Chain Rule - Complete Interactive Lesson

Part 1: Chain Rule Basics

🔗 The Chain Rule

Part 1 of 7 — Chain Rule Basics

Welcome to the Chain Rule — arguably the most important differentiation rule in calculus!

Part	Topic
1	Chain Rule Basics
2	Nested Functions & Double Chain Rule
3	Implicit Differentiation
4	Related Rates
5	Advanced Applications
6	Problem-Solving Workshop
7	Comprehensive Review

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?

$(3x + 1)^5$ — expanding this is painful
$\sin(x^2)$ — can't use basic trig rule directly
$e^{3x}$ — the exponent isn't just $x$

The Chain Rule handles ALL of these.

The Chain Rule Formula

$\boxed{\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)}$

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:

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

Key Fact: The Chain Rule is needed whenever you see a function INSIDE another function. It appears in ~80% of all derivative problems on the AP exam.

Worked Examples — Step by Step

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

Step	Work
Identify layers	Outer: $u^5$ , Inner: $u = 3x+1$
Differentiate outer	$5u^4 = 5(3x+1)^4$
Differentiate inner	$\frac{d}{dx}(3x+1) = 3$
Multiply	$5(3x+1)^4 \cdot 3 = 15(3x+1)^4$

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

Step	Work
Identify layers	Outer: $\sin(u)$ , Inner: $u = x^2$
Differentiate outer	$\cos(u) = \cos(x^2)$
Differentiate inner	$\frac{d}{dx}(x^2) = 2x$
Multiply	$\cos(x^2) \cdot 2x = 2x\cos(x^2)$

Example 3: Find $\frac{d}{dx}\sqrt{x^2 + 1}$

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	$\frac{1}{2}(x^2+1)^{-1/2} \cdot 2x = \frac{x}{\sqrt{x^2+1}}$

Example 4: Find $\frac{d}{dx}e^{-x^2}$

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

AP Tip: The most common Chain Rule error is forgetting to multiply by the inner derivative. Always ask: "Did I multiply by the derivative of what's inside?"

Apply the Chain Rule 🎯

Chain Rule Pattern Reference

Key Concept: Every basic derivative rule has a "chain rule version" where you multiply by the inner derivative.

Function	Without Chain Rule	With Chain Rule
$u^n$	$nx^{n-1}$	$n[g(x)]^{n-1} \cdot g'(x)$
$\sin u$	$\cos x$	$\cos(g(x)) \cdot g'(x)$
$\cos u$	$-\sin x$	$-\sin(g(x)) \cdot g'(x)$
$\tan u$	$\sec^2 x$	$\sec^2(g(x)) \cdot g'(x)$
$e^u$	$e^x$	$e^{g(x)} \cdot g'(x)$
$\ln u$	$\frac{1}{x}$	$\frac{g'(x)}{g(x)}$

The "Stuff" Method (Quick Shorthand)

Replace the inner function with "stuff":

Function	Derivative ("stuff" method)
$(\text{stuff})^n$	$n(\text{stuff})^{n-1} \cdot (\text{stuff})'$
$\sin(\text{stuff})$	$\cos(\text{stuff}) \cdot (\text{stuff})'$
$e^{\text{stuff}}$	$e^{\text{stuff}} \cdot (\text{stuff})'$
$\ln(\text{stuff})$	$\frac{(\text{stuff})'}{\text{stuff}}$

AP Tip: This shorthand method is how most students actually think about Chain Rule on the exam. Practice until it's automatic!

More Chain Rule Practice 🎯

Identify the Outer Function 🔍

For each composite function, select the correct outer function.

