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Chain Rule

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Chain Rule - Complete Interactive Lesson

Part 1: Chain Rule Basics

🔗 The Chain Rule

Part 1 of 7 — Chain Rule Basics

Why Do We Need the Chain Rule?

So far, you can differentiate functions like $x^3$ , $\sin x$ , or $e^x$ . But what about composite functions — functions inside other functions?

Consider $f(x) = (3x + 1)^5$ . You could expand this, but that is painful. What about $\sin(x^2)$ or ? There is no shortcut without the Chain Rule.

The Chain Rule Formula

If $y = f(g(x))$ , then:

$\frac{dy}{dx} = f'(g(x)) \cdot g'(x)$

In words: differentiate the outer function (leaving the inner function untouched), then multiply by the derivative of the inner function.

Leibniz Notation

If $y = f(u)$ where $u = g(x)$ , then:

$\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}$

Worked Example 1

Find $\frac{d}{dx}(3x+1)^5$

Step	Work
Identify layers	Outer: $u^5$ , Inner: $u = 3x+1$
Differentiate outer	$5 u^{4} = 5 (3 x + 1^{}$

Worked Example 2

Find $\frac{d}{dx}\sin(x^2)$

Step	Work
Identify layers	Outer: $\sin(u)$ , Inner: $u = x^2$
Differentiate outer	$\cos(u) = \cos(x^2)$

AP Tip: The Chain Rule appears in nearly every derivative problem on the AP exam. Master it now.

Check Your Understanding 🎯

Compute the derivative using the Chain Rule.

Worked Example 3

Find $\frac{d}{dx}\sqrt{x^2 + 1}$

Apply the Chain Rule 🎯

Identify the Outer Function 🔍

For each composite function, select the correct outer function.

Key Takeaways — Part 1

Function	Derivative
$(\text{stuff})^n$	$n(\text{stuff})^{n-1} \cdot (\text{stuff})'$

Part 2: Nested Functions & Double Chain Rule

🔗 Nested Functions & Double Chain Rule

Part 2 of 7 — Nested Functions

When the Chain Rule Applies Twice

Some functions have three or more layers. For example:

$f(x) = \sin^2(3x) = [\sin(3x)]^2$

Part 3: Implicit Differentiation

🔗 Implicit Differentiation

Part 3 of 7 — Implicit Differentiation

What Is Implicit Differentiation?

Sometimes a relationship between $x$ and $y$ is not solved for $y$ . For example:

$x^2 + y^2 = 25$

Part 4: Related Rates Intro

🔗 Related Rates

Part 4 of 7 — Related Rates Intro

What Are Related Rates?

In related rates problems, two or more quantities are changing with respect to time ( $t$ ), and they are connected by an equation. We use implicit differentiation (with respect to $t$ ) to find how fast one quantity changes given information about the other.

The Strategy

Draw a picture and label changing quantities with variables
Write an equation relating the variables
Differentiate both sides with respect to $t$ (implicit differentiation)
Substitute known values and solve for the unknown rate

Worked Example 1: Expanding Circle

A stone is dropped in a pond. The circular ripple expands so that its radius increases at ft/s. How fast is the area increasing when the radius is ft?

Part 5: Advanced Chain Rule Applications

🔗 Advanced Chain Rule Applications

Part 5 of 7 — Applications

Logarithmic Differentiation

For functions like $y = x^x$ or $y = (\sin x)^{\cos x}$ , standard rules fail. solves this:

Part 6: Mixed Chain Rule Problems

🔗 Problem-Solving Workshop

Part 6 of 7 — Mixed Chain Rule Problems

Strategy Recap

For any derivative problem:

Identify if it is a composite function (Chain Rule needed?)
Count layers — how many times must the Chain Rule be applied?
Check for Product Rule or Quotient Rule requirements as well
Apply rules carefully, and do not forget the innermost derivative

This workshop tests your ability to combine the Chain Rule with other rules in realistic AP-style problems.

AP-Style Problems — Set 1 🎯

AP-Style Problems — Set 2 🎯

Match the derivative 🔍

Select the correct derivative for each function.

Workshop Complete!

You have practiced combining the Chain Rule with:

Product Rule
Quotient Rule
Implicit differentiation
Logarithmic differentiation
Inverse trig functions

Next up: Review and comprehensive assessment.

