SECTION 1
Historical Context & Motivation
The story of differentiating transcendental functions — sines, cosines, exponentials, and logarithms — is inseparable from the story of calculus itself. When Isaac Newton and Gottfried Wilhelm Leibniz independently developed the foundations of calculus in the late seventeenth century, polynomial functions were the first to yield their derivatives. However, the physical world demanded more: planetary orbits required trigonometric functions, compound growth required exponentials, and bridging the two demanded logarithms. The quest to differentiate these four function families drove some of the most elegant mathematics of the Enlightenment era and produced results that remain indispensable on the AP Calculus AB exam today.
1665–1676 • Newton & Leibniz Develop Calculus
1748 • Euler's Introductio in Analysin Infinitorum
1750s • Trigonometric Derivatives Established
1821 • Cauchy's Rigorous Limits
The central question these mathematicians answered was deceptively simple: if a quantity oscillates sinusoidally or grows exponentially, how fast is it changing at any given instant? The four derivative formulas you will master in this lesson — for sin x, cos x, eˣ, and ln x — are the answers to that question, and they underpin virtually every applied calculus problem from modeling population growth to analyzing alternating current circuits.
SECTION 2
Core Principles & Definitions
Before diving into formulas, it is important to understand why these four functions are grouped together. They are all transcendental functions — functions that cannot be expressed as finite combinations of algebraic operations (addition, multiplication, roots). Unlike polynomials, whose derivatives always reduce the degree by one, transcendental derivatives cycle or reproduce themselves, which is precisely what makes them so powerful in modeling real-world phenomena.
d/dx (sin x) = cos x
d/dx (cos x) = −sin x
d/dx (eˣ) = eˣ
d/dx (ln x) = 1/x
KEY TAKEAWAY
A critical prerequisite for the trigonometric derivatives is the radian measure requirement: the formulas d/dx(sin x) = cos x and d/dx(cos x) = −sin x hold only when x is measured in radians. If x were in degrees, an additional conversion factor of π/180 would appear, breaking the elegant simplicity. This is the fundamental reason calculus courses insist on radian mode — it is not merely a convention but a mathematical necessity rooted in the limit lim(h→0) (sin h)/h = 1, which is true exclusively in radians.
SECTION 3
Visual Explanation — Functions and Their Derivatives
The most illuminating way to grasp these derivative relationships is to see each function graphed alongside its derivative. In the diagram below, observe how the slope of the original function at every point matches the height of the derivative curve. Where sin x reaches its peak (slope = 0), cos x crosses zero; where eˣ has a certain height, its derivative has exactly the same height.
Pay particular attention to the sin x / cos x panel. At x = 0, sin x is rising through zero at its maximum rate; correspondingly, cos(0) = 1, confirming the derivative is at its maximum. At x = π/2, sin x reaches its peak and momentarily stops increasing — its slope is zero — and indeed cos(π/2) = 0. This visual correspondence between the height of the derivative and the slope of the original function is the single most important intuition to carry forward.
SECTION 4
Mathematical Framework — Proofs from the Limit Definition
Each of the four derivative formulas can be derived rigorously from the limit definition of the derivative: f′(x) = lim(h→0) [f(x + h) − f(x)] / h. Understanding at least the key steps of these proofs deepens your conceptual grasp and prepares you for the free-response questions on the AP exam that ask you to justify derivative rules from first principles.
Derivative of sin x
LIMIT DEFINITION APPLIED TO SIN X
Expanding sin(x + h) = sin x cos h + cos x sin h via the angle-addition identity, then separating terms: = lim(h→0) [sin x (cos h − 1) + cos x sin h] / h = sin x · lim(h→0)(cos h − 1)/h + cos x · lim(h→0)(sin h)/h. Since lim(h→0)(sin h)/h = 1 and lim(h→0)(cos h − 1)/h = 0, we obtain d/dx(sin x) = cos x.
Derivative of cos x
LIMIT DEFINITION APPLIED TO COS X
Using cos(x + h) = cos x cos h − sin x sin h: = lim(h→0) [cos x (cos h − 1) − sin x sin h] / h = cos x · 0 − sin x · 1 = −sin x. The negative sign is essential and a frequent source of errors on the AP exam.