Chain Rule computation. ✍️

Key Takeaways — Part 1

$\boxed{\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)}$

Function Type	Derivative Pattern
$(\text{stuff})^n$	$n(\text{stuff})^{n-1} \cdot (\text{stuff})'$
$\sin(\text{stuff})$	$\cos(\text{stuff}) \cdot (\text{stuff})'$
$\cos(\text{stuff})$	$-\sin(\text{stuff}) \cdot (\text{stuff})'$
$e^{\text{stuff}}$	$e^{\text{stuff}} \cdot (\text{stuff})'$
$\ln(\text{stuff})$	$\frac{(\text{stuff})'}{\text{stuff}}$

The #1 Chain Rule mistake: Forgetting the inner derivative. ALWAYS multiply by $g'(x)$ !

Up Next: Part 2 — Nested Functions & the Double Chain Rule.

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:

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

Here we have three layers:

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

The general formula for nested compositions:

$\boxed{\frac{d}{dx}[f(g(h(x)))] = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)}$

Key Concept: Each layer contributes one factor. Count layers = count factors in the derivative.

Worked Examples — Layer-by-Layer

Example 1: $\frac{d}{dx}[\sin(3x)]^2$

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

Bonus: By the double-angle identity, $6\sin(3x)\cos(3x) = 3\sin(6x)$ .

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

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

Example 3: $\frac{d}{dx}\sqrt{\ln x} = \frac{d}{dx}(\ln x)^{1/2}$

Layer	Derivative
$(\cdot)^{1/2}$	$\frac{1}{2}(\ln x)^{-1/2}$
$\ln x$	$\frac{1}{x}$
Result	$\frac{1}{2x\sqrt{\ln x}}$

Example 4: $\frac{d}{dx}\sin^3(2x+1) = \frac{d}{dx}[\sin(2x+1)]^3$

Layer	Derivative
$u^3$	$3[\sin(2x+1)]^2$
$\sin(v)$	$\cos(2x+1)$
$2x+1$	$2$
Result	$6\sin^2(2x+1)\cos(2x+1)$

AP Tip: Students commonly write $2\sin(3x)\cos(3x)$ for $\frac{d}{dx}\sin^2(3x)$ — forgetting the factor of $3$ from the innermost layer. Always ask: "Is there another layer inside?"

Nested Chain Rule Practice 🎯

How Many Chain Rule Applications?

Key Fact: The number of chain rule applications = number of layers minus 1.

Function	Layers	Chain Rule Applications
$\sin(x)$	1	0
$\sin(5x)$	2 (sin, 5x)	1
$\sin^2(5x)$	3 (square, sin, 5x)	2
$e^{\sin^2(5x)}$	4 (exp, square, sin, 5x)	3

Strategy: Peel from the Outside In

$\boxed{\text{Step 1: Identify outermost operation. Step 2: Differentiate it. Step 3: Repeat for each inner layer. Step 4: Multiply all factors.}}$

Common Nested Patterns on the AP Exam

Pattern	Derivative
$\sin^n(ax)$	$na\sin^{n-1}(ax)\cos(ax)$
$e^{f(x)^2}$	$2f(x)f'(x)\,e^{f(x)^2}$
$\ln(f(g(x)))$	$\frac{g'(x) \cdot f'(g(x))}{f(g(x))}$
$[f(g(x))]^n$	$n[f(g(x))]^{n-1} \cdot f'(g(x)) \cdot g'(x)$

Multi-Layer Problems 🎯

How many Chain Rule applications? 🔍

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

Double Chain Rule computation. ✍️

Key Takeaways — Part 2

$\boxed{\frac{d}{dx}[f(g(h(x)))] = f'(g(h(x))) \cdot g'(h(x)) \cdot h'(x)}$

Mistake	Correct Approach
Stop after first layer	Multiply ALL layer derivatives
Confuse order of layers	Work outside → in
Forget innermost derivative	Always check the innermost layer
Differentiate the inner function	Leave the inner function unchanged inside the outer derivative

Up Next: Part 3 — Implicit Differentiation using the Chain Rule.

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:

$x^2 + y^2 = 25$

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.