Part 7: Chain Rule Review

🔗 Chain Rule Review

Part 7 of 7 — Review & Applications

Complete Chain Rule Summary

Scenario	Formula
Basic Chain Rule	$\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)$

5 u^{4} = 5 (3 x + 1)^{4}

cos (u) = cos (x^{2})

Rewrite: $\sqrt{x^2+1} = (x^2+1)^{1/2}$

Step	Work
Outer derivative	$\frac{1}{2}(x^2+1)^{-1/2}$
Inner derivative	$2x$
Chain Rule	$\frac{1}{2}(x^2+1)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2+1}}$

Find $\frac{d}{dx}e^{-x^2}$

Step	Work
Outer: $e^u$	$e^{-x^2}$
Inner: $u = -x^2$	$-2x$
Chain Rule	$e^{-x^2} \cdot (-2x) = -2xe^{-x^2}$

Here we have three layers:

Outermost: $u^2$
Middle: $\sin(v)$
Innermost: $v = 3x$

The Chain Rule applies at each layer, multiplying all the derivatives together.

Find $\frac{d}{dx}[\sin(3x)]^2$

$\frac{d}{dx}[\sin(3x)]^2 = 2\sin(3x) \cdot \cos(3x) \cdot 3 = 6\sin(3x)\cos(3x)$

Layer	Function	Derivative
Outer	$u^2$	$2u = 2\sin(3x)$
Middle	$\sin(v)$	$\cos(v) = \cos(3x)$
Inner	$3x$	$3$
Result		$2\sin(3x) \cdot \cos(3x) \cdot 3 = 6\sin(3x)\cos(3x)$

Find $\frac{d}{dx}e^{\cos(2x)}$

Layer	Function	Derivative
Outer	$e^u$	$e^{\cos(2x)}$
Middle	$\cos(v)$	$-\sin(2x)$
Inner	$2x$	$2$
Result		$e^{\cos(2x)} \cdot (-\sin(2x)) \cdot 2 = -2\sin(2x)e^{\cos(2x)}$

Find $\frac{d}{dx}\sqrt{\ln(x)}$

Rewrite as $(\ln x)^{1/2}$ .

Layer	Function	Derivative
Outer	$u^{1/2}$	$\frac{1}{2}(\ln x)^{-1/2}$
Inner	$\ln x$	$\frac{1}{x}$
Result		$\frac{1}{2\sqrt{\ln x}} \cdot \frac{1}{x} = \frac{1}{2x\sqrt{\ln x}}$

Differentiate these nested functions 🎯

Triple Chain Rule

Sometimes you have four layers. The process is the same — just keep peeling.

Worked Example 4

Find $\frac{d}{dx}\sin^3(2x+1)$

This means $[\sin(2x+1)]^3$ :

Layer	Derivative
$u^3$	$3[\sin(2x+1)]^2$

Common Mistake Alert ⚠️

Students often forget the innermost derivative. For $\sin^2(3x)$ , many write $2\sin(3x)\cos(3x)$ and forget the factor of $3$ .

Always ask: "Is there another layer inside that I haven't differentiated yet?"

Multi-Layer Problems 🎯

How many Chain Rule applications? 🔍

For each function, select how many times you must apply the Chain Rule.

Key Takeaways — Part 2

Nested functions require the Chain Rule applied multiple times
Multiply ALL layer derivatives together — do not stop early
Common mistake: forgetting the innermost derivative
Strategy: Write out each layer, differentiate each, then multiply

Next up: Implicit differentiation — using the Chain Rule when $y$ is a function of $x$ .

This is a circle. We cannot easily write $y$ as a single function of $x$ . But we can still find $\frac{dy}{dx}$ using the Chain Rule.

When you differentiate a term containing $y$ with respect to $x$ , treat $y$ as a function of $x$ and apply the Chain Rule:

$\frac{d}{dx}[y^2] = 2y \cdot \frac{dy}{dx}$

The $\frac{dy}{dx}$ appears because $y$ is implicitly a function of $x$ .

