Derivatives of cos x, sin x, eˣ, and ln x - AP Calculus BC

SECTION 1

Historical Context & Motivation

The quest to understand instantaneous rates of change stretches back centuries, and the derivatives of trigonometric functions, the natural exponential, and the natural logarithm sit at the very heart of that story. These four functions—sin x, cos x, eˣ, and ln x—appear throughout physics, engineering, economics, and biology, so knowing how they change is indispensable. Isaac Newton and Gottfried Wilhelm Leibniz independently developed the calculus machinery that would formalize these derivatives, but the individual functions themselves were studied by mathematicians spanning several centuries. Leonhard Euler unified the exponential and trigonometric worlds through his famous identity eⁱˣ = cos x + i sin x, revealing that these derivative formulas are not isolated facts but deeply interconnected.

1614 • Napier's Logarithms

John Napier published Mirifici Logarithmorum Canonis Descriptio, introducing logarithms as a computational tool and laying the groundwork for the natural logarithm function.

1668 • Mercator's Series for ln(1 + x)

Nicolaus Mercator published the first known infinite series for the natural logarithm, foreshadowing the limit-based approach to differentiation of ln x.

1684 • Leibniz Publishes Differential Calculus

Leibniz published his foundational paper on calculus, introducing the d/dx notation still used today and providing systematic rules for differentiation.

1748 • Euler's Introductio in Analysin Infinitorum

Leonhard Euler formalized the exponential function eˣ and connected it to trigonometric functions through his identity eⁱˣ = cos x + i sin x, unifying these derivative families.

1821 • Cauchy's Rigorous Limits

Augustin-Louis Cauchy placed derivatives on a rigorous ε-δ foundation, giving the limit proofs behind d/dx(sin x) = cos x and d/dx(eˣ) = eˣ their modern form.

The central question this lesson addresses is straightforward yet powerful: given that sin x, cos x, eˣ, and ln x are among the most frequently encountered functions in mathematics and its applications, what are their exact instantaneous rates of change, and how do we prove and apply those formulas? Answering this question will equip you with four essential derivative rules that serve as building blocks for the chain rule, product rule, and every advanced differentiation technique you will encounter in AP Calculus BC.

SECTION 2

Core Principles & Definitions

Before memorizing formulas, it is essential to understand the foundational ideas that make these derivative results both logical and inevitable. Each derivative is derived from the limit definition of the derivative: f′(x) = lim(h→0) [f(x + h) − f(x)] / h. This limit captures the slope of the tangent line at every point on the curve of f. The four functions we study here are special because their derivatives take remarkably clean forms—sin x differentiates into cos x, cos x differentiates into −sin x, eˣ differentiates into itself, and ln x differentiates into 1/x. Understanding why these results hold deepens conceptual fluency far beyond rote memorization.

The Squeeze Theorem Foundation

The derivative of sin x relies on two critical limits: lim(h→0) sin h / h = 1 and lim(h→0) (cos h − 1) / h = 0. Both are established via the Squeeze Theorem using geometric comparisons of arc length and chord length on the unit circle.

Trigonometric Duality

Sine and cosine are phase-shifted versions of each other: cos x = sin(x + π/2). Differentiation cycles through sin → cos → −sin → −cos → sin with period 4, a pattern that proves invaluable in solving higher-order differential equations.

The Exponential's Unique Self-Replication

The function eˣ is the unique function (up to scalar multiples) that equals its own derivative. This property defines the number e ≈ 2.71828 and makes eˣ central to modeling growth and decay phenomena.

Logarithmic Inversion

Since ln x is the inverse function of eˣ, its derivative follows from the inverse function theorem: if y = ln x, then x = eʸ and dx/dy = eʸ = x, so dy/dx = 1/x. This elegant argument bypasses the limit definition entirely.

✦

KEY TAKEAWAY

Think of these four derivative formulas as the periodic table of differentiation: just as all chemical compounds are built from elements, virtually every derivative you encounter in calculus is assembled from these fundamental results combined with the product, quotient, and chain rules. An engineer modeling a vibrating bridge uses d/dx(sin x) = cos x; a biologist modeling bacterial growth uses d/dx(eˣ) = eˣ; an economist measuring marginal utility uses d/dx(ln x) = 1/x. Mastering these four formulas is not just memorization—it is equipping yourself with the essential toolkit for all subsequent calculus work.