The Key Idea

$\boxed{\text{Differentiate both sides with respect to } x.\text{ Every time you differentiate } y, \text{ multiply by } \frac{dy}{dx}.}$

Why? Because $y$ is implicitly a function of $x$ , so:

$\frac{d}{dx}[y^n] = ny^{n-1} \cdot \frac{dy}{dx}$

Key Concept: Implicit differentiation is just the Chain Rule applied to $y$ , treating $y$ as a function of $x$ .

Step-by-Step Method

Step	Action
1	Differentiate every term with respect to $x$
2	Apply Chain Rule to any term with $y$ (attach $\frac{dy}{dx}$ )
3	Use Product/Quotient Rule when $x$ and $y$ are multiplied/divided
4	Collect all $\frac{dy}{dx}$ terms on one side
5	Factor out $\frac{dy}{dx}$
6	Solve for $\frac{dy}{dx}$

Worked Example 1: Circle

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

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

Notice: the derivative depends on BOTH $x$ and $y$ . This is typical for implicit differentiation.

Worked Example 2: Folium of Descartes

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 $\frac{dy}{dx}$ 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.

Practice Implicit Differentiation 🎯

Implicit Differentiation with Trig Functions

Example 3: $\sin(y) = x$

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

Key Fact: This is exactly how we derive the formula $\frac{d}{dx}[\arcsin(x)] = \frac{1}{\sqrt{1-x^2}}$ . Since $\sin(y) = x$ , we know $\cos(y) = \sqrt{1-x^2}$ .

Tangent Line Applications

Example 4: Find the slope of the tangent line to $x^2 + xy + y^2 = 7$ at $(1, 2)$ .

Step	Work
Differentiate	$2x + y + x\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$
Collect $\frac{dy}{dx}$	$(x + 2y)\frac{dy}{dx} = -2x - y$
Solve	$\frac{dy}{dx} = \frac{-2x - y}{x + 2y}$
Plug in $(1,2)$	$\frac{dy}{dx} = \frac{-2(1) - 2}{1 + 2(2)} = \frac{-4}{5}$

$\boxed{\text{Tangent line: } y - 2 = -\frac{4}{5}(x - 1)}$

Which Rules Are Needed?

Term	Rule(s) Required
$x^n$ terms	Power Rule only
$y^n$ terms	Power Rule + Chain Rule ( $\frac{dy}{dx}$ )
$xy$ products	Product Rule + Chain Rule
$\sin(y)$ , $e^y$ , etc.	Chain Rule applied to $y$

Implicit Differentiation Applications 🎯

Second Derivatives (Implicit)

On the AP exam, you may be asked to find $\frac{d^2y}{dx^2}$ implicitly.

Example: For $x^2 + y^2 = 25$ , find $\frac{d^2y}{dx^2}$ .

We already found $\frac{dy}{dx} = -\frac{x}{y}$ . Now differentiate again using quotient rule:

$\frac{d^2y}{dx^2} = \frac{d}{dx}\left(-\frac{x}{y}\right) = -\frac{y(1) - x\frac{dy}{dx}}{y^2}$

Substitute $\frac{dy}{dx} = -\frac{x}{y}$ :

$= -\frac{y - x\left(-\frac{x}{y}\right)}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2} = -\frac{\frac{y^2+x^2}{y}}{y^2} = -\frac{x^2+y^2}{y^3}$

Since $x^2+y^2 = 25$ :

$\boxed{\frac{d^2y}{dx^2} = -\frac{25}{y^3}}$

AP Tip: When finding $\frac{d^2y}{dx^2}$ , substitute the original $\frac{dy}{dx}$ expression AND use the original equation to simplify. This is a common free-response technique.

Which differentiation rule is needed? 🔍

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

Implicit Differentiation computation. ✍️

Key Takeaways — Part 3

$\boxed{\text{Every time you differentiate } y, \text{ attach } \frac{dy}{dx}}$

Step	Action
1	Differentiate both sides w.r.t. $x$
2	Chain Rule on every $y$ term
3	Product Rule when $x$ and $y$ multiply
4	Collect, factor, solve for $\frac{dy}{dx}$

Common Errors:

Forgetting $\frac{dy}{dx}$ on $y$ terms
Missing the product rule on $xy$ terms
Plugging in the point too early (always find general $\frac{dy}{dx}$ first)

Up Next: Part 4 — Related Rates (using implicit differentiation with respect to time).

