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Thursday’s lecture deck. It moves from the product and quotient rules into composite functions, and closes on the chain rule with the worked problems from section 4.3.

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Calculus 201 · Chapter 4

Lecture 8 · Derivatives

4.3 The Chain Rule

A composite function applies one rule to the result of another. Differentiate from the outside in: take the derivative of the outer function, leave the inner one untouched, then multiply by its derivative.

The chain rule

If f(x) = g(h(x)), then f′(x) = g′(h(x)) · h′(x).

In Leibniz notation, if y = g(u) and u = h(x), then dy/dx = dy/du · du/dx.

Worked example

Differentiate f(x) = (3x² + 5)⁴.

Name the two pieces.

g(u) = u⁴, u = 3x² + 5

Differentiate each one.

g′(u) = 4u³, du/dx = 6x

Multiply, then substitute back.

4(3x² + 5)³ · 6x

f′(x) = 24x(3x² + 5)³

Calculus 201 · Chain rule9

Take a look at this page first. When you’re ready, tap Explain this page below - or Next to skip ahead.

Explaining · page 9

“So the outer function here is the fourth power, and the inner one is three x squared plus five.” “Differentiate the outside, leave the inside alone, then multiply by six x.”

Summary

Differentiate the outside first and leave the inside alone. Then multiply by the derivative of that inside, here 6x.

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01Lecture 8: Derivatives

02Where we left off

03Learning goals

04Derivative, informally

05Power rule

06Product rule

07Quotient rule

08Composite functions

09The chain rule

10Worked example

11Nested functions

12Problem set 4.3

13Common mistakes

14Chain rule with roots

15Trig compositions

16Exponential growth

17Implicit differentiation

18Related rates, part 1

19Related rates, part 2

20Higher derivatives

21Summary of rules

22Practice set A

23Practice set B

24Next: integration

what counts as the inside?

3x² + 5

Whatever you would work out first if you were plugging in a number.

Outsideu⁴

Inside3x² + 5

Its derivative, 6x, is the factor you multiply by.

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Lecture 8 - Chain Rule.pdfGoogle Chrome

Problem Set 4.3.pdfPage 1 of 3 1 2 3 MATH 201 · FALL 2026 · SECTION 004 Problem Set 4.3 The Chain Rule - due Friday, September 11 at 11:59 pm 1. Differentiate f(x) = (2x + 1)⁵. 2. Find dy/dx if y = √(4x² − 3). 3. Let g(t) = sin(3t²). Find g′(t). 4. Differentiate h(x) = (x³ + 2x)⁻².

Problem Set 4.3.pdfPreview

iCloud All iCloud38 Notes21 Calculus 2019 Econ 16 Readings2 Recently Deleted Today Calculus 201 - week 89:41 AMChain rule: outside, then in Questions for office hours8:15 AMWhy multiply by the inside? Previous 7 Days Exam 1 topicsMondayLimits, continuity, derivatives Econ - elasticitySunday%ΔQ divided by %ΔP Aa September 10, 2026 at 9:41 AM Calculus 201 - week 8 Chain rule Derivative of the outside, times the derivative of the inside. (f(g(x)))′ = f′(g(x)) · g′(x) Ex: (3x² + 5)⁴ → 4(3x² + 5)³ · 6x Watch lecture 8 recording Problem set 4.3 (due Fri) Ask about page 4 in section

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Search MMom 🏠Roomies JPJonah MCMaya Chen2:21 PMlmk what it says 🙏 🧮Calc 201 study group1:58 PMJonah: does anyone get q1 on the practice set DDad11:32 AMCall me when you’re free this weekend SKSam KimYesterdayLiked “see you at the library” To:Maya Chen Today 2:09 PM did you start 4.3 yet q3 is actually impossible 😭 the sin(3t²) one? same. asking leren rn lmk what it says 🙏 iMessage

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Inbox (3) - you@university.edu× Problem Set 4.3 - Docs× The chain rule, visually× Graphing calculator× Announcements: MATH 201× Study plan - week 8× Office hours - Calendar× Lecture 7 - Implicit.pdf× Deep focus - playlist× Lecture 8 - Chain Rule.pdf× +

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Page 4 of 6 ZOOM

Below is the routine for differentiating any composite function. Work the outside first, then the inside. If a step here disagrees with your textbook, bring it to section and we will settle it there.