SECTION 3

Visual Explanation — Functions and Their Derivatives

The top-left panel shows sin x (solid cyan) alongside its derivative cos x (dashed pink). Notice that wherever sin x has a maximum, cos x crosses zero—confirming that the slope is zero at peaks. The top-right panel shows the analogous relationship between cos x and −sin x. The bottom-left panel displays eˣ (amber), which is its own derivative—the curve and its tangent slope function are identical. The bottom-right panel pairs ln x (emerald) with its derivative 1/x (violet dashed), showing how the logarithm's slope decreases as x grows.

The diagrams above illustrate the most important geometric insight for these derivatives: the derivative of a function at any point equals the slope of the tangent line to the original curve at that point. For sin x, the tangent slope is greatest (equal to 1) at x = 0, where cos 0 = 1, and the tangent slope is zero at x = π/2, where cos(π/2) = 0, corresponding to the peak of the sine wave. This phase-shift relationship between sine and cosine is a direct geometric consequence of the derivative. For eˣ, the remarkable fact that the curve is its own derivative means the slope at every point equals the function's value at that point—steep where the function is large, shallow where it is small. For ln x, the derivative 1/x is always positive for x > 0, confirming that the natural logarithm is strictly increasing, though its rate of increase perpetually diminishes.

SECTION 4

Mathematical Framework — Derivations from First Principles

Each of the four fundamental derivative formulas can be rigorously established from the limit definition. Understanding these derivations not only satisfies intellectual curiosity but also reinforces the algebraic identities and limit techniques that appear repeatedly on the AP Calculus BC exam.

Derivative of sin x

LIMIT DEFINITION FOR SIN X

d/dx (sin x) = lim(h→0) [sin(x + h) − sin x] / h

Expand sin(x + h) using the angle addition identity: sin x cos h + cos x sin h. Then the numerator becomes sin x (cos h − 1) + cos x sin h. Divide by h and apply the two critical limits: lim(h→0) sin h / h = 1 and lim(h→0) (cos h − 1) / h = 0.

RESULT

d/dx (sin x) = cos x

After applying the two limits, the first term vanishes and the second yields cos x × 1 = cos x.

Derivative of cos x

LIMIT DEFINITION FOR COS X

d/dx (cos x) = lim(h→0) [cos(x + h) − cos x] / h

Expand cos(x + h) = cos x cos h − sin x sin h. The numerator becomes cos x (cos h − 1) − sin x sin h. Dividing by h and applying the same two critical limits yields cos x × 0 − sin x × 1 = −sin x.

RESULT

d/dx (cos x) = −sin x

The negative sign is crucial: cosine decreases where sine is positive, and vice versa.

Derivative of eˣ

LIMIT DEFINITION FOR eˣ

d/dx (eˣ) = lim(h→0) [eˣ⁺ʰ − eˣ] / h = eˣ · lim(h→0) (eʰ − 1) / h

Factor eˣ from the numerator. The remaining limit lim(h→0) (eʰ − 1) / h = 1 is essentially the definition of the number e: it is the unique base for which this limit equals 1.

RESULT

d/dx (eˣ) = eˣ

The exponential function is its own derivative—the only non-trivial function (up to constant multiples) with this property.

Derivative of ln x

INVERSE FUNCTION APPROACH

If y = ln x, then x = eʸ. Differentiating: dx/dy = eʸ = x, so dy/dx = 1/x.

This follows from the inverse function theorem: dy/dx = 1 / (dx/dy). Since eʸ = x, we obtain 1/x. Domain restriction: x > 0.

RESULT

d/dx (ln x) = 1/x, x > 0

This derivative is always positive on its domain, confirming that ln x is strictly increasing for all positive x.

SECTION 5

Detailed Breakdown — The Trigonometric Derivative Cycle

One of the most elegant patterns in calculus is the cyclic behavior of successive derivatives of sine and cosine. When you differentiate sin x repeatedly, the results cycle through four phases before returning to the original function. This pattern is not merely a curiosity—it is critical for solving higher-order differential equations, computing Taylor series, and evaluating nth derivatives on the AP exam. Similarly, the exponential and logarithmic functions have their own characteristic behaviors under repeated differentiation that contrast sharply with the trigonometric cycle.

Left: The trigonometric derivative cycle. Differentiating sin x four times returns to sin x—the arrows trace the cycle sin x → cos x → −sin x → −cos x → sin x. Right-top: eˣ loops to itself under differentiation. Right-bottom: ln x produces a cascade of power functions whose nth derivative is (−1)ⁿ⁻¹(n − 1)! x⁻ⁿ.

