Derivatives of Other Trig Functions
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 3.5: “Derivatives of Trigonometric Functions” |
| Book URL | https://openstax.org/details/books/calculus-volume-1 |
Freely available and openly licensed.
Try This First (Predict, Then Derive)
You know that $\dfrac{d}{dx}[\sin x] = \cos x$ and $\dfrac{d}{dx}[\cos x] = -\sin x$.
You also know the quotient rule: $\dfrac{d}{dx}\left[\dfrac{f}{g}\right] = \dfrac{f'g - fg'}{g^2}$.
And you know $\tan x = \dfrac{\sin x}{\cos x}$.
Predict: before reading the derivation below, try to find $\dfrac{d}{dx}[\tan x]$ yourself using the quotient rule. Write your answer in simplest form.
Predicted: $\underline{\hspace{4cm}}$.
The answer is $\sec^2 x$. Check your algebra against the derivation in Section 1 below.
Prerequisite Check
Before this lesson, make sure you can do all of the following:
Quick Reference
| Function | Derivative | Domain |
|---|---|---|
| $\tan x$ | $\sec^2 x$ | $x \neq \pi/2 + n\pi$ |
| $\cot x$ | $-\csc^2 x$ | $x \neq n\pi$ |
| $\sec x$ | $\sec x \tan x$ | $x \neq \pi/2 + n\pi$ |
| $\csc x$ | $-\csc x \cot x$ | $x \neq n\pi$ |
Pattern. The “co-” functions ($\cot$, $\csc$) have a negative sign in their derivative; the non-co functions ($\tan$, $\sec$) do not.
Key Concepts
1. Derivative of $\tan x$
Since $\tan x = \dfrac{\sin x}{\cos x}$, apply the quotient rule: \[ \frac{d}{dx}[\tan x] = \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x. \]
The Pythagorean identity $\sin^2 x + \cos^2 x = 1$ is the key step.
2. Derivative of $\cot x$
$\cot x = \dfrac{\cos x}{\sin x}$. Quotient rule: \[ \frac{d}{dx}[\cot x] = \frac{(-\sin x)(\sin x) - (\cos x)(\cos x)}{\sin^2 x} = \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x} = \frac{-1}{\sin^2 x} = -\csc^2 x. \]
3. Derivative of $\sec x$
$\sec x = \dfrac{1}{\cos x}$. Quotient rule (or constant-over-$f$ via chain rule): \[ \frac{d}{dx}[\sec x] = \frac{0 \cdot \cos x - 1 \cdot (-\sin x)}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} = \sec x \tan x. \]
4. Derivative of $\csc x$
$\csc x = \dfrac{1}{\sin x}$. Quotient rule: \[ \frac{d}{dx}[\csc x] = \frac{-\cos x}{\sin^2 x} = -\frac{1}{\sin x} \cdot \frac{\cos x}{\sin x} = -\csc x \cot x. \]
5. Multiple Representations
Graphical meaning of $\sec^2 x = (\tan x)'$.
The graph of $\tan x$ has vertical asymptotes at $x = \pi/2 + n\pi$. Near these asymptotes, $\tan x$ rises (or falls) steeply, so the slope $\sec^2 x$ is large. At $x = 0$, $\tan 0 = 0$ and the curve is relatively flat; the slope is $\sec^2 0 = 1$.
Since $\sec^2 x = 1 + \tan^2 x \geq 1$ always, the slope of $\tan x$ is never less than 1. The tangent function is always increasing on its domain.
Table at key values:
| $x$ | $\tan x$ | $\sec^2 x$ (slope of $\tan x$) |
|---|---|---|
| $0$ | $0$ | $1$ |
| $\pi/4$ | $1$ | $2$ |
| $\pi/3$ | $\sqrt{3}$ | $4$ |
The slope increases as $x$ approaches $\pi/2$.
6. Ask Why: Why Is the Pattern “co-functions get a negative sign”?
The co-functions ($\cot$, $\csc$) are the “co-versions” of $\tan$ and $\sec$. In general, if $f$ and $g$ are co-functions, then $f'(x) = -g'(\pi/2 - x)$ (they are related by a reflection). The negative sign arises because the chain rule applied to the co-relationship $\cot x = \tan(\pi/2 - x)$ produces a factor of $-1$.
Explicitly: $\dfrac{d}{dx}[\cot x] = \dfrac{d}{dx}[\tan(\pi/2 - x)] = \sec^2(\pi/2 - x) \cdot (-1) = -\csc^2 x$.
Named Misconception: Forgetting the Domain
A frequent error is to differentiate $\tan x$ or $\sec x$ at a point where they are undefined. At $x = \pi/2$, $\tan x$ has a vertical asymptote; the derivative $\sec^2 x$ is also undefined there. Derivatives do not exist where the function does not exist.
Common Errors
| Error | Example | Correction |
|---|---|---|
| Sign error on $\cot$ or $\csc$ | Writing $+\csc^2 x$ for derivative of $\cot x$ | Derivative of $\cot x$ is $-\csc^2 x$; co-functions get a negative sign |
| Forgetting domain | Evaluating $\frac{d}{dx}[\tan x]$ at $x = \pi/2$ | $\tan(\pi/2)$ is undefined; the derivative does not exist there |
| Confusing $\sec^2 x$ with $\sec x$ | Writing $\sec x$ instead of $\sec^2 x$ for the derivative of $\tan x$ | The entire denominator $\cos^2 x$ becomes $\sec^2 x$ after the Pythagorean identity |
Leveled Practice
Level 1 -- Direct Differentiation
Problem 1. Differentiate: (a) $f(x) = \tan x + x$, (b) $g(x) = 3\sec x$, (c) $h(x) = \csc x - \cot x$.
