Sum and Scalar-Multiple Rules for Vector Derivatives

Why Rules for Vector Derivatives Matter

Vector-valued functions often describe several changing quantities at once. For example, a function $\mathbf r(t)$ in three dimensions has three coordinate functions, each of which may have its own formula. When vector functions are added, subtracted, or multiplied by constants, their derivatives can be found without starting over from the difference quotient each time.

The key is that vector addition and constant multiplication operate coordinate by coordinate. The familiar scalar rules therefore apply to each coordinate, and the resulting coordinate derivatives form the derivative vector.

Central idea: If the vector functions involved are differentiable at $t$, then the derivative of their sum or difference is the sum or difference of their derivatives. A constant multiple can be moved outside the derivative.

Definitions and Conditions

Let $\mathbf r$ and $\mathbf s$ be vector-valued functions defined on a common open interval $I$, with values in $\mathbb R^n$. Write their components as $\mathbf r(t)=(r_1(t),\ldots,r_n(t))$ and $\mathbf s(t)=(s_1(t),\ldots,s_n(t))$. Their sum and difference are defined coordinatewise:

(\mathbf r+\mathbf s)(t) = (r_1(t)+s_1(t),\ldots,r_n(t)+s_n(t)), $$ $$ (\mathbf r-\mathbf s)(t) = (r_1(t)-s_1(t),\ldots,r_n(t)-s_n(t)).

For a constant $c\in\mathbb R$, the constant multiple $c\mathbf r$ is also defined coordinatewise: $(c\mathbf r)(t)=(cr_1(t),\ldots,cr_n(t))$. The word constant matters: $c$ does not depend on $t$.

Definition: The derivative of $\mathbf r$ at an interior point $t$ of its domain, when the limit exists, is $$ \mathbf r'(t) = \lim_{h\to 0} \frac{\mathbf r(t+h)-\mathbf r(t)}{h}. $$ The quotient means that each component of the difference vector is divided by the scalar $h$.

A vector-valued function is differentiable at $t$ exactly when each of its component functions is differentiable there. Thus, the rules below require the functions on the right-hand side to be differentiable at the parameter value in question. If they are differentiable throughout an interval, the rules hold at every point in that interval.

The Sum, Difference, and Constant-Multiple Rules

Theorem: Suppose $\mathbf r$ and $\mathbf s$ are differentiable at $t$, and $c$ is a real constant. Then $$ (\mathbf r+\mathbf s)'(t)=\mathbf r'(t)+\mathbf s'(t), \qquad (\mathbf r-\mathbf s)'(t)=\mathbf r'(t)-\mathbf s'(t), \qquad (c\mathbf r)'(t)=c\mathbf r'(t). $$

These identities do not require the vector functions to have the same component formulas. They need only take values in the same $\mathbb R^n$, so that their coordinates can be added or subtracted. The constant-multiple rule applies to any real constant, including zero and negative constants.

Why the Rules Work

We can see the reason in each coordinate. By the componentwise characterization of vector derivatives, the $i$-th component of $\mathbf r'(t)$ is $r_i'(t)$. The $i$-th component of $\mathbf r+\mathbf s$ is $r_i+s_i$. The scalar sum rule therefore gives $(r_i+s_i)'=r_i'+s_i'$. Since this is true for every coordinate, the vector derivative is $\mathbf r'+\mathbf s'$. The scalar difference and constant-multiple rules give the other two identities in exactly the same coordinatewise way.

There is also a direct explanation from the difference quotient. For a nonzero $h$, vector subtraction and addition distribute as usual, so the quotient for a sum splits into two quotients. As $h$ approaches zero, each quotient approaches the derivative of its function; the limit law for vector sums then gives the sum of those derivatives.

\begin{aligned} \frac{(\mathbf r+\mathbf s)(t+h)-(\mathbf r+\mathbf s)(t)}{h} &= \frac{\mathbf r(t+h)-\mathbf r(t)}{h} + \frac{\mathbf s(t+h)-\mathbf s(t)}{h},\\ (\mathbf r+\mathbf s)'(t) &= \lim_{h\to 0}\frac{\mathbf r(t+h)-\mathbf r(t)}{h} + \lim_{h\to 0}\frac{\mathbf s(t+h)-\mathbf s(t)}{h}\\ &=\mathbf r'(t)+\mathbf s'(t). \end{aligned}

For a difference, the same quotient has a minus sign between the two terms. For a constant multiple, its quotient is $c$ times the quotient for $\mathbf r$. Because $c$ is constant, it can be taken outside the limit. These manipulations are valid when the relevant derivatives exist.

Worked Examples

Worked Example: Differentiate a Sum and a Difference

Let $\mathbf r(t)=(t^2,3t-1,\cos t)$ and $\mathbf s(t)=(2t+1,t^2,\sin t)$. Find the derivative of $\mathbf q(t)=\mathbf r(t)-\mathbf s(t)$, and evaluate it at $t=1$.

First subtract corresponding components:

\mathbf q(t) = (t^2-(2t+1),\ (3t-1)-t^2,\ \cos t-\sin t) = (t^2-2t-1,\ 3t-1-t^2,\ \cos t-\sin t).

Differentiate each component. The first derivative is $2t-2$, the second is $3-2t$, and the third is $-\sin t-\cos t$. Thus,

\mathbf q'(t)=(2t-2,\ 3-2t,\ -\sin t-\cos t).

At $t=1$, the first component is $2(1)-2=0$, the second is $3-2(1)=1$, and the third is $-\sin 1-\cos 1$. Therefore, $\mathbf q'(1)=(0,1,-\sin 1-\cos 1)$. As a check, differentiating the three displayed components of $\mathbf q(t)$ gives $2t-2$, $3-2t$, and $-\sin t-\cos t$, respectively, confirming the result.

