Single-Variable-Calculus · Unit 4 · Video 3 · Interactive Practice

The Chain Rule: Differentiating a Composition of Functions

IKey Formulas

Formula	Name	What you need
$\dfrac{dy}{dt} = \dfrac{dy}{dx}\cdot\dfrac{dx}{dt}$	Chain rule, Leibniz form	A named intermediate variable $x$
$\dfrac{d}{dt}f\big(g(t)\big) = f'\big(g(t)\big)\,g'(t)$	Chain rule, function form	Outer derivative evaluated at the inner function
$\dfrac{d}{dt}(\sin t)^{10} = 10\sin^9 t\,\cos t$	Power outside, sine inside	$x = \sin t,\ y = x^{10}$
$\dfrac{d}{dt}\sin(10t) = 10\cos(10t)$	Sine outside, $10t$ inside	$x = 10t,\ y = \sin x$

Key Insight: Before the limit, $\dfrac{\Delta y}{\Delta t} = \dfrac{\Delta y}{\Delta x}\cdot\dfrac{\Delta x}{\Delta t}$ is ordinary algebra — the two copies of $\Delta x$ are numbers and really do cancel. The limit turns that identity into a statement about derivatives, and the cancellation survives only as a memory aid.

IIVisualization 1 — One Value of $t$ , Passed Through Two Functions

The inner function hands its output to the outer one: $t \to x = \sin t \to y = x^{10}$ .

t 1.050

IIIVisualization 2 — Why the Derivative Is a Product

A change in $t$ causes a change in $x$ , which in turn causes a change in $y$ .

Δt 0.4500

IVVisualization 3 — The Inside's Derivative Is Not Optional

Only one candidate line is tangent: the one carrying the factor from the inside function.

Composition t 0.600

💡 Nothing stops the chain from growing: for $f\big(g(h(t))\big)$ the same argument gives a product of three derivatives, $f'\big(g(h(t))\big)\,g'\big(h(t)\big)\,h'(t)$ .

VQuiz Questions

Problem 1 · Power Outside, Sine Inside

Given: $y = (\sin t)^4$ — find $\dfrac{dy}{dt}$ .

✅ Correct! Outside first:

\frac{dy}{dx} = 4x^3

with

x = \sin t

. Inside next:

\frac{dx}{dt} = \cos t

. Their product, with

x

substituted back, is

4\sin^3 t\,\cos t

❌ Not quite. That is only the outside derivative

4x^3

with

x = \sin t

put back. The inside function still has to be differentiated, and

\frac{d}{dt}\sin t = \cos t

❌ Not quite. The two functions have swapped jobs. The power is the outside function here, so the power rule acts on

x = \sin t

; the cosine can only enter as the derivative of the inside.

❌ Close, but check the exponent. The power rule drops the exponent by one:

\frac{d}{dx}x^4 = 4x^3

, not

4x^4

❌ Not quite. Differentiate the outside power, then multiply by the derivative of the inside.

Show solution

Name the two steps: $x = \sin t$ (inside) and $y = x^4$ (outside).

\frac{dy}{dx} = 4x^3, \qquad \frac{dx}{dt} = \cos t

The chain rule multiplies them:

\frac{dy}{dt} = \frac{dy}{dx}\cdot\frac{dx}{dt} = 4x^3\cos t

The letter $x$ was ours, so substitute it back, $x = \sin t$ :

\frac{d}{dt}(\sin t)^4 = 4\sin^3 t\,\cos t = 4\sin^3 t\cos t

Problem 2 · Leave the Inside Exactly As It Is

Given: $y = \cos(7t)$ — find $\dfrac{dy}{dt}$ .

✅ Correct! The outside derivative is

-\sin x

evaluated at the untouched inside

7t

, and the inside contributes its own derivative

7

❌ Not quite. This is the classic missing factor. The inside function

x = 7t

still has to be differentiated, and

\frac{dx}{dt} = 7

❌ Close — check the sign. The magnitude is right, but

\frac{d}{dx}\cos x = -\sin x

, so the answer stays negative.

❌ Not quite. The inside expression was disturbed. The outer derivative is evaluated at the inside function, which is

7t

, not

t

❌ Not quite. Differentiate the cosine without changing what is plugged into it, then multiply by the derivative of the inside.

