Trigonometric Derivative Formulas

5 min read

Reference: Stewart §2.4

The Six Formulas That Unlock Trigonometric Calculus

Every derivative involving sine, cosine, or their relatives traces back to six core formulas. Once you know these, differentiating any trigonometric expression becomes a matter of combining them with the product rule, quotient rule, and chain rule.

The remarkable fact: the derivative of $\sin x$ is $\cos x$, and the derivative of $\cos x$ is $-\sin x$. The trig functions differentiate into each other, a beautiful cycle that makes higher derivatives predictable.

Prerequisite Map

Quick Reference

Property	Value
Concept	Derivatives
Chapter	2.4
Difficulty	Intermediate
Time	~25 minutes

The Six Derivative Formulas

The Core Table

Function	Derivative	Memory Aid
$\sin x$	$\cos x$	Positive pair
$\cos x$	$-\sin x$	Negative (cosine is “co” = contrary)
$\tan x$	$\sec^2 x$	Squares of secant
$\cot x$	$-\csc^2 x$	Negative squares
$\sec x$	$\sec x \tan x$	Sec stays, brings tan
$\csc x$	$-\csc x \cot x$	Negative, csc stays

Pattern: The “co” functions (cos, cot, csc) all have negative derivatives.

The Formulas Boxed

$$\boxed{\frac{d}{dx}(\sin x) = \cos x \qquad \frac{d}{dx}(\cos x) = -\sin x}$$

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

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

Common misconception

$\sin x$ is an algebraic expression that can be manipulated like $s \cdot i \cdot n \cdot x$, or that $\frac{d}{dx}(\sin x) = \cos x$ is a formula to memorize without meaning.

This is the trig-as-algebra-symbols error. The notation $\sin x$ does not mean a product of four letters; it means the sine function evaluated at $x$. It names a specific real-valued process: given an angle $x$ in radians, $\sin x$ returns the $y$-coordinate of the corresponding point on the unit circle. Because it is a function, the derivative formula $\frac{d}{dx}(\sin x) = \cos x$ is not an arbitrary rule -- it can be derived from the limit definition and the angle addition formula. Knowing that $\sin x$ is a function, not an algebraic expression, also explains why $\sin(x + h) \neq \sin x + \sin h$: functions do not distribute over addition the way multiplication does. Students who treat $\sin$ as an algebraic symbol will misapply every trig identity and derivative formula.

Proof: Derivative of Sine

Using the limit definition of the derivative:

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

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

$$= \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h}$$

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

Using the special trig limits:

$\lim_{h \to 0} \frac{\sin h}{h} = 1$
$\lim_{h \to 0} \frac{\cos h - 1}{h} = 0$

Therefore:

$$\frac{d}{dx}(\sin x) = \sin x \cdot 0 + \cos x \cdot 1 = \cos x$$

Proof: Derivative of Cosine

Similarly, using $\cos(x+h) = \cos x \cos h - \sin x \sin h$:

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

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

$$= \cos x \cdot 0 - \sin x \cdot 1 = -\sin x$$

Proof: Derivative of Tangent (Quotient Rule)

Since $\tan x = \frac{\sin x}{\cos x}$, apply the quotient rule:

$$\frac{d}{dx}(\tan x) = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x}$$

$$= \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x$$

The remaining formulas (cot, sec, csc) are derived similarly using the quotient rule.

The Cyclic Pattern of Higher Derivatives

The derivatives of sine and cosine repeat in a cycle of 4:

$n$	$f^{(n)}(x)$ for $f(x) = \sin x$	$f^{(n)}(x)$ for $f(x) = \cos x$
0	$\sin x$	$\cos x$
1	$\cos x$	$-\sin x$
2	$-\sin x$	$-\cos x$
3	$-\cos x$	$\sin x$
4	$\sin x$	$\cos x$
...	(repeats)	(repeats)

To find $f^{(n)}(x)$: Compute $n \mod 4$ and look up in the table.

Example: Finding the 27th Derivative of $\cos x$

Since $27 = 4 \cdot 6 + 3$, we have $27 \mod 4 = 3$.

From the table: $f^{(3)}(x) = \sin x$ when $f(x) = \cos x$.

Therefore: $\frac{d^{27}}{dx^{27}}(\cos x) = \sin x$

Practice Problems

Level 1 State the Formulas

Find the derivative of each function:

$f(x) = \sin x$
$f(x) = \cos x$
$f(x) = \tan x$

Thought Process

These are direct applications of the formulas. Just recall them from the table.

Show Answer

$f'(x) = \cos x$
$f'(x) = -\sin x$
$f'(x) = \sec^2 x$

Level 2 Simple Combinations

Differentiate $f(x) = 3\sin x - 2\cos x$.

Thought Process

Use the sum/difference rule and constant multiple rule. Differentiate each term separately.

Show Answer

$$f'(x) = 3\frac{d}{dx}(\sin x) - 2\frac{d}{dx}(\cos x)$$

$$= 3\cos x - 2(-\sin x)$$

$$= 3\cos x + 2\sin x$$

Level 3 Second Derivative

For $f(x) = \sec x$, find $f''(x)$.

Thought Process

First find $f'(x) = \sec x \tan x$.

