Derivatives of Other Trigonometric Functions

5 min read

Reference: Stewart §2.4

Beyond Sine and Cosine

Now that you know the derivatives of sine and cosine, you can find the derivatives of all six trigonometric functions. The remaining four (tangent, cotangent, secant, and cosecant) are just ratios of sine and cosine, so the quotient rule does all the work.

The results are worth memorizing because they appear frequently in calculus and its applications, but understanding how they’re derived means you can reconstruct any formula you forget.

Prerequisite Map

Quick Reference

Property	Value
Concept	Trigonometric Derivatives
Chapter	2.4
Difficulty	Intermediate
Time	~20 minutes

The Complete Table of Trig Derivatives

$$\boxed{ \begin{array}{\vert c\vert c\vert } \hline \textbf{Function} & \textbf{Derivative} \\ \hline \sin x & \cos x \\ \cos x & -\sin x \\ \tan x & \sec^2 x \\ \cot x & -\csc^2 x \\ \sec x & \sec x \tan x \\ \csc x & -\csc x \cot x \\ \hline \end{array} }$$

Memory aid: Notice the pattern in the right column:

The “co” functions (cos, cot, csc) all have negative derivatives
The derivatives of tan and cot involve squared functions
The derivatives of sec and csc involve products of two trig functions

Deriving the Formulas

Derivative of Tangent

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

$$\frac{d}{dx}(\tan x) = \frac{d}{dx}\left(\frac{\sin x}{\cos x}\right)$$

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

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

Derivative of Secant

Since $\sec x = \frac{1}{\cos x}$, we apply the quotient rule (or think of it as $(\cos x)^{-1}$):

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

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

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

Derivative of Cotangent

Since $\cot x = \frac{\cos x}{\sin x}$:

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

$$= \frac{-\sin^2 x - \cos^2 x}{\sin^2 x} = \frac{-1}{\sin^2 x} = -\csc^2 x$$

Derivative of Cosecant

Since $\csc x = \frac{1}{\sin x}$:

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

$$= \frac{-\cos x}{\sin^2 x} = -\frac{1}{\sin x} \cdot \frac{\cos x}{\sin x} = -\csc x \cot x$$

Memory Strategies

The “Co” Pattern

Every “co-function” has a negative derivative:

$(\cos x)' = -\sin x$
$(\cot x)' = -\csc^2 x$
$(\csc x)' = -\csc x \cot x$

The Structure Pattern

Function	Derivative Structure
$\tan x$	“sec squared”
$\cot x$	“negative csc squared”
$\sec x$	“itself times tan”
$\csc x$	“negative itself times cot”

Quick Derivation Check

Can’t remember if it’s $\sec x \tan x$ or $\sec^2 x$?

Test at a point: At $x = 0$:

$\sec 0 = 1$, so the function value is 1
$\tan 0 = 0$, so if the derivative is $\sec x \tan x$, then $(\sec x)'\vert _{x=0} = 1 \cdot 0 = 0$
If the derivative were $\sec^2 x$, then $(\sec x)'\vert _{x=0} = 1$

The secant function has a horizontal tangent at $x = 0$ (it’s at a local minimum), so the derivative should be 0 there. This confirms $(\sec x)' = \sec x \tan x$.

Practice Problems

Level 1 Direct Differentiation

Find $\frac{d}{dx}(3\tan x + 2\sec x)$.

Thought Process

Apply the derivative to each term separately:

Derivative of $\tan x$ is $\sec^2 x$
Derivative of $\sec x$ is $\sec x \tan x$

Constants pull out as usual.

Show Answer

$$\frac{d}{dx}(3\tan x + 2\sec x) = 3\sec^2 x + 2\sec x \tan x$$

This can be factored as $\sec x(3\sec x + 2\tan x)$.

Level 2 Product Rule with Tangent

Find $\frac{d}{dx}(x^2 \tan x)$.

Thought Process

This is a product of $f(x) = x^2$ and $g(x) = \tan x$.

Product rule: $(fg)' = f'g + fg'$

$f'(x) = 2x$
$g'(x) = \sec^2 x$

Show Answer

$$\frac{d}{dx}(x^2 \tan x) = 2x \tan x + x^2 \sec^2 x$$

This can be factored as $x(2\tan x + x\sec^2 x)$.

Level 3 Quotient with Trig Functions

Find $f'(x)$ if $f(x) = \frac{\sec x}{1 + \tan x}$.