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	x⁻¹	−x⁻²	2x⁻³	−6x⁻⁴

Successive derivatives of the four fundamental functions

The table confirms two powerful patterns. For trigonometric functions, the nth derivative of sin x equals sin(x + nπ/2), which means you can determine any higher-order derivative by computing n mod 4 and selecting the appropriate function from the cycle. For the exponential, all derivatives are identical—a fact that makes eˣ the cornerstone of solutions to linear differential equations with constant coefficients. For ln x, each successive derivative introduces a factor of (−1) and an increasing factorial in the numerator, a pattern most elegantly expressed as dⁿ/dxⁿ (ln x) = (−1)ⁿ⁻¹ (n − 1)! / xⁿ for n ≥ 1.

SECTION 6

Worked Example — Combining Fundamental Derivatives

Let us differentiate the function f(x) = 3 sin x − 2eˣ + 5 ln x + cos x. This problem requires all four fundamental derivatives together with the constant multiple rule and the sum/difference rule.

Differentiate f(x) = 3 sin x − 2eˣ + 5 ln x + cos x

1
Step 1 — Apply the Sum/Difference and Constant Multiple Rules

The derivative of a sum (or difference) is the sum (or difference) of the derivatives, and constant factors can be pulled out. Therefore: f′(x) = 3 · d/dx(sin x) − 2 · d/dx(eˣ) + 5 · d/dx(ln x) + d/dx(cos x).
2
Step 2 — Substitute Each Fundamental Derivative

Replace each derivative with its known formula: d/dx(sin x) = cos x, d/dx(eˣ) = eˣ, d/dx(ln x) = 1/x, d/dx(cos x) = −sin x. This gives f′(x) = 3 cos x − 2eˣ + 5/x + (−sin x).
3
Step 3 — Simplify the Expression

Combine like terms and write in standard form.

f′(x) = 3 cos x − sin x − 2eˣ + 5/x
4
Step 4 — Verify the Domain

The original function includes ln x, which requires x > 0. Therefore the derivative is valid only for x > 0. At x = 0 or negative values, neither f nor f′ is defined.
5
Step 5 — Evaluate at a Specific Point (x = π/2)

To check our work, evaluate f′(π/2): 3 cos(π/2) − sin(π/2) − $2e^{π/2}$ + 5/(π/2) = 3(0) − 1 − $2e^{π/2}$ + 10/π ≈ 0 − 1 − 9.621 + 3.183.

f′(π/2) ≈ −7.438

SECTION 7

Common Errors & Critical Distinctions

Even students with strong algebraic skills frequently commit specific errors when working with these four derivatives. Recognizing these pitfalls in advance—and understanding why they are wrong—is one of the most efficient ways to improve accuracy on the AP exam.

Common Error	Correct Formula	Why the Error Occurs
d/dx(cos x) = sin x (missing the negative)	d/dx(cos x) = −sin x	Students recall that sine and cosine are related but forget that cosine's derivative acquires a negative sign from the limit derivation.
d/dx(eˣ) = x·eˣ⁻¹ (applying the power rule)	d/dx(eˣ) = eˣ	Confusing the exponential function (variable in the exponent) with a power function (variable in the base). The power rule xⁿ → nxⁿ⁻¹ does NOT apply here.
d/dx(ln x) = 1/x for all x	d/dx(ln x) = 1/x for x > 0 only	Forgetting the domain restriction. The natural logarithm is defined only for positive inputs; writing the derivative without this qualifier can lead to nonsensical evaluations.
d/dx(sin x) = cos x only in radians? Yes.	d/dx(sin x) = cos x (radians only)	The formulas hold only when the argument is in radians. In degree mode, d/dx(sin x°) = (π/180) cos x°, introducing an unwanted constant factor.

Frequently tested errors on the AP Calculus BC exam

⚠

CRITICAL DISTINCTION

The single most common source of errors among these four derivatives is the sign on the derivative of cosine. A reliable mnemonic: in the derivative cycle, moving forward (differentiating) introduces a negative sign when you cross the 'south pole' of the cycle—from cos to −sin and from −cos to sin. Alternatively, remember that cos x is a decreasing function near x = 0 (from its maximum), so its derivative must be negative there: indeed, −sin(0⁺) < 0.

SECTION 8

Connection to Advanced Topics

The four derivatives studied in this lesson are not endpoints—they are launching pads for virtually every advanced technique in AP Calculus BC and beyond. Each fundamental derivative reappears inside the chain rule, integration, series expansions, and differential equations. Understanding how these basic results extend into more sophisticated territory is essential for success on the exam and for future coursework in mathematics, science, and engineering.