Part 4: Related Rates Intro

🔗 Related Rates

Part 4 of 7 — Related Rates

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.

Key Concept: Related Rates = Implicit Differentiation with respect to time instead of $x$ .

The Strategy

$\boxed{\text{Draw} \to \text{Equation} \to \text{Differentiate (w.r.t. } t\text{)} \to \text{Substitute} \to \text{Solve}}$

Step	Action	Details
1	Draw a picture	Label ALL changing quantities with variables
2	Write an equation	Relate the variables (geometry formulas, Pythagorean theorem, etc.)
3	Differentiate w.r.t. $t$	Every variable gets $\frac{d}{dt}$ (implicit diff)
4	Substitute known values	Plug in AFTER differentiating, never before!
5	Solve for unknown rate	Algebra

AP Tip (Critical): NEVER substitute numerical values before differentiating. This is the #1 related rates mistake.

Worked Example 1: Expanding Circle

A stone is dropped in a pond. The circular ripple expands so that its radius increases at $2$ ft/s. How fast is the area increasing when the radius is $5$ ft?

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	$\frac{dA}{dt} = 2\pi(5)(2) = 20\pi$ ft $^2$ /s

Worked Example 2: Ladder Problem

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

Step	Work
Setup	$x^2 + y^2 = 169$
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.

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. How fast is the water level dropping when the depth is 4 ft?

Similar triangles: $\frac{r}{h} = \frac{3}{6} = \frac{1}{2}$ , so $r = \frac{h}{2}$ .

Step	Work
Volume	$V = \frac{1}{3}\pi r^2 h = \frac{1}{3}\pi \left(\frac{h}{2}\right)^2 h = \frac{\pi h^3}{12}$
Differentiate	$\frac{dV}{dt} = \frac{\pi}{4} h^2 \frac{dh}{dt}$
Substitute	$-2 = \frac{\pi}{4}(16)\frac{dh}{dt}$
Solve	$\frac{dh}{dt} = \frac{-2}{4\pi} = -\frac{1}{2\pi}$ ft/min

Solve These Related Rates Problems 🎯

Essential Geometry Formulas for Related Rates

Shape	Formula	Differentiated
Circle area	$A = \pi r^2$	$\frac{dA}{dt} = 2\pi r\frac{dr}{dt}$
Circle circumference	$C = 2\pi r$	$\frac{dC}{dt} = 2\pi \frac{dr}{dt}$
Sphere volume	$V = \frac{4}{3}\pi r^3$	$\frac{dV}{dt} = 4\pi r^2\frac{dr}{dt}$
Sphere surface area	$S = 4\pi r^2$	$\frac{dS}{dt} = 8\pi r\frac{dr}{dt}$
Cone volume	$V = \frac{1}{3}\pi r^2 h$	Product rule needed
Right triangle	$a^2 + b^2 = c^2$	$2a\frac{da}{dt} + 2b\frac{db}{dt} = 2c\frac{dc}{dt}$
Rectangle area	$A = lw$	$\frac{dA}{dt} = l\frac{dw}{dt} + w\frac{dl}{dt}$

Common Related Rates Mistakes

Mistake	Why It's Wrong
Substituting values before differentiating	Turns variables into constants — derivative becomes 0
Forgetting to use similar triangles	Eliminates a variable (e.g., replacing $r$ with $h/2$ )
Missing the negative sign	Decreasing quantities have negative rates
Using wrong formula	Double-check: is it area, volume, or distance?

Advanced Related Rates 🎯

Related Rates Setup 🔍

Match each scenario with the correct geometric relationship.

