Basic Differentiation Rules - Complete Interactive Lesson

Part 1: The Power Rule

📐 Basic Differentiation Rules

Part 1 of 7 — The Power Rule & Foundational Rules

Welcome to differentiation! This topic covers the rules you'll use on every single calculus problem.

Part	Topic
1	Power Rule & Foundational Rules
2	Product Rule
3	Quotient Rule
4	Trigonometric Derivatives
5	Higher-Order Derivatives
6	Mixed Differentiation Problems
7	Comprehensive Review & AP Applications

The Power Rule

The single most important differentiation rule:

$\boxed{\frac{d}{dx}[x^n] = n x^{n-1} \quad \text{for any real } n}$

Key Fact: The Power Rule works for ALL real exponents — positive, negative, fractional, irrational, even $n = 0$ . When $n = 0$ : $\frac{d}{dx}[x^0] = \frac{d}{dx}[1] = 0$ .

Step-by-Step Process

Identify the exponent $n$
Bring $n$ down as a coefficient (multiply)
Subtract 1 from the exponent

Function	Exponent $n$	Derivative
$x^5$	$5$	$5x^4$
$x^{100}$	$100$	$100x^{99}$
$x^1 = x$	$1$	$1 \cdot x^0 = 1$
$x^0 = 1$	$0$	$0 \cdot x^{-1} = 0$

Constant Rule & Constant Multiple Rule

$\boxed{\frac{d}{dx}[c] = 0 \qquad \frac{d}{dx}[c \cdot f(x)] = c \cdot f'(x)}$

Constants vanish when differentiated. Constants attached to functions "pass through" the derivative.

Function	Rule Applied	Derivative
$7$	Constant Rule	$0$
$\pi^2$	Constant Rule	$0$
$7x^3$	Constant Multiple	$7 \cdot 3x^2 = 21x^2$
$-4x^5$	Constant Multiple	$-4 \cdot 5x^4 = -20x^4$
$\frac{1}{2}x^8$	Constant Multiple	$\frac{1}{2} \cdot 8x^7 = 4x^7$

⚠️ Common Mistake: Students sometimes think $\frac{d}{dx}[\pi^2] = 2\pi$ . Remember: $\pi$ is a constant, not a variable! The derivative is $0$ .

Sum & Difference Rule

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

Differentiate each term independently — linearity of the derivative.

Worked Example

Find $\frac{d}{dx}(3x^4 - 5x^2 + 7x - 2)$

Term	Power Rule	Result
$3x^4$	$3(4x^3)$	$12x^3$
$-5x^2$	$-5(2x)$	$-10x$
$7x$	$7(1)$	$7$
$-2$	constant	$0$

$\frac{d}{dx}(3x^4 - 5x^2 + 7x - 2) = 12x^3 - 10x + 7$

Apply the Power Rule 🎯

Negative and Fractional Exponents

Key Principle: Before applying the Power Rule, rewrite all roots, fractions, and radicals using exponential notation.

Rewrite Rules:

$\frac{1}{x^n} = x^{-n} \qquad \sqrt[n]{x^m} = x^{m/n}$

Original	Rewrite	Power Rule	Final Form
$\frac{1}{x^3}$	$x^{-3}$	$-3x^{-4}$	$-\frac{3}{x^4}$
$\frac{5}{x^2}$	$5x^{-2}$	$-10x^{-3}$	$-\frac{10}{x^3}$
$\sqrt{x}$	$x^{1/2}$	$\frac{1}{2}x^{-1/2}$	$\frac{1}{2\sqrt{x}}$
$\sqrt[3]{x^2}$	$x^{2/3}$	$\frac{2}{3}x^{-1/3}$	$\frac{2}{3\sqrt[3]{x}}$
$\frac{1}{\sqrt{x}}$	$x^{-1/2}$	$-\frac{1}{2}x^{-3/2}$	$-\frac{1}{2x\sqrt{x}}$

Worked Example

Find $\frac{d}{dx}\left(\frac{3}{x^2} + 4\sqrt{x} - \frac{7}{\sqrt[3]{x}}\right)$

Step 1 — Rewrite: $3x^{-2} + 4x^{1/2} - 7x^{-1/3}$

Step 2 — Differentiate: $-6x^{-3} + 2x^{-1/2} + \frac{7}{3}x^{-4/3}$

Step 3 — Simplify: $-\frac{6}{x^3} + \frac{2}{\sqrt{x}} + \frac{7}{3\sqrt[3]{x^4}}$

AP Tip: On the AP exam, you do NOT need to simplify your answer. Leaving the derivative in negative-exponent form is perfectly acceptable and saves time!