Fundamental Derivative	Advanced Extension in AP Calculus BC
d/dx(sin x) = cos x	Chain rule: d/dx(sin(g(x))) = cos(g(x)) · g′(x). Taylor series: sin x = x − x³/3! + x⁵/5! − ···, derived by differentiating term by term.
d/dx(cos x) = −sin x	Integration: ∫ sin x dx = −cos x + C (reverse the derivative). Parametric derivatives: if x = cos t, y = sin t, then dy/dx = (cos t)/(−sin t) = −cot t.
d/dx(eˣ) = eˣ	Differential equations: y′ = ky has solution y = Ceᵏˣ. Euler's method uses the derivative eˣ to approximate solutions. The Maclaurin series eˣ = 1 + x + x²/2! + x³/3! + ··· converges for all x.
d/dx(ln x) = 1/x	Logarithmic differentiation: take ln of both sides to simplify products and powers. Integration: ∫ (1/x) dx = ln \|x\| + C. The harmonic series and p-series convergence tests rely on properties of ln x.

How each fundamental derivative connects to BC-level topics

ℹ

LOOKING AHEAD

Euler's formula eⁱˣ = cos x + i sin x unifies the exponential and trigonometric derivatives into a single framework. Differentiating both sides with respect to x gives ieⁱˣ = −sin x + i cos x, confirming that d/dx(cos x) = −sin x and d/dx(sin x) = cos x emerge simultaneously from the derivative of the complex exponential. While complex analysis is beyond the AP syllabus, this connection illustrates why these four functions are inextricably linked in higher mathematics.

SECTION 9

Practice Problems

Question 1: CONCEPTUAL

Which of the following correctly states why the derivative of cos x includes a negative sign, while the derivative of sin x does not?

A. Because cos x is always negative on (0, π), its derivative must be negative.
B. Because the limit lim(h→0) sin h / h equals −1 when applied to the cos x derivation.
C. Because in the limit expansion of cos(x + h), the term −sin x · sin h produces a −sin x after dividing by h and taking the limit, while the corresponding term in the sin(x + h) expansion is +cos x · sin h.
D. Because the power rule introduces a coefficient of −1 when applied to trigonometric functions with even symmetry.
E. Because cos x = sin(−x), and the chain rule introduces a factor of −1 from the inner function −x.

Show answer (C)

C is correct. When expanding cos(x + h) = cos x cos h − sin x sin h, the critical term is −sin x · sin h. After subtracting cos x and dividing by h, this term becomes −sin x · (sin h / h), which approaches −sin x as h → 0. In contrast, the sin(x + h) expansion yields +cos x · sin h, producing +cos x. A is wrong because cos x being negative on an interval is irrelevant—the derivative's sign at a point depends on the limit process, not the function's sign. B is wrong because lim(h→0) sin h / h = +1, not −1. D is wrong because the power rule does not apply to trigonometric functions. E is wrong because cos x = sin(π/2 − x), not sin(−x); the identity cos x = sin(−x) is itself incorrect.

Question 2: BASIC CALCULATION

Find d/dx [4eˣ − 7 sin x + 2 ln x].

A. 4eˣ − 7 cos x + 2/x
B. 4eˣ + 7 cos x + 2/x
C. 4xeˣ⁻¹ − 7 cos x + 2 ln x
D. 4eˣ − 7 cos x + 2x
E. 4eˣ + 7 sin x + 2/x

Show answer (A)

A is correct. Apply the constant multiple rule and the sum rule: d/dx(4eˣ) = 4eˣ, d/dx(−7 sin x) = −7 cos x, and d/dx(2 ln x) = 2/x. Combining: 4eˣ − 7 cos x + 2/x. B is wrong because it drops the negative sign on the cos x term. C is wrong because it incorrectly applies the power rule to eˣ and fails to differentiate ln x. D is wrong because d/dx(ln x) = 1/x, not x. E is wrong because d/dx(sin x) = cos x, not sin x, and the constant −7 is lost.

Question 3: INTERMEDIATE

If f(x) = eˣ + cos x, find the value of x in [0, 2π) at which f″(x) = 0.

A. There is no solution in [0, 2π).
B. x = π
C. x = π/2 and x = 3π/2
D. x = 0
E. x = 3π/2

Show answer (D)