Thought Process

Apply the quotient rule with:

Numerator: $u = \sec x$, so $u' = \sec x \tan x$
Denominator: $v = 1 + \tan x$, so $v' = \sec^2 x$

Quotient rule: $\left(\frac{u}{v}\right)' = \frac{u'v - uv'}{v^2}$

After computing, look for simplifications using the identity $\sec^2 x = 1 + \tan^2 x$.

Show Answer

Using the quotient rule:

$$f'(x) = \frac{(\sec x \tan x)(1 + \tan x) - (\sec x)(\sec^2 x)}{(1 + \tan x)^2}$$

$$= \frac{\sec x \tan x + \sec x \tan^2 x - \sec^3 x}{(1 + \tan x)^2}$$

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

Using $\sec^2 x = 1 + \tan^2 x$:

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

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

Level 4 Horizontal Tangents

Find all values of $x$ in $[0, 2\pi)$ where the graph of $f(x) = \tan x - 2x$ has a horizontal tangent.

Thought Process

Horizontal tangent means $f'(x) = 0$.

Find $f'(x) = \sec^2 x - 2$
Set equal to zero: $\sec^2 x = 2$
This means $\cos^2 x = \frac{1}{2}$, so $\cos x = \pm \frac{1}{\sqrt{2}}$
Find all $x$ in $[0, 2\pi)$ satisfying this

Be careful to exclude values where $\tan x$ is undefined ($x = \pi/2, 3\pi/2$).

Show Answer

Step 1: Find the derivative. $$f'(x) = \sec^2 x - 2$$

Step 2: Set $f'(x) = 0$. $$\sec^2 x = 2$$ $$\cos^2 x = \frac{1}{2}$$ $$\cos x = \pm \frac{1}{\sqrt{2}} = \pm \frac{\sqrt{2}}{2}$$

Step 3: Find solutions in $[0, 2\pi)$.

$\cos x = \frac{\sqrt{2}}{2}$: $x = \frac{\pi}{4}, \frac{7\pi}{4}$

$\cos x = -\frac{\sqrt{2}}{2}$: $x = \frac{3\pi}{4}, \frac{5\pi}{4}$

Step 4: Verify these are in the domain of $f$ (not at $x = \pi/2$ or $3\pi/2$).

All four values are valid.

Answer: $x = \frac{\pi}{4}, \frac{3\pi}{4}, \frac{5\pi}{4}, \frac{7\pi}{4}$

Level 5 Deriving the Cosecant Formula

Prove that $\frac{d}{dx}(\csc x) = -\csc x \cot x$ using the quotient rule.

Then verify your answer by showing that $\frac{d}{dx}(\sin x \cdot \csc x) = 0$, which must be true since $\sin x \cdot \csc x = 1$.

Thought Process

Part 1: Write $\csc x = \frac{1}{\sin x}$ and apply the quotient rule.

Part 2: Use the product rule on $\sin x \cdot \csc x$: $$(\sin x \cdot \csc x)' = \cos x \cdot \csc x + \sin x \cdot (\csc x)'$$

This should equal 0. Substituting the formula for $(\csc x)'$ and simplifying should confirm this.

Show Answer

Part 1: Derivation

$$\frac{d}{dx}(\csc x) = \frac{d}{dx}\left(\frac{1}{\sin x}\right)$$

Quotient rule with $u = 1$, $v = \sin x$:

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

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

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

$$= -\csc x \cot x \quad \checkmark$$

Part 2: Verification

We know $\sin x \cdot \csc x = 1$ for all $x$ where $\csc x$ is defined.

By the product rule: $$\frac{d}{dx}(\sin x \cdot \csc x) = \cos x \cdot \csc x + \sin x \cdot (-\csc x \cot x)$$

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

$$= \frac{\cos x}{\sin x} - \frac{\cos x}{\sin x}$$

$$= 0 \quad \checkmark$$

This confirms our derivative formula is correct.

CCI-Style Conceptual Questions

CCI Domain Awareness

The derivative of $\tan x$ is $\sec^2 x$. This formula is valid for:

All real numbers $x$
All $x$ except where $\sin x = 0$
All $x$ except where $\cos x = 0$
All $x$ where $-1 \leq x \leq 1$

Thought Process

The derivative formula is valid wherever the original function is differentiable.