Negative & Fractional Exponents 🎯

Special Derivatives to Memorize

Beyond the Power Rule, these constants arise frequently:

$\boxed{\frac{d}{dx}[e^x] = e^x \qquad \frac{d}{dx}[\ln x] = \frac{1}{x}}$

$\boxed{\frac{d}{dx}[a^x] = a^x \ln a \qquad \frac{d}{dx}[\log_a x] = \frac{1}{x \ln a}}$

Function	Derivative	Note
$e^x$	$e^x$	Only function equal to its derivative
$\ln x$	$\frac{1}{x}$	Domain: $x > 0$
$2^x$	$2^x \ln 2$	General exponential pattern
$10^x$	$10^x \ln 10$	Common in applications
$\log_{10} x$	$\frac{1}{x \ln 10}$	Common log derivative

Key Fact: $e^x$ is the only function that equals its own derivative (up to constant multiples). This is why $e$ is so special in calculus!

Match each function to its derivative.

Find the derivative and evaluate. ✍️

Key Takeaways — Part 1

Rule	Formula
Power Rule	$\frac{d}{dx}[x^n] = nx^{n-1}$
Constant Rule	$\frac{d}{dx}[c] = 0$
Constant Multiple	$\frac{d}{dx}[cf(x)] = cf'(x)$
Sum/Difference	$\frac{d}{dx}[f \pm g] = f' \pm g'$
Exponential	$\frac{d}{dx}[e^x] = e^x$
Natural Log	$\frac{d}{dx}[\ln x] = \frac{1}{x}$

Workflow for any polynomial/power derivative:

Rewrite all roots and fractions as power expressions
Apply Power Rule term by term (bring down exponent, subtract 1)
Simplify if desired (not required on AP exam)

Up Next: Part 2 — The Product Rule, for when two functions are multiplied together.

Part 2: Product Rule

📐 The Product Rule

Part 2 of 7 — Product Rule

Why Can't We Just Multiply the Derivatives?

A common (and dangerous) mistake:

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

Quick proof it fails: $\frac{d}{dx}[x \cdot x] = \frac{d}{dx}[x^2] = 2x$ , but $1 \cdot 1 = 1 \neq 2x$ . ✗

The Product Rule

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

Memory aids:

"Derivative of first times second, plus first times derivative of second"
Short form: $(fg)' = f'g + fg'$
Leibniz form: $d(uv) = u\,dv + v\,du$

Key Fact: The Product Rule comes from the limit definition. The "extra" term $fg'$ accounts for the fact that both factors are changing simultaneously.

Worked Examples — Product Rule

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

Component	Value
$f = x^2$	$f' = 2x$
$g = \sin x$	$g' = \cos x$
$f'g + fg'$	$2x \sin x + x^2 \cos x$

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

Example 2: Find $\frac{d}{dx}[e^x \ln x]$

Component	Value
$f = e^x$	$f' = e^x$
$g = \ln x$	$g' = \frac{1}{x}$
$f'g + fg'$	$e^x \ln x + e^x \cdot \frac{1}{x}$

$= e^x\left(\ln x + \frac{1}{x}\right)$

AP Tip: Factor common terms in your final answer when possible. Graders appreciate clean answers, and factoring helps with sign analysis later.

Example 3: Find $\frac{d}{dx}[xe^x]$ — the most commonly tested product!

$\frac{d}{dx}[xe^x] = (1)e^x + x(e^x) = e^x(1 + x)$

At $x = 0$ : $f'(0) = e^0(1+0) = 1$ . At $x = -1$ : $f'(-1) = e^{-1}(0) = 0$ (this is a critical point!)

Apply the Product Rule 🎯

When to Use Product Rule vs. Expand First

Situation	Strategy	Why
Both factors are polynomials	Expand first, then Power Rule	Faster and simpler
One factor is $e^x$ , $\sin x$ , $\ln x$	Product Rule (must use)	Can't combine unlike functions
One factor is a constant	Pull constant out	Constant Multiple Rule suffices
Very complex product	Product Rule	Expansion would be unwieldy

Examples:

Expression	Best Strategy
$(2x+1)(x^2-3)$	Expand: $2x^3 + x^2 - 6x - 3$
$x^5 e^x$	Product Rule (can't expand)
$\sin x \cdot \cos x$	Product Rule (or use $\frac{1}{2}\sin 2x$ )
$3(x^4 + 2x)$	Constant Multiple: $3(4x^3 + 2)$
$(x^2+1)^2$	Expand: $x^4 + 2x^2 + 1$

Product Rule with Tables (AP Exam Favorite!)

If $f(2) = 3$ , $f'(2) = -1$ , $g(2) = 4$ , $g'(2) = 5$ , find $\frac{d}{dx}[f(x)g(x)]$ at $x = 2$ :

$f'(2)g(2) + f(2)g'(2) = (-1)(4) + (3)(5) = -4 + 15 = 11$

AP Tip: Table-based derivative problems appear on almost EVERY AP exam. Practice reading values from tables and plugging into Product Rule.