Show answer
(a) $\sec^2 x + 1$.
(b) $3\sec x\tan x$.
(c) $-\csc x\cot x - (-\csc^2 x) = \csc^2 x - \csc x\cot x = \csc x(\csc x - \cot x)$.
Problem 2. Find the equation of the tangent line to $y = \tan x$ at $x = \pi/4$.
Show answer
$y(\pi/4) = 1$. $y'(\pi/4) = \sec^2(\pi/4) = (\sqrt{2})^2 = 2$.
Tangent: $y - 1 = 2(x - \pi/4)$, i.e., $y = 2x - \pi/2 + 1$.
Level 2 -- Combining Rules
Problem 3. Differentiate $f(x) = x^2\sec x$.
Show answer
Product rule: $f'(x) = 2x\sec x + x^2 \sec x\tan x = x\sec x(2 + x\tan x)$.
Problem 4. Differentiate $g(x) = \dfrac{\tan x}{1 + \sin x}$.
Show answer
Quotient rule: $g'(x) = \dfrac{\sec^2 x(1+\sin x) - \tan x \cdot \cos x}{(1+\sin x)^2}$.
Simplify numerator: $\sec^2 x(1+\sin x) - \sin x$ (since $\tan x\cos x = \sin x$).
$= \dfrac{(1+\sin x)/\cos^2 x - \sin x}{(1+\sin x)^2}$.
This can be left in terms of $\sec$ and $\tan$; the exact form depends on how much simplification is expected.
Level 3 -- Low-Floor-High-Ceiling Extension
Problem 5 (Extension).
(a) (Floor) Verify $\dfrac{d}{dx}[\cot x] = -\csc^2 x$ using the identity $\cot x = \cos x / \sin x$ and the quotient rule.
(b) (Mid) Prove the identity $\sec^2 x = 1 + \tan^2 x$, then use it to write the derivative of $\tan x$ in two forms: as $\sec^2 x$ and as $1 + \tan^2 x$.
(c) (Ceiling) Let $f(x) = e^x \tan x$. Find $f'(x)$ and determine all critical points in $(0, \pi/2)$ (where $f' = 0$ or $f'$ is undefined). Discuss whether any exist.
Show answer
(a) $\frac{d}{dx}[\cos x / \sin x] = \frac{-\sin x \cdot \sin x - \cos x \cdot \cos x}{\sin^2 x} = \frac{-1}{\sin^2 x} = -\csc^2 x$.
(b) $1 + \tan^2 x = 1 + \sin^2x/\cos^2 x = (\cos^2 x + \sin^2 x)/\cos^2 x = 1/\cos^2 x = \sec^2 x$.
So the derivative of $\tan x$ can be written as $\sec^2 x$ or equivalently $1 + \tan^2 x$.
(c) $f'(x) = e^x\tan x + e^x\sec^2 x = e^x(\tan x + \sec^2 x)$.
Since $e^x > 0$ always, $f'(x) = 0$ requires $\tan x + \sec^2 x = 0$. Using $\sec^2 x = 1 + \tan^2 x$: need $\tan x + 1 + \tan^2 x = 0$, i.e., $\tan^2 x + \tan x + 1 = 0$. The discriminant is $1 - 4 = -3 < 0$: no real solutions. So $f'(x) \neq 0$ on $(0, \pi/2)$. Also $f'$ is undefined at $x = \pi/2$ (endpoint, not interior). No critical points.
Common Misconceptions
$\sec x$ is the inverse cosine function. The secant function is the reciprocal of cosine: $\sec x = 1/\cos x$. The inverse cosine is written $\cos^{-1} x$ or $\arccos x$ and is a completely different function. The notation $\sec x$ and $\cos^{-1} x$ are not interchangeable. This matters for derivatives: $(\sec x)' = \sec x \tan x$, while $(\arccos x)' = -1/\sqrt{1-x^2}$.
differentiating $\tan(3x)$ gives $\sec^2(3x)$. The formula $(\tan u)' = \sec^2 u$ applies only when $u$ itself is the variable being differentiated. When $u = 3x$, the chain rule requires multiplying by the derivative of the inner function: $\frac{d}{dx}[\tan(3x)] = \sec^2(3x) \cdot 3$. Omitting the chain rule factor is one of the most common errors with trig derivatives.
Mastery Checklist
You have mastered this skill when you can do all of the following without referring to notes:
Mental Model
All four derivative formulas here follow from the same recipe: write the function in terms of $\sin$ and $\cos$, apply the quotient rule, use $\sin^2 + \cos^2 = 1$ to simplify the numerator, and factor back into trig notation.
The pattern “co-functions get a negative sign” is a compact memory rule grounded in the fact that co-functions are related by replacing $x$ with $\pi/2 - x$, which introduces a factor of $-1$ from the chain rule.
Connections
Within Calculus I (MATH161)
- Chain rule: Once you know these four derivatives, the chain rule lets you differentiate $\tan(g(x))$, $\sec(g(x))$, etc. for any differentiable $g$.
- Integration: $\int \sec^2 x\,dx = \tan x + C$ and $\int \sec x\tan x\,dx = \sec x + C$ are the direct antiderivative facts from this lesson.
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