Unit 2Derivatives of Trigonometric Functions

Concept 4.102Built for May 2027 AP Exam

Derivatives of Trigonometric Functions

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

Two derivatives carry this topic: $\frac{d}{dx}\sin x = \cos x$ and $\frac{d}{dx}\cos x = -\sin x$ . The minus belongs to the cosine row only, and it is the part students forget under time pressure. The sign is not arbitrary. It follows from the special limit $\lim_{h \to 0}\frac{\sin h}{h} = 1$ together with $\lim_{h \to 0}\frac{\cos h - 1}{h} = 0$ , both of which hold only in radians. When the angle is not a bare $x$ , the pair still applies but an inner factor appears, so $\sin 3x$ differentiates to $3\cos 3x$ . Read every trigonometric derivative as two decisions: which row of the pair, and what the inner angle contributes.

On the AP Exam

Trigonometric derivatives are almost always a vehicle rather than the target. They appear inside an acceleration, a tangent-line slope, or an implicit derivative, and the scored step is the derivative expression itself. Because the cosine row carries a sign, this is the most common place in Unit 2 for a correct method to end in a wrong answer.

What you’ll see: A velocity or position model built from sine and cosine, a curve defined implicitly with a trigonometric term, or a product mixing a polynomial with a trigonometric factor.
What you’ll do: Identify the row, apply the sign, multiply by any inner constant, and evaluate in radian mode.
Memory hook: Cosine is the one with the minus. Sine behaves; cosine pays a tax every time it is differentiated.
Use it on the exam: For MCQs: Check the sign first and the coefficient second. Most wrong options differ from the right one by exactly one of those two. For FRQs: When a rate is modelled by a sum such as $20 + 15\sin\left(\frac{\pi t}{6}\right)$ , the derivative of the sine term carries the factor $\frac{\pi}{6}$ from the inner angle, and dropping it changes the answer by more than half.

Concept Review

Figure 1

The pair is asymmetric on purpose. The three chips show the same rule carrying a coefficient, an inner factor, and a product partner without changing its shape.

SINE AND COSINE TRADE A SIGN

The Pair, and the Blocks Built From It

Function	Derivative	Sign
$\sin x$	$\cos x$	Positive
$\cos x$	$-\sin x$	Negative
$\sin(kx)$	$k\cos(kx)$	Positive, scaled by $k$
$\cos(kx)$	$-k\sin(kx)$	Negative, scaled by $k$
$\tan x$	$\sec^2 x$	Positive (BC extension)
$\sec x$	$\sec x \tan x$	Positive (BC extension)

The last two rows are worth knowing but are not the centre of this topic. Both follow from the quotient rule applied to $\frac{\sin x}{\cos x}$ and $\frac{1}{\cos x}$ , which is why they arrive later in most courses.

Where the Two Special Limits Come From

The derivative of sine is a limit before it is a formula:

$\frac{d}{dx}\sin x = \lim_{h\to 0}\frac{\sin(x+h) - \sin x}{h} = \lim_{h\to 0}\left[\sin x \cdot \frac{\cos h - 1}{h} + \cos x \cdot \frac{\sin h}{h}\right]$

The expansion uses the addition formula for sine. The first bracket vanishes and the second equals $1$ , leaving $\cos x$ . Two facts did the work: $\frac{\sin h}{h} \to 1$ and $\frac{\cos h - 1}{h} \to 0$ . Both are false if $h$ is measured in degrees, which is the entire reason the calculus of trigonometry is written in radians.

The Cycle

Differentiate sine four times and the original function returns: $\sin x \to \cos x \to -\sin x \to -\cos x \to \sin x$ . The cycle has period four, and it explains why the second derivative of a sine is a negative sine. A spring model $y = A\sin(\omega t)$ therefore satisfies $y'' = -\omega^2 y$ , which is the differential equation behind simple harmonic motion.

Common Misconception Check

Myth 1: "The derivative of $\cos x$ is $\sin x$ ." Truth: The sign is negative. A quick check at $x = 0$ settles it: the cosine graph is falling there, so its slope must be negative, and $\sin 0 = 0$ while $-\sin 0 = 0$ gives no help, so check at $x = \frac{\pi}{2}$ instead, where cosine falls steeply and $-\sin\frac{\pi}{2} = -1$ .
Myth 2: "Degrees and radians give the same derivative." Truth: In degrees the derivative of $\sin x$ picks up a factor of $\frac{\pi}{180}$ . Calculus requires radians, and the AP exam states this in the directions.
Myth 3: "The inner angle does not affect the derivative." Truth: Every inner constant comes out as a multiplier. $\sin(3x)$ differentiates to $3\cos(3x)$ , not to $\cos(3x)$ .

Topic-Specific Study Method

Row, Sign, Factor

Borrowed idea: The trig derivatives come in a pair with a sign attached to only one of them. Sine differentiates into cosine and stops there. Cosine differentiates into sine with a minus in front. Where does the minus come from? It comes from the shape of the cosine graph at the origin, which falls while the sine graph climbs, and that asymmetry is the entire content of the topic. Everything else is bookkeeping. The angle must be in radians, and an inner constant such as the $3$ in $\sin 3x$ must be pulled out as a factor before the pair is applied.