Product Rule Challenge 🎯

Extended Product Rule — Three or More Factors

For three functions:

$\frac{d}{dx}[f \cdot g \cdot h] = f'gh + fg'h + fgh'$

Pattern: Each factor takes a turn being differentiated while the others stay.

Example: Find $\frac{d}{dx}[x^2 \sin x \cdot e^x]$

$= 2x \sin x \cdot e^x + x^2 \cos x \cdot e^x + x^2 \sin x \cdot e^x$

$= e^x(2x\sin x + x^2 \cos x + x^2 \sin x)$

Key Concept: You can also apply the two-factor Product Rule twice: treat $(fg)$ as one function and apply Product Rule with $h$ . This gives the same result.

Product Rule with given values.

Product Rule computation. ✍️

Key Takeaways — Part 2

Concept	Detail
Product Rule	$(fg)' = f'g + fg'$
Common Error	$(fg)' \neq f'g'$ — NEVER multiply derivatives
Strategy	Expand polynomials first when possible
Factoring	Factor common terms ( $e^x$ , $x$ , etc.) for clean answers
Table Problems	Plug given values directly into the formula
Triple Product	$f'gh + fg'h + fgh'$ — each factor takes a turn

Up Next: Part 3 — The Quotient Rule, for derivatives of fractions.

Part 3: Quotient Rule

📐 The Quotient Rule

Part 3 of 7 — Quotient Rule

The Quotient Rule

$\boxed{\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right] = \frac{f'(x)\,g(x) - f(x)\,g'(x)}{[g(x)]^2}}$

Memory aids:

"Low d-High minus High d-Low, all over Low squared"
Short form: $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$

⚠️ Critical Warning: The minus sign in the numerator is the #1 source of errors. The order matters — it's $f'g$ MINUS $fg'$ , not the other way around. Unlike the Product Rule, the Quotient Rule is NOT symmetric!

Comparison: Product Rule vs. Quotient Rule

Rule	Formula	Sign
Product	$f'g + fg'$	Plus between terms
Quotient	$\frac{f'g - fg'}{g^2}$	Minus between terms

Worked Examples

Example 1: Find $\frac{d}{dx}\frac{x^2}{\sin x}$

Component	Value
$f = x^2$	$f' = 2x$
$g = \sin x$	$g' = \cos x$
$\frac{f'g - fg'}{g^2}$	$\frac{2x\sin x - x^2\cos x}{\sin^2 x}$

$\frac{d}{dx}\frac{x^2}{\sin x} = \frac{2x\sin x - x^2\cos x}{\sin^2 x}$

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

$= \frac{e^x(x+1) - e^x(1)}{(x+1)^2} = \frac{e^x(x+1-1)}{(x+1)^2} = \frac{xe^x}{(x+1)^2}$

AP Tip: Always look for common factors in the numerator after applying the Quotient Rule. Simplifying makes it easier to find critical points and sign analysis.

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

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

Key Fact: When $\frac{d}{dx}\frac{x}{x+c}$ yields a positive constant over a square, the function is always increasing. This is useful for sign analysis!

Apply the Quotient Rule 🎯

When to Avoid the Quotient Rule

The Quotient Rule is powerful but often overkill. Use smarter alternatives when possible:

Situation	Better Strategy	Example
Denominator is a constant	Constant Multiple Rule	$\frac{x^3+2x}{5} = \frac{1}{5}(3x^2+2)$
Denominator is a power of $x$	Rewrite as negative exponent	$\frac{3}{x^4} = 3x^{-4} \to -12x^{-5}$
Can split the fraction	Divide term by term	$\frac{x^3+x}{x^2} = x + x^{-1}$
Numerator is a constant	Rewrite as negative exponent	$\frac{5}{x+1}$ — must use Q.R. here

Splitting Fractions — A Powerful Technique

$\frac{x^3 + 6x^2 - 2}{x^2} = x + 6 - 2x^{-2}$

Now differentiate term by term: $1 + 0 + 4x^{-3} = 1 + \frac{4}{x^3}$

Compare to using Quotient Rule on the original — much more work for the same answer!

Deriving Trig Derivatives via Quotient Rule

$\frac{d}{dx}[\tan x] = \frac{d}{dx}\frac{\sin x}{\cos x} = \frac{\cos x \cdot \cos x - \sin x(-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x$

Key Concept: The Quotient Rule is how we derive the derivatives of $\tan x$ , $\cot x$ , $\sec x$ , and $\csc x$ from $\sin x$ and $\cos x$ .