STEPDetails

Step 1outer
inner
checkName the outer function: u⁴
Name the inner: u = 3x² + 5
The inner sits inside the outer

Step 2outer
innerDifferentiate the outer: 4u³
Leave the inner exactly as it is

Step 3multiply
resultMultiply by the inner’s derivative: 6x
f′(x) = 24x(3x² + 5)³

Wed 9/9MyLabSection 3.4 Problems 1, 3, 5, 9, 12 & 17 (due 9/14, 11:59pm)

Fri 9/11Section
QuizbCourses: Week 3 Quiz – Chain Rule
due 9/16, 11:59pm

Mon 9/14topic
readingImplicit Differentiation
Read Section 3.5

Wed 9/16topic
reading
MyLabRelated Rates
Read Section 3.9
Section 3.9 Problems 2, 4, 7, 11 & 15 (due 9/21, 11:59pm)

Fri 9/18Section
HomeworkbCourses: Week 4 Section HW – Related Rates
due 9/17, 11:59pm

Mon 9/21topic
reading
MyLabLinear Approximation
Read Section 3.10
Section 3.10 Problems 1, 5, 8 & 13 (due 9/23, 11:59pm)

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Hi! Where do we start?

walk me through the chain rule on page 4

Walk me through the chain rule on page 4.

Differentiate the outside, then multiply by the derivative of the inside.

1Outer: u⁴ → 4u³
2Inner: u = 3x² + 5 → 6x
3Multiply the two together.

f′(x) = 24x(3x² + 5)³

why do i multiply by the 6x at the end

Why do I multiply by the 6x at the end?

Because the inside is moving too, so its rate has to count.

1Inside: du/dx = 6x
2Outside: dy/du = 4u³
3One feeds the other, so the rates multiply.

dy/dx = dy/du · du/dx

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Q1.Differentiate a composite and read off its rate of changeCase study

A lab models the height of a bouncing ball as h(t) = (2t² + 3)⁵ centimetres, with t in seconds. The team tracks the inner quantity u = 2t² + 3 separately, and needs the rate of change of height at one second.

1)Write h′(t) using the chain rule.

Your answer…5(2t² + 3)⁴ · 4t

2)Evaluate h′(1).

Your answer…12 500 cm/s

Q2.Fill in the blank

For f(x) = (5x − 2)³, the inner function is #15x − 2 and its derivative is #25.

Q3.Which is the derivative of f(x) = √(4x + 1)?Multiple choice

A1 / (2√(4x + 1))

B2 / √(4x + 1)

C4 / √(4x + 1)

D1 / √(4x + 1)

Q4.Put the chain rule steps in the best orderPut in order

1Name the outer function and the inner function.

2Differentiate the inner function on its own.

3Differentiate the outer, leaving the inner alone.

4Multiply the two derivatives together.

5Substitute the inner function back in.

Use the arrows to arrange the steps from first to last.

Q5.Complete the derivative tableComplete the table

FunctionInner udu/dxf′(x)

(3x² + 5)⁴3x² + 56x…24x(3x² + 5)³

(5x − 2)³5x − 25…15(5x − 2)²

sin(2x)2x2…2cos(2x)

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Summary: Lecture 8 introduces the chain rule for differentiating composite functions: differentiate the outer function, leave the inner function unchanged, then multiply by the derivative of the inner function. Your Live Tutor session on problem 4.3 explains why that last multiplication is needed.

Document Details

Course: Calculus 201
Lecture: 8 - Derivatives
Section: 4.3, The Chain Rule
Date: September 10, 2026
Sources: Lecture notes and one Live Tutor session

The Chain Rule

A composite function puts one function inside another, like g(h(x)).
To differentiate it, work from the outside in:
- Differentiate the outer function and leave the inner one unchanged
- Multiply by the derivative of the inner function

f′(x) = g′(h(x)) · h′(x)

In Leibniz notation: dy/dx = dy/du · du/dx

Worked Example

Differentiate f(x) = (3x² + 5)⁴.
Name the two pieces:
- Outer: u⁴, with derivative 4u³
- Inner: u = 3x² + 5, with derivative 6x
Multiply, then put the inner function back in:

f′(x) = 4(3x² + 5)³ · 6x = 24x(3x² + 5)³

From Your Live Tutor Session

You asked why the 6x is multiplied at the end.
The inner function is changing too, so its rate of change has to count:
- Outer rate: dy/du = 4u³
- Inner rate: du/dx = 6x
One rate feeds into the other, so the two multiply.

Common Mistakes

Mistake	How to Avoid It
Forgetting the inner derivative	Finish by multiplying by du/dx
Putting the inner derivative inside	Write 4(3x² + 5)³, not 4(6x)³
Expanding the power first	Skip it - the chain rule gets there in one step

Before the Exam

Rework problem set 4.3 without looking at your notes
Practice nested functions and roots from pages 11 and 14
Build a short Exam Practice set on the chain rule

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