Exam disguise	What you see on the page	What your brain must translate	Required execution
A rate model	$20 + 15\sin\left(\frac{\pi t}{6}\right)$	A sum whose second term has an inner factor	Differentiate to $\frac{15\pi}{6}\cos\left(\frac{\pi t}{6}\right)$
An acceleration request	"find the acceleration at $t = 3$ "	The derivative of a trigonometric velocity	Differentiate, then substitute
An implicit curve	$2y^2 - 6 = y\sin x$	A product on the right, a trig term inside	Differentiate both sides, keep $\cos x$
A product with trig	$x^2\sin x$	Two rules at once	Product rule, with the cosine row inside

① Differentiate $\sin x$ , $\cos x$ , $\sin(-x)$ , and $\cos(-x)$ and compare the four results. The sign pattern becomes obvious once the even and odd symmetry of the originals is on the page.

② For any model built from sine or cosine, write the derivative and then evaluate it at one input by hand. Doing the arithmetic catches a missing inner factor faster than checking the formula.

③ On every released problem that involves a trigonometric rate, cover the given answer and compute the derivative yourself first. The comparison trains the sign reflex that the exam rewards.

Training & Error Diagnosis

Core Archetype (Simplest Form): Differentiate $f(x) = 3\sin x - 2\cos x$ .

Standard Execution:

Split. Two terms, two independent derivatives.

First term. The coefficient $3$ stays and the sine row gives cosine: $3\cos x$ .

Second term. The coefficient $-2$ stays and the cosine row gives negative sine: $-2 \cdot (-\sin x) = 2\sin x$ .

Assemble. $f'(x) = 3\cos x + 2\sin x$ .

Check one value. At $x = 0$ the derivative is $3$ , and the graph of $f$ is indeed rising at that rate.

Exam Variation (Context / Parameter Shift): A particle moves along a line with velocity $v(t) = 4\cos\left(\frac{\pi t}{3}\right)$ for $0 \le t \le 6$ . Find the acceleration at $t = 1$ and determine whether the speed is increasing or decreasing there.

Solution: $a(t) = v'(t) = 4 \cdot \left(-\sin\left(\frac{\pi t}{3}\right)\right) \cdot \frac{\pi}{3} = -\frac{4\pi}{3}\sin\left(\frac{\pi t}{3}\right)$ . At $t = 1$ , $a(1) = -\frac{4\pi}{3}\sin\left(\frac{\pi}{3}\right) = -\frac{4\pi}{3} \cdot \frac{\sqrt3}{2} = -\frac{2\pi\sqrt3}{3}$ . Meanwhile $v(1) = 4\cos\left(\frac{\pi}{3}\right) = 2 > 0$ . Velocity and acceleration have opposite signs, so the speed is decreasing. Two signs, one conclusion, and the inner factor $\frac{\pi}{3}$ never disappeared.

Common Traps & Failure Signals:

The missing minus (Signal: a cosine term arriving as positive sine): the sign of the cosine row was dropped.
The frozen angle (Signal: a derivative with no inner factor): the chain contribution of $kx$ was ignored.
The degree habit (Signal: an answer containing $180$ or $\pi/180$ ): the angle was converted where the rule expects radians.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Sign Swap $\rightarrow$ Write the minus before you write the sine.
Degree Reflex: You used degrees out of habit.
Frozen Angle: You differentiated $\sin 3x$ as $\cos 3x$ and left the inner factor of $3$ behind, so the answer came out a third of its true size and every evaluation after that carried the same missing multiplier.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Trig derivatives are scored as the inner step of an acceleration, a tangent slope, or an implicit derivative, rarely as a standalone item
10-Year Total Count: 6 of 9 Exams · 8 FRQ Parts
FRQ Years Appeared: 2017 (Q6a), 2018 (Q2a, Q5b), 2019 (Q1d), 2021 (Q5a), 2023 (Q1d, Q2b), 2026 (Q5A)

Practice Examples

MCQ 1

The Sign

If $f(x) = 5\cos x$ , what is $f'\left(\frac{\pi}{2}\right)$ ?

MCQ 2

The Inner Factor

What is the derivative of $g(x) = \sin(4x)$ ?

MCQ 3

Two Rules at Once

What is the derivative of $h(x) = x\cos x$ ?

FRQ 4

Critical Points of a Trig Model

Let $f$ be the function defined by $f(x) = 2\sin x - x$ on the closed interval $[0, 2\pi]$ .

(a) Find $f'(x)$ .

(b) Find all values of $x$ in $(0, 2\pi)$ at which $f$ has a critical point.

(c) Determine whether $f$ has a relative maximum or a relative minimum at each critical point found in part (b). Justify your answer.

(d) Write an equation for the line tangent to the graph of $f$ at $x = \pi$ .

Cheatsheet