Quotient Rule Mastery 🎯

Quotient Rule with Tables (AP Exam Staple)

Given:

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

Find $\frac{d}{dx}\left[\frac{f(x)}{g(x)}\right]$ at $x = 1$ :

$\frac{f'(1)g(1) - f(1)g'(1)}{[g(1)]^2} = \frac{(-2)(4) - (3)(5)}{4^2} = \frac{-8 - 15}{16} = \frac{-23}{16}$

Find $\frac{d}{dx}\left[\frac{g(x)}{f(x)}\right]$ at $x = 2$ :

$\frac{g'(2)f(2) - g(2)f'(2)}{[f(2)]^2} = \frac{(-3)(-1) - (2)(6)}{(-1)^2} = \frac{3 - 12}{1} = -9$

AP Tip: Watch for problems that ask for $\frac{d}{dx}\left[\frac{g}{f}\right]$ instead of $\frac{d}{dx}\left[\frac{f}{g}\right]$ — swapping the roles of $f$ and $g$ is a common trap!

Choose the best differentiation strategy.

Quotient Rule computation. ✍️

Key Takeaways — Part 3

Concept	Detail
Quotient Rule	$\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$
Order matters	$f'g - fg'$ (NOT $fg' - f'g$ )
Avoid when possible	Rewrite as negative exponents or split fractions
Constant denominator	Just use Constant Multiple Rule
Table problems	Plug values directly into formula
Trig connection	Derives $\tan, \cot, \sec, \csc$ derivatives

Decision Tree: Which Rule?

Expression Type	Rule to Use
$f \cdot g$	Product Rule
$\frac{f}{g}$ (both non-trivial)	Quotient Rule
$\frac{f}{\text{constant}}$	Constant Multiple
$\frac{\text{constant}}{x^n}$	Power Rule rewrite
$\frac{\text{polynomial}}{x^n}$	Split, then Power Rule

Up Next: Part 4 — Trigonometric Derivatives in depth.

Part 4: Trig Derivatives

📐 Trigonometric Derivatives

Part 4 of 7 — Trig Derivatives

The Six Trigonometric Derivatives

These must be memorized perfectly for the AP exam:

$\boxed{\frac{d}{dx}[\sin x] = \cos x \qquad \frac{d}{dx}[\cos x] = -\sin x}$

$\boxed{\frac{d}{dx}[\tan x] = \sec^2 x \qquad \frac{d}{dx}[\cot x] = -\csc^2 x}$

$\boxed{\frac{d}{dx}[\sec x] = \sec x \tan x \qquad \frac{d}{dx}[\csc x] = -\csc x \cot x}$

Pattern Recognition — The Negative Sign Rule

Key Fact: The co-functions (cos, cot, csc) ALL have negative derivatives. The regular functions (sin, tan, sec) have positive derivatives.

Regular Function	Derivative (Positive)	Co-Function	Derivative (Negative)
$\sin x$	$\cos x$	$\cos x$	$-\sin x$
$\tan x$	$\sec^2 x$	$\cot x$	$-\csc^2 x$
$\sec x$	$\sec x \tan x$	$\csc x$	$-\csc x \cot x$

Another Pattern — Squared vs. Product

Function	Derivative Type
$\tan x$ → $\sec^2 x$	Squared function
$\cot x$ → $-\csc^2 x$	Squared function
$\sec x$ → $\sec x \tan x$	Product of two trig functions
$\csc x$ → $-\csc x \cot x$	Product of two trig functions

Worked Examples — Basic Trig Derivatives

Problem	Solution	Rule Used
$\frac{d}{dx}(3\sin x + 2\cos x)$	$3\cos x - 2\sin x$	Constant Multiple + Sum
$\frac{d}{dx}(x^2 + \tan x)$	$2x + \sec^2 x$	Power + Trig
$\frac{d}{dx}(5\sec x)$	$5\sec x \tan x$	Constant Multiple
$\frac{d}{dx}(-\csc x + \pi)$	$\csc x \cot x$	Trig + Constant

Key Angle Values Reference

Angle	$\sin$	$\cos$	$\tan$	$\sec$
$0$	$0$	$1$	$0$	$1$
$\frac{\pi}{6}$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{3}}{3}$	$\frac{2\sqrt{3}}{3}$
$\frac{\pi}{4}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$	$1$	$\sqrt{2}$
$\frac{\pi}{3}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$	$2$
$\frac{\pi}{2}$	$1$	$0$	undef	undef

AP Tip: You need instant recall of trig values at these angles. The derivative questions almost always evaluate at one of these special angles.

