Chain Rule for Composed Vector Functions

Why Composition Changes a Derivative

A vector-valued function can describe a curve using a parameter $u$. Sometimes that parameter is itself determined by another variable $t$. To differentiate the resulting function of $t$, we must account for both changes: how the vector function changes with $u$, and how $u$ changes with $t$.

For example, suppose a curve is described by $\mathbf r(u)$, but the parameter is chosen as $u=g(t)$. The new vector-valued function is $\mathbf r(g(t))$. Its derivative is not generally just $\mathbf r'(g(t))$; the rate of change of the inner parameter $g(t)$ must also be included.

Central idea: Differentiate the outer vector function at the current inner parameter, then multiply by the derivative of that inner parameter.

Definition and Conditions

Let $I$ and $J$ be intervals, let $\mathbf r:I\to\mathbb R^n$ be a vector-valued function, and let $g:J\to\mathbb R$ be a scalar-valued function. The composition $\mathbf r\circ g$ is defined at those $t\in J$ for which $g(t)\in I$, and its value is $\mathbf r(g(t))$.

Theorem: Chain rule for a vector-valued function composed with a scalar function. Suppose $g$ is differentiable at $t$, and $\mathbf r$ is differentiable at $g(t)$. Then the composition is differentiable at $t$, and $$ \frac{d}{dt}\mathbf r(g(t)) = \mathbf r'(g(t))\,g'(t). $$ Here $\mathbf r'(g(t))$ is a vector and $g'(t)$ is a scalar, so the right-hand side is scalar multiplication of that vector.

In three dimensions, if $\mathbf r(u)=\langle x(u),y(u),z(u)\rangle$, then the formula means

\frac{d}{dt}\mathbf r(g(t)) = \left\langle x'(g(t))g'(t),\; y'(g(t))g'(t),\; z'(g(t))g'(t) \right\rangle.

The input to each outer derivative is $g(t)$, not $t$. The factor $g'(t)$ appears in every component because every component depends on $t$ through the same inner function.

Why the Chain Rule Works

The formula follows from the definition of differentiability. Fix $t$, and write $s=g(t)$. For a small increment $h$, let

\delta=g(t+h)-g(t).

Since $g$ is differentiable at $t$, $\delta/h$ approaches $g'(t)$ as $h$ approaches zero. Since $\mathbf r$ is differentiable at $s$, its change for a small increment $\delta$ can be written as its derivative times $\delta$, together with a remainder that is small compared with $|\delta|$.

\mathbf r(s+\delta)-\mathbf r(s) = \mathbf r'(s)\delta+\mathbf e(\delta), \qquad \frac{\|\mathbf e(\delta)\|}{|\delta|}\longrightarrow 0 \quad\text{as }\delta\longrightarrow 0,\ \delta\ne 0.

If $\delta\ne 0$, divide this change by $h$:

\frac{\mathbf r(g(t+h))-\mathbf r(g(t))}{h} = \mathbf r'(s)\frac{\delta}{h} + \frac{\mathbf e(\delta)}{h}.

The first term approaches $\mathbf r'(s)g'(t)$. For the remainder, its magnitude satisfies

\left\|\frac{\mathbf e(\delta)}{h}\right\| = \frac{\|\mathbf e(\delta)\|}{|\delta|} \left|\frac{\delta}{h}\right|.

The first factor approaches zero, while the second approaches the finite number $|g'(t)|$. Thus the remainder term approaches the zero vector. If $\delta=0$, then $g(t+h)=g(t)$, so the numerator involving $\mathbf r$ is zero for that increment; this causes no additional term. Therefore the difference quotient approaches $\mathbf r'(s)g'(t)$, proving the formula.

This argument explains the two factors in the rule: for a small change in $t$, the inner function changes by approximately $g'(t)h$, and the outer function responds to that change at the rate $\mathbf r'(g(t))$.

Worked Examples

Worked Example: A Polynomial Curve with a Polynomial Parameter

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

First differentiate the outer vector function with respect to $u$:

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

The inner function and its derivative are

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

Apply the chain rule and substitute $u=g(t)=t^2+1$:

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

At $t=2$, the inner value is $g(2)=2^2+1=5$, and $g'(2)=4$. Hence

\mathbf r'(5)=\langle 10,75,2\rangle, \qquad \frac{d}{dt}\mathbf r(g(t))\bigg|_{t=2} = \langle 10,75,2\rangle(4) = \langle 40,300,8\rangle.

Check by composing first: $\mathbf r(g(t))=\langle (t^2+1)^2,(t^2+1)^3,2t^2+1\rangle$. Differentiating these components gives $\langle 4t(t^2+1),6t(t^2+1)^2,4t\rangle$, which at $t=2$ is $\langle40,300,8\rangle$, as required.

Worked Example: A Trigonometric Parameter Inside a Vector Function

Let $\mathbf r(u)=\langle u^2,\cos(\pi u),\sin(2u)\rangle$ and $u=g(t)=\sin t$. Find the derivative at $t=\pi/6$.

Differentiate the outer function:

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

For the inner function, $g'(t)=\cos t$. At $t=\pi/6$,

g\left(\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{6}\right)=\frac12, \qquad g'\left(\frac{\pi}{6}\right) = \cos\left(\frac{\pi}{6}\right)=\frac{\sqrt3}{2}.