Find $\frac{dy}{dx}$ for $x^2 + y^2 = 25$

Step	Work
Differentiate both sides	$2x + 2y\frac{dy}{dx} = 0$
Isolate $\frac{dy}{dx}$	$2y\frac{dy}{dx} = -2x$
Solve	$\frac{dy}{dx} = -\frac{x}{y}$

Find $\frac{dy}{dx}$ for $x^3 + y^3 = 6xy$

Step	Work
Differentiate	$3x^2 + 3y^2\frac{dy}{dx} = 6y + 6x\frac{dy}{dx}$
Group $dy/dx$ terms	$3y^2\frac{dy}{dx} - 6x\frac{dy}{dx} = 6y - 3x^2$
Factor	$\frac{dy}{dx}(3y^2 - 6x) = 6y - 3x^2$
Solve	$\frac{dy}{dx} = \frac{6y - 3x^2}{3y^2 - 6x} = \frac{2y - x^2}{y^2 - 2x}$

AP Tip: Implicit differentiation appears frequently on the AP exam, especially when finding slopes of tangent lines to curves defined implicitly.

Practice Implicit Differentiation 🎯

Implicit Differentiation with Trig Functions

Worked Example 3

Find $\frac{dy}{dx}$ for $\sin(y) = x$

Step	Work
Differentiate	$\cos(y) \cdot \frac{dy}{dx} = 1$
Solve

This also equals $\sec(y)$ , which makes sense since $y = \arcsin(x)$ and $\frac{d}{dx}\arcsin(x) = \frac{1}{\sqrt{1-x^2}} = \sec(\arcsin(x))$ .

Worked Example 4: Finding a Tangent Line

Find the slope of the tangent line to $x^2 + xy + y^2 = 7$ at $(1, 2)$ .

Step	Work
Differentiate	$2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$

The tangent line has slope $-\frac{4}{5}$ at $(1, 2)$ .

Implicit Differentiation Applications 🎯

Which differentiation rule is needed? 🔍

For each term, select the rule needed when differentiating with respect to $x$ .

Key Takeaways — Part 3

Implicit differentiation uses the Chain Rule: every time you differentiate $y$ , multiply by $\frac{dy}{dx}$
Steps: differentiate both sides → collect $\frac{dy}{dx}$ terms → factor → solve
Product rule is often needed when $x$ and $y$ appear together (like $xy$ )
Tangent lines: plug in the point after finding the general $\frac{dy}{dx}$ formula

Next up: Related rates — using implicit differentiation with respect to time.

Step	Work
Known	$\frac{dr}{dt} = 2$ ft/s, $r = 5$ ft
Find	$\frac{dA}{dt}$
Equation	$A = \pi r^2$
Differentiate	$\frac{dA}{dt} = 2\pi r \cdot \frac{dr}{dt}$
Substitute	$\frac{dA}{dt} = 2\pi(5)(2) = 20\pi \approx 62.8$ ft/s

Worked Example 2: Ladder Problem

A 13-ft ladder leans against a wall. The bottom slides away from the wall at 2 ft/s. How fast is the top sliding down when the bottom is 5 ft from the wall?

Step	Work
Setup	$x^2 + y^2 = 169$ (Pythagorean theorem)
Known	$\frac{dx}{dt} = 2$ ft/s, $x = 5$
Find $y$	$y = \sqrt{169 - 25} = 12$
Differentiate	$2x\frac{dx}{dt} + 2y\frac{dy}{dt} = 0$
Substitute	$2(5)(2) + 2(12)\frac{dy}{dt} = 0$
Solve	$\frac{dy}{dt} = -\frac{20}{24} = -\frac{5}{6}$ ft/s

The negative sign means the top is sliding down at $\frac{5}{6}$ ft/s.

Solve These Related Rates Problems 🎯

Worked Example 3: Conical Tank

Water drains from a conical tank (vertex down) at 2 ft $^3$ /min. The cone has radius 3 ft and height 6 ft at the top. How fast is the water level dropping when the depth is 4 ft?

Since the cone is similar: $\frac{r}{h} = \frac{3}{6} = \frac{1}{2}$ , so $r = \frac{h}{2}$ .

Step	Work
Volume formula	$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{\pi h^3}{12}$

AP Tip: Related rates problems are a staple of AP Calculus free-response questions. Always clearly state what each variable represents and what rate you are finding.

More Related Rates 🎯

Key Takeaways — Part 4

Related rates = implicit differentiation with respect to time
Steps: draw → equation → differentiate → substitute → solve
Key equations to know: Pythagorean theorem, area/volume formulas, similar triangles
Watch signs: negative rates mean decreasing quantities

Next up: More Chain Rule applications including logarithmic differentiation.