D is correct. Given f(x) = eˣ + cos x, we differentiate twice. The first derivative is f′(x) = eˣ − sin x, and the second derivative is f″(x) = eˣ − cos x. Setting f″(x) = 0 gives the equation eˣ = cos x. At x = 0: f″(0) = e⁰ − cos 0 = 1 − 1 = 0, so x = 0 is a solution. To confirm this is the only solution in [0, 2π), note that for all x > 0, eˣ > 1. Meanwhile, cos x ≤ 1 for all x, with cos x = 1 only at x = 0, 2π, …. Since eˣ > 1 ≥ cos x for every x in (0, 2π), we have f″(x) = eˣ − cos x > 0 throughout that interval. Therefore x = 0 is the unique solution in [0, 2π). A is wrong because x = 0 is in [0, 2π) and satisfies f″(0) = 0. B is wrong because f″(π) = eᵖ − cos π = eᵖ + 1 ≈ 24.14, which is not zero. C is wrong because f″(π/2) =

e^{π/2}

− cos(π/2) =

e^{π/2}

− 0 ≈ 4.81 > 0, and f″(3π/2) =

e^{3π/2}

− cos(3π/2) =

e^{3π/2}

− 0 ≈ 111.32 > 0. Neither value is zero. E is wrong because f″(3π/2) =

e^{3π/2}

− cos(3π/2) =

e^{3π/2}

− 0 ≈ 111.32 > 0, which is not zero.

Question 4: APPLIED

4 pts

A particle moves along the x-axis with position function s(t) = 5 sin t + 2eᵗ − 3 ln t for t > 0. Find the velocity v(t) and the acceleration a(t). Then determine whether the particle is speeding up or slowing down at t = π.

Show answer

Point 1 (1 pt) — Velocity: v(t) = s′(t) = 5 cos t + 2eᵗ − 3/t. This follows from d/dx(sin t) = cos t, d/dx(eᵗ) = eᵗ, and d/dx(ln t) = 1/t, combined with constant multiple and sum rules. Point 2 (1 pt) — Acceleration: a(t) = v′(t) = −5 sin t + 2eᵗ + 3/t². This follows from d/dx(cos t) = −sin t, d/dx(eᵗ) = eᵗ, and d/dx(−3/t) = d/dx(−3t⁻¹) = 3t⁻² = 3/t². Point 3 (1 pt) — Evaluate at t = π: v(π) = 5 cos π + 2eᵖ − 3/π = 5(−1) + 2eᵖ − 3/π = −5 + 2(23.1407) − 0.9549 ≈ −5 + 46.28 − 0.955 ≈ 40.33. a(π) = −5 sin π + 2eᵖ + 3/π² = 0 + 2(23.1407) + 0.304 ≈ 46.59. Point 4 (1 pt) — Speeding up/slowing down: Since v(π) > 0 and a(π) > 0, velocity and acceleration have the same sign. Therefore the particle is speeding up at t = π.

Question 5: CRITICAL THINKING

3 pts

Prove that the function g(x) = eˣ − x − 1 satisfies g(x) ≥ 0 for all real x, and explain how the derivative of eˣ is central to your argument.

Show answer

Point 1 (1 pt) — Find the critical point: g′(x) = eˣ − 1. Setting g′(x) = 0 gives eˣ = 1, so x = 0. This step relies directly on the fundamental derivative d/dx(eˣ) = eˣ. Point 2 (1 pt) — Classify the critical point: g″(x) = eˣ (again using d/dx(eˣ) = eˣ). Since g″(0) = e⁰ = 1 > 0, the second derivative test confirms that x = 0 is a local minimum. Moreover, since g″(x) = eˣ > 0 for all x, g is concave up everywhere, so x = 0 is a global minimum. Point 3 (1 pt) — Evaluate the minimum and conclude: g(0) = e⁰ − 0 − 1 = 1 − 0 − 1 = 0. Since the global minimum value is 0, we have g(x) ≥ 0 for all real x. This is equivalent to the well-known inequality eˣ ≥ x + 1 for all x ∈ ℝ. The entire proof hinges on the fact that d/dx(eˣ) = eˣ, which allows us to find the critical point, and d²/dx²(eˣ) = eˣ > 0, which establishes global convexity.

SUMMARY

Summary & Key Formulas

Summary

This lesson established the four foundational derivative formulas that every calculus student must know: d/dx(sin x) = cos x, d/dx(cos x) = −sin x, d/dx(eˣ) = eˣ, and d/dx(ln x) = 1/x (for x > 0). The trigonometric derivatives were derived using the angle addition identities together with the Squeeze Theorem limits lim(h→0) sin h / h = 1 and lim(h→0) (cos h − 1) / h = 0. The exponential derivative follows from the defining property of the number e, and the logarithmic derivative follows from the inverse function theorem.

Key patterns to remember include the period-4 trigonometric derivative cycle (sin → cos → −sin → −cos → sin), the self-replicating property of eˣ under differentiation, and the domain restriction x > 0 for ln x. Watch for the common error of dropping the negative sign in d/dx(cos x) and the trap of applying the power rule to eˣ. These four formulas combine with the sum, constant multiple, product, quotient, and chain rules to differentiate virtually any function encountered in AP Calculus BC.