Related Rates computation. ✍️

Key Takeaways — Part 4

$\boxed{\text{Related Rates = Implicit Differentiation with respect to } t}$

Problem Type	Key Formula
Expanding/contracting circle	$A = \pi r^2$
Sliding ladder	$x^2 + y^2 = L^2$
Filling/draining cone	$V = \frac{1}{3}\pi r^2 h$ + similar triangles
Balloon inflation	$V = \frac{4}{3}\pi r^3$
Separating objects	$d^2 = a^2 + b^2$

Remember: Differentiate FIRST, substitute AFTER!

Up Next: Part 5 — Advanced Chain Rule Applications (Logarithmic Differentiation & Inverse Trig).

Part 5: Advanced Chain Rule Applications

🔗 Advanced Chain Rule Applications

Part 5 of 7 — Logarithmic Differentiation & Inverse Trig

Logarithmic Differentiation

For functions like $y = x^x$ or $y = (\sin x)^{\cos x}$ , standard rules fail because both the base AND exponent depend on $x$ . Logarithmic differentiation handles these:

$\boxed{\text{Step 1: } \ln(\text{both sides}) \quad \text{Step 2: Simplify} \quad \text{Step 3: Differentiate implicitly} \quad \text{Step 4: Solve for } \frac{dy}{dx}}$

Key Concept: When do you need log differentiation?

Variable base AND variable exponent: $f(x)^{g(x)}$

Products/quotients of many factors (to simplify)

When to Use Each Technique

Situation	Technique
$[f(x)]^n$ (constant exponent)	Power Rule + Chain Rule
$a^{f(x)}$ (constant base)	$a^{f(x)} \ln(a) \cdot f'(x)$
$f(x)^{g(x)}$ (both variable)	Logarithmic Differentiation
Complex products/quotients	Logarithmic Differentiation (optional but easier)

Worked Example 1: $\frac{d}{dx}x^x$

Step	Work
Let $y = x^x$	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	$\frac{dy}{dx} = y(\ln x + 1) = x^x(\ln x + 1)$

$\boxed{\frac{d}{dx}x^x = x^x(\ln x + 1)}$

Worked Example 2: Simplifying Complex Products

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

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

AP Tip: Log differentiation is rarely tested directly on AP Calc AB, but it's an important tool for AP Calc BC and is excellent for building understanding.

Logarithmic Differentiation 🎯

Chain Rule with Inverse Trig Functions

The inverse trig derivatives all require the Chain Rule when the argument is a composite:

Function	Derivative (with Chain Rule)
$\arcsin(u)$	$\frac{u'}{\sqrt{1-u^2}}$
$\arccos(u)$	$\frac{-u'}{\sqrt{1-u^2}}$
$\arctan(u)$	$\frac{u'}{1+u^2}$
$\text{arccot}(u)$	$\frac{-u'}{1+u^2}$
$\text{arcsec}(u)$	$\frac{u'}{
$\text{arccsc}(u)$	$\frac{-u'}{

Key Fact: On the AP exam, $\arctan$ is the most commonly tested inverse trig function. Know its derivative cold.

Worked Examples

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

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

Example 5: $\frac{d}{dx}\arctan(e^x) = \frac{e^x}{1+(e^x)^2} = \frac{e^x}{1+e^{2x}}$

Inverse Trig Derivatives 🎯

Match the derivative technique 🔍

For each function, select the best approach.

Inverse trig computation. ✍️

Key Takeaways — Part 5

Technique	When to Use	Formula
Log Differentiation	$f(x)^{g(x)}$	$\ln$ both sides → implicit diff
Inverse Trig + Chain Rule	$\arcsin(u)$ , $\arctan(u)$ , etc.	Standard formulas × $u'$
Exponential ( $a^{f(x)}$ )	Constant base	$a^{f(x)} \ln(a) \cdot f'(x)$

$\boxed{\frac{d}{dx}[\arctan(u)] = \frac{u'}{1+u^2} \qquad \frac{d}{dx}[\arcsin(u)] = \frac{u'}{\sqrt{1-u^2}}}$

Up Next: Part 6 — Problem-Solving Workshop with mixed Chain Rule problems.