Trig Derivatives 🎯

Combining Trig Derivatives with Product & Quotient Rules

Example 1: $\frac{d}{dx}(e^x \sin x)$

Product Rule: $e^x \sin x + e^x \cos x = e^x(\sin x + \cos x)$

At $x = 0$ : $e^0(0 + 1) = 1$

Example 2: $\frac{d}{dx}\left(\frac{\tan x}{x}\right)$

Quotient Rule: $\frac{\sec^2 x \cdot x - \tan x \cdot 1}{x^2} = \frac{x\sec^2 x - \tan x}{x^2}$

Example 3: $\frac{d}{dx}(x^2 \sec x)$

Product Rule: $2x \sec x + x^2 \sec x \tan x = x\sec x(2 + x\tan x)$

Key Concept: When combining trig derivatives with Product/Quotient Rule, always set up the table ( $f, f', g, g'$ ) to stay organized and avoid sign errors.

Mixed Trig Problems 🎯

Where Do Trig Derivatives Come From?

The derivatives of $\sin x$ and $\cos x$ come from the limit definition:

$\frac{d}{dx}[\sin x] = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h}$

Using the angle addition formula $\sin(x+h) = \sin x \cos h + \cos x \sin h$ :

$= \lim_{h \to 0} \frac{\sin x(\cos h - 1) + \cos x \sin h}{h} = \sin x \cdot 0 + \cos x \cdot 1 = \cos x$

This relies on the special limits: $\lim_{h \to 0}\frac{\sin h}{h} = 1$ and $\lim_{h \to 0}\frac{\cos h - 1}{h} = 0$ .

The other four come from $\sin x$ and $\cos x$ :

Derivative	Derived Using
$\frac{d}{dx}[\tan x] = \sec^2 x$	Quotient Rule on $\frac{\sin x}{\cos x}$
$\frac{d}{dx}[\cot x] = -\csc^2 x$	Quotient Rule on $\frac{\cos x}{\sin x}$
$\frac{d}{dx}[\sec x] = \sec x \tan x$	Quotient Rule on $\frac{1}{\cos x}$
$\frac{d}{dx}[\csc x] = -\csc x \cot x$	Quotient Rule on $\frac{1}{\sin x}$

Complete the derivative.

Trig derivative evaluation. ✍️

Key Takeaways — Part 4

Must Memorize	Derivative
$\sin x$	$\cos x$
$\cos x$	$-\sin x$
$\tan x$	$\sec^2 x$
$\cot x$	$-\csc^2 x$
$\sec x$	$\sec x \tan x$
$\csc x$	$-\csc x \cot x$

Memory checklist:

Co-functions → negative sign (cos, cot, csc)
tan and cot → squared results ( $\sec^2$ , $\csc^2$ )
sec and csc → product results (sec·tan, csc·cot)
Know exact trig values at $0, \frac{\pi}{6}, \frac{\pi}{4}, \frac{\pi}{3}, \frac{\pi}{2}$

Up Next: Part 5 — Higher-Order Derivatives.

Part 5: Higher-Order Derivatives

📐 Higher-Order Derivatives

Part 5 of 7 — Higher-Order Derivatives

What Are Higher-Order Derivatives?

The second derivative is the derivative of the derivative:

$\boxed{f''(x) = \frac{d^2 y}{dx^2} = \frac{d}{dx}\left[\frac{dy}{dx}\right]}$

Notation Comparison

Order	Prime Notation	Leibniz Notation	Other
1st	$f'(x)$	$\frac{dy}{dx}$	$\dot{y}$ (physics)
2nd	$f''(x)$	$\frac{d^2y}{dx^2}$	$\ddot{y}$ (physics)
3rd	$f'''(x)$	$\frac{d^3y}{dx^3}$	—
$n$ th	$f^{(n)}(x)$	$\frac{d^ny}{dx^n}$	—

Key Fact: For $n \geq 4$ , we write $f^{(n)}(x)$ with parentheses to avoid confusion with powers: $f^{(4)}(x)$ is the 4th derivative, not $[f(x)]^4$ .

Physical Interpretation — Motion

Derivative	In Motion Context	Units (if position in meters, time in seconds)
$s(t)$	Position	meters
$s'(t) = v(t)$	Velocity	m/s
$s''(t) = a(t)$	Acceleration	$m/s^{2}$
$s'''(t) = j(t)$	Jerk	$m/s^{3}$

Worked Examples

Example 1: Find all derivatives of $f(x) = x^5 - 3x^3 + 2x$

Derivative	Computation	Result
$f'(x)$	$5x^4 - 9x^2 + 2$	Polynomial degree 4
$f''(x)$	$20x^3 - 18x$	Polynomial degree 3
$f'''(x)$	$60x^2 - 18$	Polynomial degree 2
$f^{(4)}(x)$	$120x$	Polynomial degree 1
$f^{(5)}(x)$	$120$	Constant!
$f^{(6)}(x)$	$0$	Zero forever

Key Principle: Any polynomial of degree $n$ has $f^{(n+1)}(x) = 0$ . The $n$ th derivative of $x^n$ is $n!$ (n factorial).

