rangle\) and \(\mathbf{s}(t)=\langle 2t+1,t^2\rangle\). Find \((\mathbf{r}+\mathbf{s})'(t)\) by forming the difference quotient and taking its limit component by component.
First add corresponding components:
For an increment \(h\), subtract the value at \(t\) from the value at \(t+h\), then divide each component by \(h\):
Taking \(h\to 0\) component by component gives:
Check: Differentiating the two components of the sum gives \(\frac{d}{dt}(t^2+2t+1)=2t+2\) and \(\frac{d}{dt}(t^2+3t-1)=2t+3\), matching the limit result.
Worked Example: Apply the Difference Rule
Let \(\mathbf{u}(t)=\langle t^2,\sin t\rangle\) and \(\mathbf{v}(t)=\langle 3t,\cos t\rangle\). Find \((\mathbf{u}-\mathbf{v})'(t)\), and then evaluate it at \(t=0\).
Both functions are differentiable for all real \(t\), and their components can be subtracted:
Using the difference rule and differentiating each component, we obtain:
At \(t=0\), use \(2(0)-3=-3\), \(\cos 0=1\), and \(\sin 0=0\):
Check: The subtracted function is \(\langle t^2-3t,\sin t-\cos t\rangle\). Its component derivatives are \(2t-3\) and \(\cos t+\sin t\), so its derivative agrees with the result; at zero, those components are \(-3\) and \(1\).
Worked Example: Differentiate a Constant Linear Combination
Let \(\mathbf{f}(t)=\langle t,t^2\rangle\), \(\mathbf{g}(t)=\langle 1-t,2t+3\rangle\), and define \(\mathbf{w}(t)=3\mathbf{f}(t)-2\mathbf{g}(t)\). Find \(\mathbf{w}'(t)\) and evaluate it at \(t=2\).
The coefficients \(3\) and \(-2\) are constants, so the constant-multiple and difference rules apply:
At \(t=2\), the first component is \(5\), and the second is \(6(2)-4=12-4=8\). Therefore \(\mathbf{w}'(2)=\langle 5,8\rangle\).
Check: First combine the original components: \(\mathbf{w}(t)=\langle 3t-2(1-t),3t^2-2(2t+3)\rangle=\langle 5t-2,3t^2-4t-6\rangle\). Their derivatives are \(5\) and \(6t-4\), giving \(\langle 5,8\rangle\) at \(t=2\), as found above.
Common Pitfall: Treating a Variable Multiplier as a Constant
The constant-multiple rule applies only when the multiplier does not change with \(t\). For example, the following calculation is not valid if \(a(t)\) is a nonconstant scalar function:
This incorrectly treats \(a(t)\) like a fixed number. A simple counterexample shows the problem. Let \(a(t)=t\) and \(\mathbf{r}(t)=\langle 1,0\rangle\). Then \(a(t)\mathbf{r}(t)=\langle t,0\rangle\), whose derivative is \(\langle 1,0\rangle\). But the incorrect calculation gives \(t\mathbf{r}'(t)=t\langle 0,0\rangle=\langle 0,0\rangle\), which is different. The constant-multiple rule cannot be used here. Differentiating a product of a scalar function and a vector function requires a separate rule.
Another error is to combine components in the wrong order. Vector addition and subtraction require matching components: first component with first component, second with second, and so on. The operations are not performed by mixing coordinates.
Using Linearity Efficiently
When differentiating a vector expression, first check its structure. If it consists of sums, differences, and constant multiples of differentiable vector functions, apply the linearity rules to simplify the task. Then differentiate the component functions. The reasoning depends on two conditions: the vector functions must have the same number of components for sums and differences, and each function being differentiated must be differentiable at the point in question.
Linearity does not say that every operation on vector functions can be passed through a derivative. It covers the three stated operations only. In particular, a multiplier that varies with the parameter is not a constant, and it cannot be pulled outside the derivative using this result.
Check Your Understanding
Use the derivative definition, the linearity rules, and component differentiation to answer these questions.
- State the condition on the ranges of \(\mathbf{u}\) and \(\mathbf{v}\) that allows \(\mathbf{u}+\mathbf{v}\) to be defined.
- For differentiable \(\mathbf{p}\) and \(\mathbf{q}\), write the difference rule and briefly identify the limit law used to justify it.
- If \(\mathbf{r}'(t)=\langle 2t,3\rangle\), find the derivative of \(-4\mathbf{r}(t)\).
- Explain why the constant-multiple proof cannot simply factor \(a(t)\) out of the limit when \(a\) depends on \(t\).
- For \(\mathbf{a}(t)=\langle t^2,1-t\rangle\) and \(\mathbf{b}(t)=\langle 2t,t^2\rangle\), find \((\mathbf{a}+\mathbf{b})'(t)\).
