Chain Rule for Composed Vector Functions

Why a Vector Chain Rule Is Needed

A vector-valued function can describe several quantities that change together as one input changes. For example, its components might be position coordinates expressed in terms of a parameter $u$. If that parameter is itself a function $g(t)$, the resulting function is a composition: the vector function receives the output of $g$, rather than $t$ directly.

The derivative of this composition must account for both changes: how each component of the vector function changes with its input, and how that input changes with $t$. The rule is not a new operation on vectors. It follows by applying the ordinary scalar chain rule separately to every component.

Key idea: Differentiate the vector function with respect to its own input, evaluate that derivative at $g(t)$, and multiply every component by the scalar $g'(t)$.

Definitions and the Chain Rule

Let $\mathbf r$ be a vector-valued function of a scalar input $u$, with $n$ components, and let $g$ be a scalar-valued function of $t$. Write

\mathbf r(u) = \langle r_1(u),r_2(u),\ldots,r_n(u)\rangle, \qquad \mathbf R(t)=\mathbf r(g(t)).

The composition $\mathbf R$ is defined wherever $g(t)$ belongs to the domain of $\mathbf r$. The derivative of $\mathbf r$ with respect to its scalar input is the vector of derivatives of its components:

\mathbf r'(u) = \langle r_1'(u),r_2'(u),\ldots,r_n'(u)\rangle.

Here the prime on $\mathbf r$ means that each component is differentiated with respect to $u$. The derivative $\mathbf R'(t)$, in contrast, is with respect to $t$.

Theorem: Chain rule for a composed vector function. Suppose $g$ is differentiable at $t$, and each component $r_i$ is differentiable at $u=g(t)$. Then $\mathbf R(t)=\mathbf r(g(t))$ is differentiable at $t$, and $$ \mathbf R'(t)=\mathbf r'(g(t))g'(t) = \langle r_1'(g(t))g'(t),\ldots,r_n'(g(t))g'(t)\rangle. $$ The factor $g'(t)$ is a scalar, so it multiplies every component of the vector.

The input to $\mathbf r'$ must be $g(t)$, not $t$ unless $g(t)=t$. The scalar multiplier must be $g'(t)$, because the inner input changes at that rate.

Why the Rule Works

The justification is component by component. For each $i$, the $i$th component of the composition is the scalar function $r_i(g(t))$. Since $r_i$ and $g$ are differentiable at the relevant inputs, the ordinary scalar chain rule gives its derivative:

\frac{d}{dt}r_i(g(t)) = r_i'(g(t))g'(t).

The derivative of a vector-valued function is defined by differentiating its components. Placing these scalar derivatives together therefore gives the stated vector formula. This argument also explains why the same scalar $g'(t)$ appears in every component: all components share the same inner input $g(t)$.

Differentiate the outer components.
Find $r_1'(u),r_2'(u),\ldots,r_n'(u)$, keeping the outer input named $u$.

Evaluate at the inner function.
Replace every $u$ in those derivatives by $g(t)$.

Multiply by the inner derivative.
Multiply every component by $g'(t)$.

Substitute the requested parameter value.
Evaluate $g(t)$ and $g'(t)$ at that value, and simplify each component.

Worked Examples

Worked Example: Polynomial, Trigonometric, and Exponential Components

Let $\mathbf r(u)=\langle u^2+1,\sin u,e^u\rangle$ and $g(t)=3t-1$. Find the derivative of $\mathbf r(g(t))$ at $t=1$.

First differentiate the components of $\mathbf r$ with respect to $u$:

\mathbf r'(u)=\langle 2u,\cos u,e^u\rangle.

Next evaluate the inner function and its derivative at $t=1$:

g(1)=3(1)-1=2, \qquad g'(t)=3, \qquad g'(1)=3.

Evaluate the outer derivative at $g(1)=2$, then multiply by $g'(1)$:

\mathbf r'(g(1))g'(1) = \langle 2(2),\cos 2,e^2\rangle(3) = \langle 4,\cos 2,e^2\rangle(3) = \langle 12,3\cos 2,3e^2\rangle.

Check by differentiating the composed components directly. The composition is $\langle(3t-1)^2+1,\sin(3t-1),e^{3t-1}\rangle$, whose derivative is $\langle6(3t-1),3\cos(3t-1),3e^{3t-1}\rangle$. At $t=1$, this is $\langle12,3\cos2,3e^2\rangle$, as required.

Worked Example: A Square Root and a Polynomial Component

Let $\mathbf r(u)=\langle\sqrt{u+4},u^3-2u\rangle$ and $g(t)=t^2+4$. Find the derivative of the composition at $t=2$.

The square-root component is defined for $u\geq -4$, and $g(2)=8$ is in that domain. Differentiate the outer components:

\mathbf r'(u) = \left\langle \frac{1}{2\sqrt{u+4}}, 3u^2-2 \right\rangle.

Evaluate the inner function and its derivative:

g(2)=2^2+4=8, \qquad g'(t)=2t, \qquad g'(2)=4.

Now evaluate and multiply component by component:

\mathbf r'(8) = \left\langle \frac{1}{2\sqrt{8+4}}, 3(8)^2-2 \right\rangle = \left\langle \frac{1}{2\sqrt{12}}, 192-2 \right\rangle = \left\langle \frac{1}{4\sqrt{3}}, 190 \right\rangle, $$ $$ \mathbf r'(8)g'(2) = \left\langle \frac{1}{4\sqrt{3}},190 \right\rangle(4) = \left\langle \frac{1}{\sqrt{3}},760 \right\rangle.