Then differentiate again using the product rule on $\sec x \cdot \tan x$.

Show Answer

First derivative: $f'(x) = \sec x \tan x$

Second derivative (using product rule):

$$f''(x) = \frac{d}{dx}(\sec x) \cdot \tan x + \sec x \cdot \frac{d}{dx}(\tan x)$$

$$= \sec x \tan x \cdot \tan x + \sec x \cdot \sec^2 x$$

$$= \sec x \tan^2 x + \sec^3 x$$

$$= \sec x(\tan^2 x + \sec^2 x)$$

Alternative form using $\sec^2 x = 1 + \tan^2 x$:

$$= \sec x(2\tan^2 x + 1) \quad \text{or} \quad \sec x(2\sec^2 x - 1)$$

Level 4 Higher Derivative Pattern

Find the 100th derivative of $f(x) = \sin x$.

Thought Process

The derivatives of sine cycle with period 4: $\sin x \to \cos x \to -\sin x \to -\cos x \to \sin x \to \ldots$

Find $100 \mod 4$ to determine where we are in the cycle.

Show Answer

Since the derivatives of $\sin x$ repeat every 4 steps:

$f^{(0)}(x) = \sin x$
$f^{(1)}(x) = \cos x$
$f^{(2)}(x) = -\sin x$
$f^{(3)}(x) = -\cos x$
$f^{(4)}(x) = \sin x$ (back to start)

Calculate $100 \div 4 = 25$ with remainder $0$.

Since $100 \mod 4 = 0$, we’re at position 0 in the cycle.

$$f^{(100)}(x) = \sin x$$

Level 5 Prove the Secant Derivative

Prove that $\frac{d}{dx}(\sec x) = \sec x \tan x$ using the quotient rule and the known derivative of cosine.

Thought Process

Write $\sec x = \frac{1}{\cos x}$ and apply the quotient rule.

The numerator is constant (1), so its derivative is 0.

Then simplify using trig identities.

Show Answer

Write $\sec x = \frac{1}{\cos x}$ and apply the quotient rule:

$$\frac{d}{dx}(\sec x) = \frac{d}{dx}\left(\frac{1}{\cos x}\right)$$

$$= \frac{\cos x \cdot \frac{d}{dx}(1) - 1 \cdot \frac{d}{dx}(\cos x)}{\cos^2 x}$$

$$= \frac{\cos x \cdot 0 - 1 \cdot (-\sin x)}{\cos^2 x}$$

$$= \frac{\sin x}{\cos^2 x}$$

$$= \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x}$$

$$= \sec x \tan x \quad \blacksquare$$

Conceptual Check (CCI-Style)

Conceptual The Negative Sign Pattern

Notice that:

$\frac{d}{dx}(\sin x) = \cos x$ (positive)
$\frac{d}{dx}(\cos x) = -\sin x$ (negative)

Why does the derivative of cosine have a negative sign but the derivative of sine doesn’t?

Thought Process

Think about what the graphs of sine and cosine look like near $x = 0$.

At $x = 0$:

$\sin x$ is increasing (slope positive), and $\cos 0 = 1 > 0$ ✓
$\cos x$ is decreasing (slope negative), and $-\sin 0 = 0$...

Look at $x = \pi/2$ instead:

$\cos x$ at $x = \pi/2$ is decreasing, and $-\sin(\pi/2) = -1 < 0$ ✓

Show Answer

Graphical interpretation:

At $x = 0$:

The sine curve is at a minimum-to-maximum crossing, going up. Positive slope matches $\cos 0 = 1 > 0$. ✓

At $x = 0$:

The cosine curve is at its maximum, about to go down. The slope is zero, matching $-\sin 0 = 0$. ✓

At $x = \pi/2$:

The cosine curve is decreasing (going from 1 to 0 to -1). The slope is negative, matching $-\sin(\pi/2) = -1$. ✓

The negative sign reflects that cosine decreases where sine is positive.

The derivatives encode the phase relationship: cosine leads sine by $\pi/2$, so when sine is positive (going up), cosine is going down (negative derivative).

Mastery Checklist

I can state all six trig derivative formulas from memory
I know the “negative pattern”: co-functions (cos, cot, csc) have negative derivatives
I can prove $\frac{d}{dx}(\sin x) = \cos x$ using the limit definition
I can derive tan, sec, cot, csc derivatives using the quotient rule
I can find high-order derivatives using the period-4 cycle
I understand why the formulas require radian measure

Mental Model

The trig derivative cycle:

sin x  →  cos x
  ↑         ↓
-cos x ← -sin x

Every four derivatives, you return to where you started. The negative signs enter at positions 2 and 3 of the cycle.

For the other four: Remember that “co” functions (those with “co” in the name) always have a negative sign in their derivatives.

Connections

Looking back:

Special Trig Limits provides the key limits for the proofs
Quotient Rule derives tan, cot, sec, csc

Looking ahead:

Differentiating Trig Expressions combines these with product/quotient rules
Chain Rule extends these to compositions like $\sin(x^2)$

Real-world connections:

Simple harmonic motion: position, velocity, acceleration are trig functions
Wave equations in physics involve trig derivatives

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Special Trig Limits	Skills Index	Differentiating Trig Expressions

Last updated: 2026-01-22