Derivative of eˣ
LIMIT DEFINITION APPLIED TO eˣ
Factor out eˣ since it does not depend on h. The remaining limit, lim(h→0)(eʰ − 1)/h, equals 1 — this is in fact one way to define the number e. Therefore d/dx(eˣ) = eˣ · 1 = eˣ.
Derivative of ln x
LIMIT DEFINITION APPLIED TO LN X
Rewrite as lim(h→0) (1/h) · ln(1 + h/x). Let u = h/x so h = ux and as h→0, u→0: = lim(u→0) (1/(ux)) · ln(1 + u) = (1/x) · lim(u→0) ln(1 + u)/u. Since lim(u→0) ln(1 + u)/u = 1, the result is 1/x for x > 0.
Two Critical Limits
SECTION 5
Detailed Breakdown — Patterns and Sign Behavior
A deeper look at these four derivatives reveals elegant patterns that simplify memorization and deepen understanding. The trigonometric derivatives cycle through a four-step pattern, the exponential is uniquely self-referential, and the logarithmic derivative bridges transcendental and algebraic worlds.
| Function f(x) | f′(x) | f″(x) | f‴(x) | f⁽⁴⁾(x) |
|---|---|---|---|---|
| sin x | cos x | −sin x | −cos x | sin x |
| cos x | −sin x | −cos x | sin x | cos x |
| eˣ | eˣ | eˣ | eˣ | eˣ |
| ln x | 1/x = x⁻¹ | −x⁻² | 2x⁻³ | −6x⁻⁴ |
Successive derivatives of the four key functions
The table above reveals that the trigonometric derivatives repeat with period 4 — a fact frequently exploited in AP problems asking for the 50th or 101st derivative of sin x. To find the nth derivative of sin x, compute n mod 4: if the remainder is 0, the answer is sin x; if 1, cos x; if 2, −sin x; if 3, −cos x. The exponential function eˣ is invariant under all orders of differentiation, and the successive derivatives of ln x follow the power rule once you express 1/x as x⁻¹, generating factorials with alternating signs: (−1)ⁿ⁺¹(n − 1)! · x⁻ⁿ.
SECTION 6
Worked Example
Let us work through a multi-step problem that combines several of our derivative rules. This mirrors the level of complexity you will encounter in both the multiple-choice and free-response portions of the AP exam.
Differentiate f(x) = 3 sin x − 2eˣ + 5 ln x
- 1
Step 1 — Identify the Structure
The function f(x) = 3 sin x − 2eˣ + 5 ln x is a linear combination of three transcendental functions. By the constant multiple rule and the sum/difference rule, we can differentiate each term separately and preserve the coefficients. - 2
Step 2 — Differentiate 3 sin x
Applying d/dx(sin x) = cos x with the constant multiple rule: d/dx(3 sin x) = 3 cos x.3 cos x - 3
Step 3 — Differentiate −2eˣ
Since d/dx(eˣ) = eˣ: d/dx(−2eˣ) = −2eˣ.−2eˣ - 4
Step 4 — Differentiate 5 ln x
Since d/dx(ln x) = 1/x: d/dx(5 ln x) = 5/x.5/x - 5
Step 5 — Combine Results
Assembling all three terms: f′(x) = 3 cos x − 2eˣ + 5/x. This expression is valid for all x > 0 (the domain restriction comes from the ln x term, which requires x > 0).f′(x) = 3 cos x − 2eˣ + 5/x
Common Pitfall
SECTION 7
Common Errors & Comparisons
Mastering these four derivatives means not only knowing the correct formulas but also understanding where students most commonly go wrong. The table below contrasts the correct derivative with the most frequent incorrect responses seen on the AP exam, along with the conceptual error behind each mistake.
| Function | Correct Derivative | Common Error | Why It's Wrong |
|---|---|---|---|
| sin x | cos x | −cos x | Confusing the sign pattern; the negative appears when differentiating cos x, not sin x. |
| cos x | −sin x | sin x | Dropping the negative sign. Remember: cos x is decreasing when sin x is positive. |
| eˣ | eˣ | xeˣ⁻¹ | Misapplying the power rule. The power rule applies to xⁿ, not to aˣ. The base is a constant and the exponent is the variable. |
| ln x | 1/x | 1/(ln x) | Confusing the function with its derivative. The ln disappears entirely upon differentiation. |
Common derivative errors and their sources
KEY TAKEAWAY
SECTION 8
Connection to Advanced Theory
The four derivative rules you have learned form the foundation upon which more sophisticated techniques are built. In particular, the chain rule will extend each of these formulas to composite functions: d/dx(sin(g(x))) = cos(g(x)) · g′(x), d/dx() = · g′(x), and so on. The table below shows how each basic rule generalizes once the chain rule is introduced, a topic you will encounter very soon in your AP Calculus AB course.