Example 2: Higher derivatives of $e^{2x}$

$\frac{d^n}{dx^n}[e^{2x}] = 2^n e^{2x}$

Each derivative multiplies by 2 (Chain Rule): $y' = 2e^{2x}$ , $y'' = 4e^{2x}$ , $y''' = 8e^{2x}$ , ...

Example 3: The Trig Cycle

$n$	$\frac{d^n}{dx^n}[\sin x]$	$\frac{d^n}{dx^n}[\cos x]$
0	$\sin x$	$\cos x$
1	$\cos x$	$-\sin x$
2	$-\sin x$	$-\cos x$
3	$-\cos x$	$\sin x$
4	$\sin x$ ← repeats!	$\cos x$ ← repeats!

$\boxed{\frac{d^n}{dx^n}[\sin x] = \sin\left(x + \frac{n\pi}{2}\right) \qquad \frac{d^n}{dx^n}[\cos x] = \cos\left(x + \frac{n\pi}{2}\right)}$

Find Higher-Order Derivatives 🎯

Concavity and the Second Derivative

The second derivative provides crucial information about the shape of a graph:

Condition	Meaning	Graph Shape
$f''(x) > 0$	Concave up	Holds water (∪)
$f''(x) < 0$	Concave down	Spills water (∩)
$f''(x) = 0$	Possible inflection point	Concavity may change

⚠️ Critical Warning: $f''(c) = 0$ does NOT guarantee an inflection point! You must verify that $f''$ actually changes sign at $c$ . Example: $f(x) = x^4$ has $f''(0) = 0$ but NO inflection point (concave up on both sides).

The Second Derivative Test

At a critical point where $f'(c) = 0$ :

$f''(c)$	Conclusion
$f''(c) > 0$	Local minimum
$f''(c) < 0$	Local maximum
$f''(c) = 0$	Inconclusive — use First Derivative Test

$\boxed{f'(c) = 0 \text{ and } f''(c) > 0 \implies \text{local minimum at } x = c}$

Worked Example

Find where $f(x) = x^3 - 3x$ is concave up.

$f''(x) = 6x$ .

Concave up when $f''(x) > 0$ : $6x > 0 \implies x > 0$ .

So $f$ is concave up on $(0, \infty)$ and concave down on $(-\infty, 0)$ with an inflection point at $x = 0$ .

Second Derivative Applications 🎯

Connecting $f$ , $f'$ , and $f''$ — The Big Picture

If you know...	Then you can determine...
$f'(c) = 0$	Critical point (possible max/min)
$f'(c) > 0$	$f$ is increasing at $c$
$f'(c) < 0$	$f$ is decreasing at $c$
$f''(c) > 0$	$f$ is concave up; $f'$ is increasing
$f''(c) < 0$	$f$ is concave down; $f'$ is decreasing
$f'(c) = 0$ and $f''(c) > 0$	Local minimum
$f'(c) = 0$ and $f''(c) < 0$	Local maximum

AP Tip: The AP exam frequently gives you a graph of $f'(x)$ and asks about $f(x)$ or $f''(x)$ . Remember: the derivative of $f'$ IS $f''$ , so where $f'$ is increasing, $f'' > 0$ (concave up for $f$ ).

Analyze concavity and extrema.

Higher-order derivative computation. ✍️

Key Takeaways — Part 5

Concept	Formula / Fact
Second Derivative	$f''(x) = \frac{d}{dx}[f'(x)]$
Motion	Position → Velocity → Acceleration
Concave up	$f''(x) > 0$
Concave down	$f''(x) < 0$
Inflection point	$f''$ changes sign
2nd Deriv Test	$f'(c)=0$ : $f''(c)>0$ → min; $f''(c)<0$ → max
Polynomials	Degree $n$ → $(n+1)$ th derivative is 0
Trig cycle	Repeats every 4 derivatives
Exponential	$\frac{d^n}{dx^n}[e^{kx}] = k^n e^{kx}$

Up Next: Part 6 — Mixed Differentiation Problems workshop.

Part 6: Mixed Differentiation Problems

📐 Problem-Solving Workshop

Part 6 of 7 — Mixed Differentiation Problems

The Decision Framework

Before differentiating, ask yourself: What structure does this expression have?

Structure	Rule to Use	Example
Single term $cx^n$	Power Rule	$7x^4$
Sum/difference	Term-by-term	$x^3 + \sin x$
Product $f \cdot g$	Product Rule	$x^2 e^x$
Quotient $f/g$	Quotient Rule (or rewrite)	$\frac{e^x}{x+1}$
Composition $f(g(x))$	Chain Rule	$\sin(x^2)$
Constant ÷ power	Rewrite as negative exponent	$\frac{5}{x^3}$
Polynomial ÷ monomial	Split fraction	$\frac{x^3+x}{x^2}$

Key Strategy: Always simplify first when possible. Rewriting can eliminate the need for Product or Quotient Rule entirely.