Part 6: Mixed Chain Rule Problems

🔗 Problem-Solving Workshop

Part 6 of 7 — Mixed Chain Rule Problems

Decision Framework

Every derivative problem begins with the same question: What is the outermost operation?

$\boxed{\text{Identify outermost operation} \to \text{Apply rule} \to \text{Chain Rule for inner layers}}$

Outermost Operation	Primary Rule	Then Apply
Sum/difference	Sum Rule	Chain Rule to each term
Product $f \cdot g$	Product Rule	Chain Rule inside each factor
Quotient $f/g$	Quotient Rule	Chain Rule inside each part
Composition $f(g(x))$	Chain Rule directly	Continue peeling layers

Key Strategy: Work from the OUTSIDE IN. The outermost operation determines which rule to start with.

Combining Chain Rule with Other Rules

Example 1: Product + Chain

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

Example 2: Quotient + Chain

$\frac{d}{dx}\left[\frac{e^{2x}}{x+1}\right] = \frac{2e^{2x}(x+1) - e^{2x}}{(x+1)^2} = \frac{e^{2x}(2x+1)}{(x+1)^2}$

Example 3: Chain on $\ln$

$\frac{d}{dx}[\ln(\cos x)] = \frac{-\sin x}{\cos x} = -\tan x$

Important: $\frac{d}{dx}\frac{\sin x}{e^x}$ (Shortcut)

$\frac{d}{dx}\left[\frac{\sin x}{e^x}\right] = \frac{d}{dx}[\sin x \cdot e^{-x}] = \cos x \cdot e^{-x} + \sin x \cdot (-e^{-x}) = \frac{\cos x - \sin x}{e^x}$

AP Tip: Sometimes rewriting $\frac{f(x)}{e^x}$ as $f(x) \cdot e^{-x}$ and using Product Rule is easier than Quotient Rule.

AP-Style Problems — Set 1 🎯

Table-Based Chain Rule Problems

On the AP exam, you may be given a table of values and asked to compute a composite function's derivative.

Key Fact: If $h(x) = f(g(x))$ , then $h'(a) = f'(g(a)) \cdot g'(a)$ . You need: $g(a)$ from the table, then $f'$ at that value, then $g'(a)$ .

$x$	$f(x)$	$f'(x)$	$g(x)$	$g'(x)$
1	3	$-2$	2	4
2	5	1	1	$-3$
3	$-1$	6	3	2

Example: Find $h'(1)$ where $h(x) = f(g(x))$ .

$h'(1) = f'(g(1)) \cdot g'(1) = f'(2) \cdot 4 = 1 \cdot 4 = 4$

Example: Find $k'(2)$ where $k(x) = g(f(x))$ .

$k'(2) = g'(f(2)) \cdot f'(2) = g'(5) \cdot 1$

But $g'(5)$ is not in the table — insufficient information!

AP Tip: Always check that the required values are in the table before computing. If $g(a)$ gives a value not in the table, you can't find $f'(g(a))$ .

Table-Based & Mixed Problems 🎯

Use the table from the previous section.

Subtle Distinctions

These look similar but have very different derivatives:

Function	Derivative	Rule Used
$(5x)^3 = 125x^3$	$375x^2$	Chain Rule (or expand)
$5x^3$	$15x^2$	Constant Multiple
$5^x$	$5^x \ln 5$	Exponential
$x^5$	$5x^4$	Power Rule
$5^3 = 125$	$0$	Constant

Key Principle: Know the difference between: constant exponent (Power Rule), constant base (Exponential Rule), both variable (Log Differentiation).

FTC Part 1 + Chain Rule

$\boxed{\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x)) \cdot g'(x)}$

Example: $\frac{d}{dx}\int_0^{x^2} \sin(t)\,dt = \sin(x^2) \cdot 2x = 2x\sin(x^2)$

Identify the correct derivative 🔍

Match each function to its derivative.

Table-based Chain Rule computation. ✍️

Workshop Complete!