Show solution

Inside $x = 7t$ , outside $y = \cos x$ .

\frac{dy}{dx} = -\sin x, \qquad \frac{dx}{dt} = 7

\frac{dy}{dt} = -\sin x \cdot 7 = -7\sin x

Substituting $x = 7t$ back:

\frac{d}{dt}\cos(7t) = -7\sin(7t)

Both classic mistakes are avoided here: the inside $7t$ is left exactly as it was, and the factor $7$ is not dropped.

Problem 3 · Both Factors, Then Substitute Back

Given: $y = (2t+1)^6$ , with the intermediate variable named $x = 2t+1$ so that $y = x^6$ .

What is $\dfrac{dy}{dx}$ ?

What is $\dfrac{dy}{dt}$ , written in terms of $t$ ?

✅ Correct!

\frac{dy}{dx} = 6x^5

and

\frac{dx}{dt} = 2

, so

\frac{dy}{dt} = 12x^5 = 12(2t+1)^5

❌ Not yet. The factor

2

belongs to the inside function;

\frac{dy}{dx}

asks for the derivative of

y = x^6

with respect to its own variable only.

❌ Check the power rule. Differentiating

x^6

with respect to

x

multiplies by the exponent and lowers it by one.

❌ Not quite. The outside derivative is right, but

\frac{dx}{dt} = 2

has not been multiplied in.

❌ Check the product. Multiply

\frac{dy}{dx} = 6x^5

\frac{dx}{dt} = 2

, then replace

x

2t+1

— the exponent does not change at this step.

Show solution

Step 1 — outside, with respect to its own variable:

\frac{dy}{dx} = \frac{d}{dx}x^6 = 6x^5

Step 2 — inside:

\frac{dx}{dt} = \frac{d}{dt}(2t+1) = 2

Step 3 — multiply:

\frac{dy}{dt} = 6x^5\cdot 2 = 12x^5

Step 4 — substitute back $x = 2t+1$ :

\frac{d}{dt}(2t+1)^6 = 12(2t+1)^5

Check at $t = 0$ : the formula gives $12(1)^5 = 12$ . Expanding first, $(2t+1)^6 = 1 + 12t + \dots$ , whose derivative at $t = 0$ is also $12\ \checkmark$

Problem 4 · Roles Reversed

Given: $y = \sin(t^2)$ — the sine is now the outside function and the squaring is inside. Find $\dfrac{dy}{dt}$ .

✅ Correct! The cosine keeps the inside

t^2

untouched, and the inside contributes

\frac{d}{dt}t^2 = 2t

❌ Not quite. The outside derivative is right and the inside was left alone, but the factor

\frac{dx}{dt} = 2t

is missing.

❌ Not quite. The factor

2t

is right, but the inside expression was changed: the outer derivative must be evaluated at

t^2

, giving

\cos(t^2)

❌ Not quite. Here the inside has been differentiated inside the cosine. The inside stays as it is inside the outer function; its derivative appears as a separate factor out front.

❌ Not quite. Differentiate the sine, leave

t^2

exactly where it is, and multiply by the derivative of

t^2

Show solution

Inside $x = t^2$ , outside $y = \sin x$ — the opposite arrangement to Problem 1.

\frac{dy}{dx} = \cos x, \qquad \frac{dx}{dt} = 2t

\frac{dy}{dt} = \cos x \cdot 2t

Substituting $x = t^2$ back, and writing the inside derivative in front:

\frac{d}{dt}\sin(t^2) = 2t\cos(t^2)

Compare with $\dfrac{d}{dt}(\sin t)^2 = 2\sin t\,\cos t$ : the same two ingredients composed in the other order give a different function and a different derivative.

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The Chain Rule: Differentiating a Composition of Functions

IKey Formulas

IIVisualization 1 — One Value of ttt, Passed Through Two Functions

IIIVisualization 2 — Why the Derivative Is a Product

IVVisualization 3 — The Inside's Derivative Is Not Optional

VQuiz Questions

IIVisualization 1 — One Value of $t$ , Passed Through Two Functions