Simplification Strategies

Before	After	Rule Avoided
$\frac{x^3 + 1}{x}$	$x^2 + x^{-1}$	Quotient Rule
$x^2(x+3)$	$x^3 + 3x^2$	Product Rule
$\frac{5}{x^2}$	$5x^{-2}$	Quotient Rule
$(x+1)^2$	$x^2 + 2x + 1$	Chain Rule

Identify and Apply 🎯

Worked Examples — Multi-Rule Problems

Example 1: Find $\frac{d}{dx}\left[\frac{x^2 \sin x}{e^x}\right]$

Strategy: This is a quotient where the numerator is itself a product. Use Quotient Rule with $f = x^2 \sin x$ and $g = e^x$ .

First, find $f'$ using Product Rule: $f' = 2x\sin x + x^2 \cos x$

Then Quotient Rule: $\frac{(2x\sin x + x^2\cos x)e^x - x^2\sin x \cdot e^x}{e^{2x}} = \frac{2x\sin x + x^2\cos x - x^2\sin x}{e^x}$

Example 2: Find the tangent line to $y = \frac{x^2 + 1}{x - 1}$ at $x = 2$

Step 1: $y(2) = \frac{5}{1} = 5$ . Point: $(2, 5)$ .

Step 2: $y' = \frac{2x(x-1) - (x^2+1)}{(x-1)^2} = \frac{x^2-2x-1}{(x-1)^2}$

Step 3: $y'(2) = \frac{4-4-1}{1} = -1$ . Slope: $m = -1$ .

Step 4: Tangent line: $y - 5 = -1(x - 2)$ → $y = -x + 7$

AP Tip: Tangent line questions combine differentiation with algebra. Always clearly state the point and slope before writing the equation.

Particle Motion — A Complete Analysis

Problem: A particle moves along the $x$ -axis with position $s(t) = t^3 - 6t^2 + 9t + 2$ for $t \geq 0$ .

Question	Computation	Answer
Velocity	$v(t) = 3t^2 - 12t + 9 = 3(t-1)(t-3)$	—
At rest when?	$v(t) = 0$	$t = 1$ and $t = 3$
Moving right when?	$v(t) > 0$	$0 < t < 1$ or $t > 3$
Moving left when?	$v(t) < 0$	$1 < t < 3$
Acceleration	$a(t) = 6t - 12$	—
Speeding up when?	$v$ and $a$ same sign	$1 < t < 2$ or $t > 3$
Slowing down when?	$v$ and $a$ opposite sign	$0 < t < 1$ or $2 < t < 3$

Key Concept: "Speeding up" means $|v(t)|$ is increasing, which happens when velocity and acceleration have the same sign. This is different from "accelerating" (which just means $a > 0$ ).

Speed vs. Velocity

$\boxed{\text{Speed} = |v(t)| \qquad \text{Velocity} = v(t) \text{ (signed)}}$

Speed is always non-negative. The particle speeds up when $v(t) \cdot a(t) > 0$ .

Motion & Mixed Problems 🎯

Choose the best strategy for each derivative.

Mixed problem. ✍️

Key Takeaways — Part 6

Strategy	When to Use
Simplify first	Polynomial ÷ monomial, expandable products
Product Rule	Products with unlike functions ( $xe^x$ , $x\sin x$ )
Quotient Rule	True fractions with unlike functions
Rewrite	Constants over powers → negative exponents
Multiple rules	Nested structures (quotient of products, etc.)

Particle Motion Checklist:

At rest: $v(t) = 0$
Direction: sign of $v(t)$
Speeding up: $v(t) \cdot a(t) > 0$
Slowing down: $v(t) \cdot a(t) < 0$

Up Next: Part 7 — Comprehensive Review & AP Exam preparation.

Part 7: Comprehensive Review

📐 Review & Applications

Part 7 of 7 — Comprehensive Review & AP Exam Preparation

Complete Derivative Reference Table

Rule	Formula
Power	$\frac{d}{dx}[x^n] = nx^{n-1}$
Constant	$\frac{d}{dx}[c] = 0$
Constant Multiple	$\frac{d}{dx}[cf] = cf'$
Sum/Difference	$\frac{d}{dx}[f \pm g] = f' \pm g'$
Product	$(fg)' = f'g + fg'$
Quotient	$\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$

Special Function Derivatives

Function	Derivative	Domain Note
$e^x$	$e^x$	All reals
$\ln x$	$\frac{1}{x}$	$x > 0$
$a^x$	$a^x \ln a$	$a > 0, a \neq 1$
$\log_a x$	$\frac{1}{x \ln a}$	$x > 0$

Trig Derivatives (Must Memorize!)