Skill	Practiced
Product Rule + Chain Rule	$x^2\sin(3x)$
Quotient Rule + Chain Rule	$\frac{e^{2x}}{x+1}$
Chain on logarithms	$\ln(\cos x) = -\tan x$
Table-based derivatives	$h(x) = f(g(x))$
FTC + Chain Rule	$\frac{d}{dx}\int_0^{g(x)} f(t)\,dt$
Subtle distinctions	$(5x)^3$ vs $5x^3$ vs $5^x$

Up Next: Part 7 — Comprehensive Review & Final Assessment.

Part 7: Chain Rule Review

🔗 Chain Rule — Comprehensive Review

Part 7 of 7 — Review & Final Assessment

Complete Chain Rule Summary

$\boxed{\frac{d}{dx}[f(g(x))] = f'(g(x)) \cdot g'(x)}$

Scenario	Technique	Key Formula
Basic composition	Chain Rule	$f'(g(x)) \cdot g'(x)$
Nested (3+ layers)	Repeated Chain Rule	Multiply ALL layer derivatives
Implicit ( $y$ as fn of $x$ )	Implicit Differentiation	Attach $\frac{dy}{dx}$ to every $y$ term
Rates changing w.r.t. time	Related Rates	Differentiate w.r.t. $t$
$f(x)^{g(x)}$	Log Differentiation	$\ln$ both sides → implicit diff
$\arcsin(u)$ , $\arctan(u)$	Inverse Trig + Chain	Standard formula $\times u'$
FTC Part 1	FTC + Chain	$f(g(x)) \cdot g'(x)$

AP Exam Frequency

Key Fact: The Chain Rule appears in ~80% of all derivative problems on the AP exam. It is embedded in:

All implicit differentiation problems

All related rates problems

FTC Part 1 with variable upper limit

Most trig, exponential, and logarithmic derivatives

Quick Reference — All Chain Rule Patterns

Function	Derivative
$[g(x)]^n$	$n[g(x)]^{n-1} \cdot g'(x)$
$\sin(g(x))$	$\cos(g(x)) \cdot g'(x)$
$\cos(g(x))$	$-\sin(g(x)) \cdot g'(x)$
$\tan(g(x))$	$\sec^2(g(x)) \cdot g'(x)$
$e^{g(x)}$	$e^{g(x)} \cdot g'(x)$
$a^{g(x)}$	$a^{g(x)} \ln(a) \cdot g'(x)$
$\ln(g(x))$	$\frac{g'(x)}{g(x)}$
$\arcsin(g(x))$	$\frac{g'(x)}{\sqrt{1-[g(x)]^2}}$
$\arctan(g(x))$	$\frac{g'(x)}{1+[g(x)]^2}$

Common Errors to Avoid

Error	Example	Correct
Forgetting inner derivative	$\frac{d}{dx}\sin(3x) = \cos(3x)$	$3\cos(3x)$
Confusing constant vs variable exponent	Treating $2^x$ like $x^2$	$2^x\ln 2$ vs $2x$
Missing Product Rule	$\frac{d}{dx}[x\sin(x)] = \cos(x)$	$\sin(x) + x\cos(x)$
Forgetting $\frac{dy}{dx}$ in implicit	$2y$ instead of $2y\frac{dy}{dx}$	Always attach $\frac{dy}{dx}$
Substituting before differentiating	In related rates	Always differentiate first

Comprehensive Assessment — Part A 🎯

No hints — test your mastery.