$\tan x = \frac{\sin x}{\cos x}$ is undefined where $\cos x = 0$, which happens at $x = \pm\frac{\pi}{2}, \pm\frac{3\pi}{2}, \ldots$

At these points, $\tan x$ has vertical asymptotes and is not differentiable.

Show Answer

Answer: (C) All $x$ except where $\cos x = 0$

The tangent function (and therefore its derivative) is defined everywhere except at odd multiples of $\frac{\pi}{2}$, which is exactly where $\cos x = 0$.

CCI Sign Patterns

On the interval $(0, \frac{\pi}{2})$, which of the following is true about $\frac{d}{dx}(\sec x)$?

It is always negative
It is always positive
It changes from negative to positive
It equals zero at $x = \frac{\pi}{4}$

Thought Process

$\frac{d}{dx}(\sec x) = \sec x \tan x$

On $(0, \frac{\pi}{2})$:

$\sec x = \frac{1}{\cos x} > 0$ (since $\cos x > 0$)
$\tan x = \frac{\sin x}{\cos x} > 0$ (since both $\sin x > 0$ and $\cos x > 0$)

So the product $\sec x \tan x > 0$ throughout the interval.

This makes sense: $\sec x$ is increasing on $(0, \frac{\pi}{2})$, going from 1 toward infinity.

Show Answer

Answer: (B) It is always positive

On $(0, \frac{\pi}{2})$, both $\sec x > 0$ and $\tan x > 0$, so their product is positive.

Geometrically, this matches the fact that $\sec x = \frac{1}{\cos x}$ increases from 1 to $+\infty$ as $x$ goes from 0 to $\frac{\pi}{2}$.

Common Mistakes

Mistake	Correction
Writing $(\tan x)' = \sec^2 x \tan x$	$(\tan x)' = \sec^2 x$ (no extra $\tan x$)
Forgetting the negative: $(\csc x)' = \csc x \cot x$	$(\csc x)' = -\csc x \cot x$
Confusing $(\sec x)'$ and $(\tan x)'$	sec → “self times tan”; tan → “sec squared”
Applying formulas where trig function is undefined	Check domain: $\tan, \sec$ undefined at $x = \frac{\pi}{2} + n\pi$

Common Misconceptions

Common misconception

the negative sign in $(\csc x)'$ and $(\cot x)'$ can be omitted because the functions are “just reciprocals.” The co-functions csc, cot, and cos all carry a negative sign in their derivatives. A useful pattern is that every derivative of a “co” function (cosine, cotangent, cosecant) picks up a negative sign. Omitting this sign changes the answer entirely: $(\csc x)' = -\csc x \cot x$, not $\csc x \cot x$.

Common misconception

$\tan x$ and $\sec x$ are defined for all real $x$. Both functions are undefined wherever $\cos x = 0$, that is, at $x = \pi/2 + n\pi$ for integer $n$. Applying derivative formulas at these points or including them in a domain without restriction produces undefined results. Every problem involving $\tan$, $\sec$, $\cot$, or $\csc$ requires attention to domain.

Mastery Checklist

I can state the derivatives of all six trig functions
I can derive $(\tan x)'$ and $(\sec x)'$ using the quotient rule
I recognize the “co = negative” pattern
I can differentiate combinations like $a\tan x + b\sec x$
I can apply the product and quotient rules with these functions
I know the domain restrictions for each trig derivative

Mental Model

The Quotient Rule Does It All:

Think of tan, cot, sec, and csc as “built from” sin and cos:

           sin x                    cos x
tan x = ─────────      cot x = ─────────
           cos x                    sin x

              1                        1
sec x = ─────────      csc x = ─────────
           cos x                    sin x

If you ever forget a formula, just apply the quotient rule to the appropriate ratio. The key ingredients are:

$(\sin x)' = \cos x$
$(\cos x)' = -\sin x$
The identity $\sin^2 x + \cos^2 x = 1$

Connections

Looking back:

Derivatives of sine and cosine provide the building blocks
The quotient rule is the essential tool for derivation

Looking ahead:

The chain rule lets you handle $\tan(3x)$, $\sec(x^2)$, etc.
Implicit differentiation often involves these derivatives
These formulas are essential for integration techniques later

Real-world connections:

$\sec^2 x$ appears in projectile motion with air resistance
These derivatives are essential in analyzing periodic signals in engineering

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Derivatives of Sine and Cosine	Skills Index	The Chain Rule

Last updated: 2026-01-22