Logarithmic differentiation

Take $\ln$ of both sides
Use log properties to simplify
Differentiate implicitly
Solve for $\frac{dy}{dx}$

Find $\frac{d}{dx}x^x$ for $x > 0$

Step	Work
Take ln	$\ln y = x \ln x$
Differentiate	$\frac{1}{y}\frac{dy}{dx} = \ln x + x \cdot \frac{1}{x} = \ln x + 1$
Solve	$\frac{dy}{dx} = y(\ln x + 1) = x^x(\ln x + 1)$

Worked Example 2: Simplifying Products

Find $\frac{d}{dx}\frac{x^2\sqrt{x+1}}{(2x-3)^4}$

Without log differentiation, you would need the quotient rule, product rule, and chain rule combined. With logs:

Step	Work
Take ln	$\ln y = 2\ln x + \frac{1}{2}\ln(x+1) - 4\ln(2x-3)$
Differentiate	$\frac{y'}{y} = \frac{2}{x} + \frac{1}{2(x+1)} - \frac{8}{2x-3}$
Solve	$y' = \frac{x^2\sqrt{x+1}}{(2x-3)^4}\left(\frac{2}{x} + \frac{1}{2(x+1)} - \frac{8}{2x-3}\right)$

Logarithmic Differentiation 🎯

Chain Rule with Inverse Trig Functions

Recall the derivatives of inverse trig functions — they all involve the Chain Rule when the argument is not just $x$ :

Function	Derivative
$\arcsin(u)$	$\frac{u'}{\sqrt{1-u^2}}$
$\arccos(u)$	$\frac{-u'}{\sqrt{1-u^2}}$
$\arctan(u)$	$\frac{u'}{1+u^2}$

Worked Example 3

Find $\frac{d}{dx}\arctan(3x)$

$\frac{d}{dx}\arctan(3x) = \frac{3}{1+(3x)^2} = \frac{3}{1+9x^2}$

Worked Example 4

Find $\frac{d}{dx}\arcsin(x^2)$

$\frac{d}{dx}\arcsin(x^2) = \frac{2x}{\sqrt{1-(x^2)^2}} = \frac{2x}{\sqrt{1-x^4}}$

Inverse Trig Derivatives 🎯

Key Takeaways — Part 5

Logarithmic differentiation handles $u^v$ where both base and exponent are functions of $x$
Steps: $\ln$ both sides → simplify → differentiate implicitly → solve
Inverse trig + Chain Rule: always multiply by the derivative of the inner function
The Chain Rule is truly everywhere in calculus — almost no real problem avoids it

Next up: Problem-Solving Workshop with mixed Chain Rule problems.

The Chain Rule appears in:

~80% of all derivative problems (it is rarely absent)
All implicit differentiation problems
All related rates problems
FTC Part 1 problems involving $\frac{d}{dx}\int_a^{g(x)} f(t)\,dt$

Comprehensive Assessment 🎯

No hints — test your mastery.

Final Challenge 🎯

Chain Rule — Complete! ✅

You have mastered:

✅ Basic Chain Rule with single composition
✅ Nested functions requiring multiple applications
✅ Implicit differentiation
✅ Related rates
✅ Logarithmic differentiation
✅ Inverse trig with Chain Rule
✅ Combining Chain Rule with Product and Quotient Rules

You are ready to tackle any Chain Rule problem on the AP exam!

\frac{d y}{d x} = \frac{1}{c o s ( y )}

(x2+2(x+1)1−2x−38)

Chain Rule

Worked Example 4

Worked Example 1

Worked Example 2

Worked Example 3

Triple Chain Rule

Worked Example 4

Common Mistake Alert ⚠️

Key Takeaways — Part 2

The Key Idea

Worked Example 1

Worked Example 2

Implicit Differentiation with Trig Functions

Worked Example 3

Worked Example 4: Finding a Tangent Line

Key Takeaways — Part 3

Worked Example 2: Ladder Problem

Worked Example 3: Conical Tank

Key Takeaways — Part 4

Worked Example 1

Worked Example 2: Simplifying Products

Chain Rule with Inverse Trig Functions

Worked Example 3

Worked Example 4

Key Takeaways — Part 5

AP Exam Frequency

Chain Rule — Complete! ✅