Comprehensive Assessment — Part B 🎯

FTC Part 1 + Chain Rule

This is one of the most important AP Calculus formulas:

$\boxed{\frac{d}{dx}\int_a^{g(x)} f(t)\,dt = f(g(x)) \cdot g'(x)}$

Example 1: $\frac{d}{dx}\int_0^{x^2} \sin(t)\,dt = \sin(x^2) \cdot 2x = 2x\sin(x^2)$

Example 2: $\frac{d}{dx}\int_1^{e^x} \frac{1}{t}\,dt = \frac{1}{e^x} \cdot e^x = 1$

Example 3: Both limits variable:

$\frac{d}{dx}\int_{x}^{x^2} t^3\,dt = (x^2)^3 \cdot 2x - x^3 \cdot 1 = 2x^7 - x^3$

AP Tip: FTC + Chain Rule appears on nearly every AP exam. The key is: "plug in the upper limit for $t$ , then multiply by the derivative of that upper limit."

Final Matching 🔍

Select the correct derivative.

Final Challenge ✍️

Chain Rule — Complete! ✅

Topic	Mastered
Basic Chain Rule	$f'(g(x)) \cdot g'(x)$
Nested Functions	Multiply all layer derivatives
Implicit Differentiation	Attach $\frac{dy}{dx}$ to $y$ terms
Related Rates	Differentiate w.r.t. $t$
Log Differentiation	$\ln$ both sides for $f(x)^{g(x)}$
Inverse Trig + Chain	Standard formulas $\times u'$
FTC + Chain Rule	$f(g(x)) \cdot g'(x)$
Table-Based Problems	Look up $g(a)$ , then $f'(g(a))$

The Chain Rule is the single most important differentiation technique. You are now ready to tackle any derivative problem on the AP exam!

Chain Rule

Chain Rule - Complete Interactive Lesson

Part 1: Chain Rule Basics

🔗 The Chain Rule

Why Do We Need the Chain Rule?

The Chain Rule Formula

Leibniz Notation

Worked Examples — Step by Step

Chain Rule Pattern Reference

The "Stuff" Method (Quick Shorthand)

Key Takeaways — Part 1

Part 2: Nested Functions & Double Chain Rule

🔗 Nested Functions & Double Chain Rule

When the Chain Rule Applies Twice

Worked Examples — Layer-by-Layer

How Many Chain Rule Applications?

Strategy: Peel from the Outside In

Common Nested Patterns on the AP Exam

Key Takeaways — Part 2

Part 3: Implicit Differentiation

🔗 Implicit Differentiation

What Is Implicit Differentiation?

The Key Idea

Step-by-Step Method

Worked Example 1: Circle

Worked Example 2: Folium of Descartes

Implicit Differentiation with Trig Functions

Tangent Line Applications

Which Rules Are Needed?

Second Derivatives (Implicit)

Key Takeaways — Part 3

Part 4: Related Rates Intro

🔗 Related Rates

What Are Related Rates?

The Strategy

Worked Example 1: Expanding Circle

Worked Example 2: Ladder Problem

Worked Example 3: Conical Tank

Essential Geometry Formulas for Related Rates

Common Related Rates Mistakes

Key Takeaways — Part 4

Part 5: Advanced Chain Rule Applications

🔗 Advanced Chain Rule Applications

Logarithmic Differentiation

When to Use Each Technique

Worked Example 1: ddxxx\frac{d}{dx}x^xdxd​xx

Worked Example 2: Simplifying Complex Products

Chain Rule with Inverse Trig Functions

Worked Examples

Key Takeaways — Part 5

Part 6: Mixed Chain Rule Problems

🔗 Problem-Solving Workshop

Decision Framework

Combining Chain Rule with Other Rules

Important: ddxsin⁡xex\frac{d}{dx}\frac{\sin x}{e^x}dxd​exsinx​ (Shortcut)

Table-Based Chain Rule Problems

Subtle Distinctions

FTC Part 1 + Chain Rule

Workshop Complete!

Part 7: Chain Rule Review

🔗 Chain Rule — Comprehensive Review

Complete Chain Rule Summary

AP Exam Frequency

Quick Reference — All Chain Rule Patterns

Common Errors to Avoid

FTC Part 1 + Chain Rule

Chain Rule — Complete! ✅

Worked Example 1: $\frac{d}{dx}x^x$

Important: $\frac{d}{dx}\frac{\sin x}{e^x}$ (Shortcut)