Positive Derivatives	Negative Derivatives
$\frac{d}{dx}[\sin x] = \cos x$	$\frac{d}{dx}[\cos x] = -\sin x$
$\frac{d}{dx}[\tan x] = \sec^2 x$	$\frac{d}{dx}[\cot x] = -\csc^2 x$
$\frac{d}{dx}[\sec x] = \sec x \tan x$	$\frac{d}{dx}[\csc x] = -\csc x \cot x$

AP Exam Question Types for Basic Differentiation

Type	What They Ask	Key Skill
Direct computation	"Find $f'(x)$ "	Apply correct rule
Evaluate at a point	"Find $f'(2)$ "	Differentiate then substitute
From a table	Given $f(a), f'(a), g(a), g'(a)$	Plug into Product/Quotient Rule
Tangent line	"Equation of tangent at $x = c$ "	Need point + slope
Normal line	"Equation of normal at $x = c$ "	Slope = $-\frac{1}{f'(c)}$
Horizontal tangent	"Where is tangent horizontal?"	Solve $f'(x) = 0$
Particle motion	"When at rest? Direction?"	Analyze $v(t) = s'(t)$

Table-Based Problems — Complete Strategy

Given this table:

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

Find each of the following at $x = 1$ :

Expression	Formula	Computation	Answer
$(f+g)'(1)$	$f'(1)+g'(1)$	$-2+5$	$3$
$(fg)'(1)$	$f'g+fg'$	$(-2)(3)+(4)(5)$	$14$
$(f/g)'(1)$	$\frac{f'g-fg'}{g^2}$	$\frac{(-2)(3)-(4)(5)}{9}$	$-\frac{26}{9}$
$(3f)'(1)$	$3f'(1)$	$3(-2)$	$-6$

Comprehensive Assessment 🎯

Tangent & Normal Lines — AP Exam Template

Tangent Line at $x = c$ :

$\boxed{y - f(c) = f'(c)(x - c)}$

Normal Line at $x = c$ (perpendicular to tangent):

$\boxed{y - f(c) = -\frac{1}{f'(c)}(x - c)}$

Complete Worked Example

Find the tangent and normal lines to $y = x^3 - 4x$ at $x = 2$ .

Step	Tangent	Normal
Point: $y(2) = 8-8 = 0$	$(2, 0)$	$(2, 0)$
$y' = 3x^2 - 4$	$y'(2) = 8$	slope $= -\frac{1}{8}$
Equation	$y = 8(x-2) = 8x-16$	$y = -\frac{1}{8}(x-2)$

Horizontal & Vertical Tangent Lines

Type	Condition	Meaning
Horizontal tangent	$f'(c) = 0$	Critical point candidate
Vertical tangent	$f'(c)$ is undefined, $f$ continuous	Cusp or vertical tangent point

AP Tip: When asked "for what values of $x$ is the tangent horizontal?", you are being asked to solve $f'(x) = 0$ . Always check that $f$ is defined at those points!

AP-Style Final Problems 🎯

Common Errors to Avoid on the AP Exam

Error	Wrong	Correct
Multiplying derivatives	$(fg)' = f'g'$	$(fg)' = f'g + fg'$
Forgetting negative in QR	$\frac{f'g + fg'}{g^2}$	$\frac{f'g - fg'}{g^2}$
Co-function sign	$\frac{d}{dx}[\cos x] = \sin x$	$\frac{d}{dx}[\cos x] = -\sin x$
Constant derivative	$\frac{d}{dx}[\pi^2] = 2\pi$	$\frac{d}{dx}[\pi^2] = 0$
Forgetting to rewrite	$\frac{d}{dx}[\sqrt{x}] = \frac{1}{2}\sqrt{x-1}$	$x^{1/2} \to \frac{1}{2}x^{-1/2}$
Wrong evaluation	Computing $f'(x)$ but forgetting to plug in $x = c$	Always substitute AFTER differentiating

Quick fire — identify the derivative.

Final challenge problem. ✍️

Basic Differentiation Rules — Complete! ✅

You have mastered:

✅ Power Rule (including negative/fractional exponents)
✅ Constant, Constant Multiple, and Sum/Difference Rules
✅ Product Rule: $(fg)' = f'g + fg'$
✅ Quotient Rule: $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$
✅ All six trigonometric derivatives
✅ Higher-order derivatives and concavity
✅ Particle motion analysis
✅ Tangent and normal lines
✅ Table-based derivative problems

What's Next?

Next Topic	What You'll Learn
Chain Rule	Derivatives of compositions: $f(g(x))$
Implicit Differentiation	When $y$ is not explicitly solved
Related Rates	How quantities change together

The Chain Rule is arguably the most important rule in calculus — it extends everything you've learned to composite functions!

Basic Differentiation Rules