Worked Example: Differentiate a Constant Linear Combination

Let $\mathbf r(t)=(t^2,\sin t,2t)$, $\mathbf s(t)=(e^t,t^2,\cos t)$, and $\mathbf A(t)=-2\mathbf r(t)+3\mathbf s(t)$. Find $\mathbf A'(t)$, then evaluate it at $t=0$.

Apply the constant-multiple rule to each term, then differentiate each vector component:

\begin{aligned} \mathbf A'(t) &=-2\mathbf r'(t)+3\mathbf s'(t)\\ &=-2(2t,\cos t,2)+3(e^t,2t,-\sin t)\\ &=(-4t,-2\cos t,-4)+(3e^t,6t,-3\sin t)\\ &=(-4t+3e^t,\ -2\cos t+6t,\ -4-3\sin t). \end{aligned}

Substituting $t=0$, use $e^0=1$, $\cos 0=1$, and $\sin 0=0$. The components are $-4(0)+3(1)=3$, $-2(1)+6(0)=-2$, and $-4-3(0)=-4$. Hence $\mathbf A'(0)=(3,-2,-4)$.

To check, first combine the original functions: $\mathbf A(t)=(-2t^2+3e^t,\ -2\sin t+3t^2,\ -4t+3\cos t)$. Differentiating these components gives $(-4t+3e^t,\ -2\cos t+6t,\ -4-3\sin t)$, the same derivative as above. Its value at zero is indeed $(3,-2,-4)$.

Worked Example: Combine Rules and Evaluate at a Parameter

Suppose $\mathbf p(t)=(t^2,2t,t-4)$ and $\mathbf q(t)=(3t,t^2-1,2t^2)$. Find the derivative of $\mathbf F(t)=2\mathbf p(t)-3\mathbf q(t)$ at $t=1$.

Differentiate the two vector functions componentwise: $\mathbf p'(t)=(2t,2,1)$ and $\mathbf q'(t)=(3,2t,4t)$. The constants stay attached to their respective derivatives, and subtraction retains its minus sign:

\begin{aligned} \mathbf F'(t) &=2\mathbf p'(t)-3\mathbf q'(t)\\ &=2(2t,2,1)-3(3,2t,4t)\\ &=(4t,4,2)-(9,6t,12t)\\ &=(4t-9,\ 4-6t,\ 2-12t). \end{aligned}

At $t=1$, the components are $4(1)-9=-5$, $4-6(1)=-2$, and $2-12(1)=-10$. Therefore, $\mathbf F'(1)=(-5,-2,-10)$.

For a direct check, combine the coordinates first: $\mathbf F(t)=(2t^2-9t,\ 4t-3t^2+3,\ 2t-8-6t^2)$. Their derivatives are $4t-9$, $4-6t$, and $2-12t$. At $t=1$, these are $-5$, $-2$, and $-10$, as found using the rules.

Common Mistake: Losing a Constant Factor

A constant multiplying a vector function multiplies every component derivative. For example, let $\mathbf r(t)=(t^2,\sin t,t)$. A common incorrect calculation is

(3\mathbf r)'(t)=(6t,\cos t,1).

This result differentiates $\mathbf r$ but fails to multiply its derivative by $3$. The correct calculation is

\mathbf r'(t)=(2t,\cos t,1), \qquad (3\mathbf r)'(t)=3\mathbf r'(t)=(6t,3\cos t,3).

For a check, $3\mathbf r(t)=(3t^2,3\sin t,3t)$. Differentiating these components gives $6t$, $3\cos t$, and $3$, exactly the corrected result.

A related sign error can occur with differences: the rule is $(\mathbf r-\mathbf s)'=\mathbf r'-\mathbf s'$, not $\mathbf r'+\mathbf s'$. Keep the subtraction sign when applying the rule, then calculate each component carefully.

Using the Rules Efficiently

These rules let you organize a derivative before doing the coordinate calculations. For a finite expression made from vector functions using addition, subtraction, and constant multiples, differentiate each vector function and retain the same addition, subtraction, and constant factors. Then simplify each coordinate if useful.

Identify the operations.
Separate the expression into sums, differences, and constant multiples of vector functions.

Check differentiability.
Confirm that each vector function is differentiable at the parameter value or on the interval under consideration.

Apply the vector rules.
Keep every constant multiplier and preserve each plus or minus sign.

Differentiate and verify components.
Differentiate coordinate functions and check the resulting vector by differentiating the combined coordinates or by comparing components.

In applications, this organization is useful whenever several coordinate functions are combined to describe a changing vector. It reduces repetitive work while making signs and constant factors visible. The rules do not replace component differentiation; they summarize how component derivatives fit together.

Takeaway: Vector derivatives respect vector addition, subtraction, and multiplication by real constants. The reason is that each operation and each derivative is defined coordinatewise, where the ordinary scalar rules apply.

Check Your Understanding

For each question, use the componentwise definition of the operations and the vector derivative rules.

If $\mathbf u$ and $\mathbf v$ are differentiable, what is the derivative of $4\mathbf u-2\mathbf v$?
Differentiate $\mathbf w(t)=(t^3,\cos t,5t)$ and then find $(-\mathbf w)'(t)$.
Why must the same plus or minus sign be retained when differentiating a sum or difference?
For $\mathbf a(t)=(t, t^2)$ and $\mathbf b(t)=(t^2,3t)$, find $(\mathbf a+\mathbf b)'(t)$.
What differentiability condition is needed to apply the sum rule at a parameter value $t$?