For a check, the first composed component is $\sqrt{t^2+8}$, with derivative $t/\sqrt{t^2+8}$. At $t=2$, this is $2/\sqrt{12}=1/\sqrt{3}$. The second is $(t^2+4)^3-2(t^2+4)$, with derivative $(3(t^2+4)^2-2)(2t)$. At $t=2$, its value is $(3(8)^2-2)(4)=(192-2)(4)=760$. Both components agree.

Worked Example: A Trigonometric Component with a Nonlinear Inner Function

Let $\mathbf r(u)=\langle u^2-3u,\cos(2u)\rangle$ and $g(t)=t^3+1$. Find the derivative of the composition at $t=1$.

Differentiate with respect to the outer input $u$. In particular, differentiating $\cos(2u)$ requires the scalar chain rule:

\mathbf r'(u) = \langle 2u-3,-2\sin(2u)\rangle.

At the requested parameter value,

g(1)=1^3+1=2, \qquad g'(t)=3t^2, \qquad g'(1)=3.

Therefore,

\mathbf r'(g(1))g'(1) = \langle 2(2)-3,-2\sin(2(2))\rangle(3) = \langle 1,-2\sin4\rangle(3) = \langle 3,-6\sin4\rangle.

Check directly: the composition is $\langle(t^3+1)^2-3(t^3+1),\cos(2(t^3+1))\rangle$. Its derivative is $\langle(2(t^3+1)-3)(3t^2),-6t^2\sin(2(t^3+1))\rangle$. Substituting $t=1$ gives $\langle(4-3)3,-6\sin4\rangle=\langle3,-6\sin4\rangle$, confirming the result.

A Common Mistake: Forgetting the Inner Derivative

A frequent error is to differentiate the outer vector function and evaluate it at $g(t)$, but then omit the factor $g'(t)$. For the first example, that mistake at $t=1$ would give

\mathbf r'(g(1)) = \mathbf r'(2) = \langle4,\cos2,e^2\rangle,

as though the input $u$ changed at a rate of $1$. But $u=g(t)=3t-1$ changes at rate $g'(1)=3$. The correct calculation is

\mathbf r'(g(1))g'(1) = \langle4,\cos2,e^2\rangle(3) = \langle12,3\cos2,3e^2\rangle.

A second related error is to evaluate $\mathbf r'$ at $t$ instead of at $g(t)$. The outer function receives $g(t)$, so its derivative must be evaluated at that same input. A reliable check is to write down $g(t)$ and $g'(t)$ separately before substituting the requested value.

When the Rule Applies

The formula requires differentiability of the inner function at the parameter value and differentiability of every outer component at the resulting inner input. It is not enough for $g'(t)$ to exist if an outer component is not differentiable at $g(t)$. The composition may sometimes be differentiable in special cases despite a failed hypothesis, but this chain-rule result does not establish that; such a case needs separate analysis.

When the hypotheses hold, the rule applies to any finite number of components. For a three-component function, for example, the calculation has exactly the same structure:

\mathbf r(u)=\langle x(u),y(u),z(u)\rangle \quad\Longrightarrow\quad \frac{d}{dt}\mathbf r(g(t)) = \langle x'(g(t))g'(t),y'(g(t))g'(t),z'(g(t))g'(t)\rangle.

The formula is useful whenever several dependent quantities share one changing input. It keeps the components organized while ensuring that the rate of change of the shared input is included in each component.

Check Your Understanding

For each question, identify the inner input and apply the chain rule component by component.

If $\mathbf r(u)=\langle u^2,\sin u\rangle$ and $g(t)=2t+1$, write a formula for $\frac{d}{dt}\mathbf r(g(t))$.
For $\mathbf r(u)=\langle u^3,1/u\rangle$ and $g(t)=t^2+2$, find $\frac{d}{dt}\mathbf r(g(t))$ at $t=1$.
State the differentiability conditions needed to apply the vector chain rule at a particular $t$.
Explain why the derivative $\mathbf r'(g(t))$ is multiplied by $g'(t)$.
Describe the two distinct evaluation errors that can occur when using the chain rule for a composition $\mathbf r(g(t))$.

Chain Rule for Composed Vector Functions

What You'll Learn

Why a Vector Chain Rule Is Needed

Definitions and the Chain Rule

Why the Rule Works

Worked Examples

Worked Example: Polynomial, Trigonometric, and Exponential Components

Worked Example: A Square Root and a Polynomial Component

Worked Example: A Trigonometric Component with a Nonlinear Inner Function

A Common Mistake: Forgetting the Inner Derivative

When the Rule Applies

Check Your Understanding

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Vol. 2: Solutions Manual

Chain Rule for Composed Vector Functions

What You'll Learn

Why a Vector Chain Rule Is Needed

Definitions and the Chain Rule

Why the Rule Works

Worked Examples

Worked Example: Polynomial, Trigonometric, and Exponential Components

Worked Example: A Square Root and a Polynomial Component

Worked Example: A Trigonometric Component with a Nonlinear Inner Function

A Common Mistake: Forgetting the Inner Derivative

When the Rule Applies

Check Your Understanding

Want all 1,114 practice problems?

Get the Calc Pass App

Vol. 1: 1,114 Questions

Vol. 2: Solutions Manual

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