| Basic Rule (this lesson) | With Chain Rule (upcoming) | Example |
|---|---|---|
| d/dx(sin x) = cos x | d/dx(sin u) = cos u · du/dx | d/dx(sin 3x) = cos 3x · 3 = 3 cos 3x |
| d/dx(cos x) = −sin x | d/dx(cos u) = −sin u · du/dx | d/dx(cos x²) = −sin x² · 2x |
| d/dx(eˣ) = eˣ | d/dx(eᵘ) = eᵘ · du/dx | d/dx(e⁻ˣ) = e⁻ˣ · (−1) = −e⁻ˣ |
| d/dx(ln x) = 1/x | d/dx(ln u) = (1/u) · du/dx | d/dx(ln(x² + 1)) = 2x/(x² + 1) |
Basic rules vs. chain-rule extensions
Beyond the chain rule, these derivatives connect to deep structural ideas in mathematics. Euler's formula, eⁱˣ = cos x + i sin x, reveals that the exponential and trigonometric functions are secretly unified through complex numbers. This unity explains why their derivatives share such elegant properties — they are all manifestations of the same underlying exponential structure. In AP Calculus BC and later in differential equations, you will see how the derivative d/dx(eˣ) = eˣ makes exponential functions the natural solutions to first-order linear ODEs, while the cycling behavior of sin and cos derivatives makes them solutions to the second-order equation y″ + y = 0.
Looking Ahead: Integration
SECTION 9
Practice Problems
Question 1: CONCEPTUAL
- A. Because the power rule requires subtracting 1 from the exponent of cos x
- B. Because cos x is the horizontal reflection of sin x, and reflections reverse signs
- C. Because when sin x is positive, cos x is decreasing, so the slope of cos x must be negative in that region, which is captured by −sin x
- D. Because the limit lim(h→0)(sin h)/h = −1 introduces the negative sign
- E. Because cos x = sin(−x), and replacing x with −x always introduces a negative sign in the derivative
Show answer (C)
Question 2: BASIC CALCULATION
- A. 4eˣ + 7 sin x + 1/x
- B. 4eˣ − 7 sin x + 1/x
- C. 4xeˣ⁻¹ + 7 sin x + 1/x
- D. 4eˣ + 7 sin x + 1/(ln x)
- E. 4eˣ − 7 sin x + x
Show answer (A)
Question 3: INTERMEDIATE
- A. sin x
- B. cos x
- C. −sin x
- D. −cos x
- E. 83 cos x
Show answer (D)
Question 4: APPLIED
5 ptsShow answer
Question 5: CRITICAL THINKING
3 ptsShow answer
SUMMARY
Lesson Summary
Summary
This lesson established the four essential transcendental derivative formulas: d/dx(sin x) = cos x, d/dx(cos x) = −sin x, d/dx(eˣ) = eˣ, and d/dx(ln x) = 1/x. Each was derived from the limit definition of the derivative, relying on critical prerequisite limits: lim(h→0)(sin h)/h = 1 and lim(h→0)(cos h − 1)/h = 0 for the trigonometric derivatives, lim(h→0)(eʰ − 1)/h = 1 for the exponential, and the logarithmic limit identity for ln x.
Key structural insights include the period-4 cycle of successive trigonometric derivatives (useful for finding the nth derivative), the self-replicating property of eˣ (it is invariant under differentiation), and the domain restriction of ln x (x > 0). Remember that all trigonometric derivatives require radian measure, and avoid confusing the power rule (variable base, fixed exponent) with the exponential derivative (fixed base, variable exponent). These four formulas, combined with the chain rule you will learn next, will enable you to differentiate any combination of trigonometric, exponential, and logarithmic functions on the AP Calculus AB exam.