Evaluate the outer derivative at $u=1/2$:

\mathbf r'\left(\frac12\right) = \left\langle 1,-\pi\sin\left(\frac{\pi}{2}\right),2\cos(1) \right\rangle = \langle 1,-\pi,2\cos(1)\rangle.

The chain rule now gives

\frac{d}{dt}\mathbf r(\sin t)\bigg|_{t=\pi/6} = \langle 1,-\pi,2\cos(1)\rangle\frac{\sqrt3}{2} = \left\langle \frac{\sqrt3}{2},-\frac{\pi\sqrt3}{2},\sqrt3\cos(1) \right\rangle.

Check directly from the composed function: $\mathbf r(\sin t)=\langle \sin^2t,\cos(\pi\sin t),\sin(2\sin t)\rangle$. Differentiating gives $\langle 2\sin t\cos t,-\pi\sin(\pi\sin t)\cos t,2\cos(2\sin t)\cos t\rangle$. At $t=\pi/6$, this is $\langle\sqrt3/2,-\pi\sqrt3/2,\sqrt3\cos(1)\rangle$, agreeing with the chain-rule result.

Worked Example: A Polynomial and Trigonometric Function Nested Together

Let $\mathbf r(u)=\langle u,u^2\rangle$ and $u=g(t)=\sin(2t)$. Find the derivative at $t=\pi/6$. This composition has a trigonometric inner function whose argument is itself $2t$.

The outer derivative is

\mathbf r'(u)=\langle 1,2u\rangle.

For the inner function, differentiating $\sin(2t)$ gives

g'(t)=2\cos(2t).

At $t=\pi/6$, $2t=\pi/3$, so

g\left(\frac{\pi}{6}\right) = \sin\left(\frac{\pi}{3}\right)=\frac{\sqrt3}{2}, \qquad g'\left(\frac{\pi}{6}\right) = 2\cos\left(\frac{\pi}{3}\right)=1.

Therefore

\frac{d}{dt}\mathbf r(\sin(2t))\bigg|_{t=\pi/6} = \left\langle 1,2\left(\frac{\sqrt3}{2}\right)\right\rangle(1) = \langle 1,\sqrt3\rangle.

Check by composing first: $\mathbf r(\sin(2t))=\langle\sin(2t),\sin^2(2t)\rangle$. Its derivative is $\langle2\cos(2t),4\sin(2t)\cos(2t)\rangle$. At $t=\pi/6$, this is $\langle1,4(\sqrt3/2)(1/2)\rangle=\langle1,\sqrt3\rangle$, as obtained above.

Nested Compositions: Apply the Rule at Each Layer

An inner parameter may itself be a composition. For example, suppose $u=g(h(t))$, with $h$ scalar-valued. First differentiate $g(h(t))$ using the scalar chain rule, obtaining $g'(h(t))h'(t)$. Then use the vector chain rule for $\mathbf r(u)$. The result, provided the functions are differentiable at the relevant inputs, is

\frac{d}{dt}\mathbf r(g(h(t))) = \mathbf r'(g(h(t)))\,g'(h(t))\,h'(t).

The order of the factors reflects the layers from outside to inside. The vector derivative is evaluated at the final inner input; each scalar derivative accounts for how one parameter changes with the next.

A Common Mistake: Forgetting the Inner Derivative

A frequent error is to differentiate the outer function at $g(t)$ but omit $g'(t)$. For example, let $\mathbf r(u)=\langle u^2,\sin u\rangle$ and $u=3t$. An incorrect calculation is

\frac{d}{dt}\mathbf r(3t) \ \text{(incorrect)} = \langle 2(3t),\cos(3t)\rangle.

This omits the rate $du/dt=3$. The correct result is

\mathbf r'(u)=\langle 2u,\cos u\rangle, \qquad \frac{d}{dt}\mathbf r(3t) = \langle 6t,3\cos(3t)\rangle.

At $t=0$, the incorrect answer is $\langle0,1\rangle$, while the correct answer is $\langle0,3\rangle$. A direct check confirms the correction: the second component of the composition is $\sin(3t)$, whose derivative at zero is $3\cos(0)=3$.

When the Rule Is Useful

The chain rule lets us change the parameter used to describe a vector-valued function without differentiating every composed component from scratch. It is especially helpful when a curve is first described in a convenient parameter $u$, then $u$ is specified as a function of another variable. Polynomial substitutions, trigonometric substitutions, and combinations of the two all use the same rule.

Before applying it, check two things: the inner value $g(t)$ must be in the domain of $\mathbf r$, and both $g$ at $t$ and $\mathbf r$ at $g(t)$ must be differentiable. Then evaluate $\mathbf r'$ at the inner value and multiply the entire vector by $g'(t)$. The chain rule does not assert differentiability where these conditions fail.

Check Your Understanding

For each question, identify the outer derivative, the inner derivative, and where each is evaluated.

If $\mathbf r(u)=\langle u^2,3u\rangle$ and $g(t)=t^2$, write the formula for $\frac{d}{dt}\mathbf r(g(t))$.
For $\mathbf r(u)=\langle\cos u,\sin u\rangle$ and $g(t)=2t+1$, find the derivative at $t=0$.
State the differentiability and domain conditions needed to apply the vector chain rule at a given $t$.
In $\mathbf r(g(t))$, why is $\mathbf r'$ evaluated at $g(t)$, rather than at $t$?
Find $\frac{d}{dt}\langle \sin(t^2),(\sin(t^2))^2\rangle$ by identifying the nested scalar parameter.