Derivatives of Tangent, Cotangent, Secant, and Cosecant (AP Calculus AB 2.10)

Tangent, cotangent, secant and cosecant are all built from sine and cosine by division: $\tan x = \frac{\sin x}{\cos x}$ , $\cot x = \frac{\cos x}{\sin x}$ , $\sec x = \frac{1}{\cos x}$ , $\csc x = \frac{1}{\sin x}$ . So their derivatives need nothing new: the quotient rule (topic 2.9), the derivatives of sine and cosine (topic 2.7), and the identity $\sin^2 x + \cos^2 x = 1$ . Deriving them once shows you where each formula comes from, and the same rewriting habit lets you simplify many trig expressions before differentiating.

Tangent and cotangent

For $\tan x = \frac{\sin x}{\cos x}$ , the top has derivative $\cos x$ and the bottom has derivative $-\sin x$ . The quotient rule gives

\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 Pythagorean identity is what collapses the numerator to 1. This holds wherever $\cos x \neq 0$ , which is exactly where $\tan x$ is defined.

The result says something about the graph. Since $\cos^2 x \le 1$ , its reciprocal $\sec^2 x$ is always at least 1. So every tangent line to $y = \tan x$ has slope at least 1: the graph rises on every interval of its domain, with its gentlest slope, exactly 1, at $x = 0, \pm\pi, \ldots$ , where $\cos^2 x = 1$ . Near $x = \pm\frac{\pi}{2}$ , $\cos x \to 0$ and the slope grows without bound, matching the vertical asymptotes.

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

\frac{d}{dx}\cot x = \frac{(-\sin x)\sin x - \cos x\cos x}{\sin^2 x} = \frac{-\left(\sin^2 x + \cos^2 x\right)}{\sin^2 x} = -\frac{1}{\sin^2 x} = -\csc^2 x,

valid wherever $\sin x \neq 0$ . The minus sign means cotangent falls on every interval of its domain.

Secant and cosecant

Both are reciprocals, so the reciprocal form of the quotient rule, $\left(\frac{1}{g}\right)' = -\frac{g'}{g^2}$ , does the work. For $\sec x = \frac{1}{\cos x}$ :

\frac{d}{dx}\sec x = -\frac{-\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.

The last step splits $\cos^2 x$ into two factors so the answer can be written with the familiar functions. For $\csc x = \frac{1}{\sin x}$ :

\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.

The pattern makes these easy to keep straight. Each "co" function's derivative is its partner's derivative with every function swapped for its co-function and a minus sign in front: $\tan \to \cot$ , $\sec \to \csc$ , and a negative sign. That is the same pattern as $\sin x \to \cos x$ versus $\cos x \to -\sin x$ .

Equivalent forms are all correct. Because $1 + \tan^2 x = \sec^2 x$ (divide $\sin^2 x + \cos^2 x = 1$ by $\cos^2 x$ ), the derivative of $\tan x$ can also be written $1 + \tan^2 x$ or $\frac{1}{\cos^2 x}$ . Exam answers in any equivalent form earn credit, and multiple-choice questions sometimes list the derivative in one of the less familiar forms.

Rewrite with identities first

These four rules plus the ones from topics 2.6 to 2.9 can differentiate any combination of trig functions, but rewriting first often turns a long calculation into a short one. The tools are the definitions and the Pythagorean identities:

\tan x = \frac{\sin x}{\cos x}, \quad \sec x = \frac{1}{\cos x}, \quad \sin^2 x + \cos^2 x = 1, \quad 1 + \tan^2 x = \sec^2 x, \quad 1 + \cot^2 x = \csc^2 x.

Some quick cases:

$\cos x\tan x = \cos x\cdot\frac{\sin x}{\cos x} = \sin x$ (where $\cos x \neq 0$ ), so its derivative is $\cos x$ , without the product rule.
$\frac{\sec x}{\tan x} = \frac{1}{\cos x}\cdot\frac{\cos x}{\sin x} = \csc x$ , so its derivative is $-\csc x\cot x$ , without the quotient rule.
$\sec^2 x - \tan^2 x = 1$ , a constant, so its derivative is 0.

The rewritten function equals the original only where the original is defined. $\cos x\tan x$ is undefined at $x = \frac{\pi}{2}$ even though $\sin x$ is not, so the derivative $\cos x$ applies only where $\cos x \neq 0$ .

Evaluating these derivatives at special angles is where arithmetic errors happen. Get each value from sine and cosine:

$x$	$\sin x$	$\cos x$	$\tan x$	$\sec x$	$\sec^2 x$
$\frac{\pi}{6}$	$\frac{1}{2}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{\sqrt{3}}$	$\frac{2}{\sqrt{3}}$	$\frac{4}{3}$
$\frac{\pi}{4}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{2}}{2}$	$1$	$\sqrt{2}$	$2$
$\frac{\pi}{3}$	$\frac{\sqrt{3}}{2}$	$\frac{1}{2}$	$\sqrt{3}$	$2$	$4$

Worked examples

Worked exampleExample 1: evaluating at a special angle

Let $f(x) = 3\tan x - 2\sec x$ . Find $f'\left(\dfrac{\pi}{3}\right)$ .

Show solution

Differentiate term by term with the constant multiple rule:

f'(x) = 3\sec^2 x - 2\sec x\tan x.

At $x = \frac{\pi}{3}$ , $\cos x = \frac{1}{2}$ , so $\sec x = 2$ and $\sec^2 x = 4$ , and $\tan x = \sqrt{3}$ :

f'\left(\tfrac{\pi}{3}\right) = 3(4) - 2(2)\left(\sqrt{3}\right) = 12 - 4\sqrt{3}.

Worked exampleExample 2: simplify, then differentiate

Find $\dfrac{dy}{dx}$ for $y = \dfrac{1 - \cos^2 x}{\sin x\cos x}$ .

Show solution

Differentiating this as it stands needs the quotient rule and a product rule in the denominator. Rewrite first. By the Pythagorean identity the numerator is $\sin^2 x$ , so wherever $\sin x \neq 0$ and $\cos x \neq 0$ ,

y = \frac{\sin^2 x}{\sin x\cos x} = \frac{\sin x}{\cos x} = \tan x.

Therefore

\frac{dy}{dx} = \sec^2 x,

for every $x$ where the original function is defined (that is, where neither $\sin x$ nor $\cos x$ is 0).

Worked exampleExample 3: a product with secant

Write an equation for the line tangent to the graph of $h(x) = x\sec x$ at $x = \dfrac{\pi}{3}$ .

Show solution

The point: $h\left(\frac{\pi}{3}\right) = \frac{\pi}{3}\cdot 2 = \frac{2\pi}{3}$ .

The slope: $h$ is a product, so

h'(x) = 1\cdot\sec x + x\sec x\tan x.

At $x = \frac{\pi}{3}$ , with $\sec x = 2$ and $\tan x = \sqrt{3}$ ,

h'\left(\tfrac{\pi}{3}\right) = 2 + \frac{\pi}{3}(2)\left(\sqrt{3}\right) = 2 + \frac{2\sqrt{3}\,\pi}{3}.

The tangent line is

y = \frac{2\pi}{3} + \left(2 + \frac{2\sqrt{3}\,\pi}{3}\right)\left(x - \frac{\pi}{3}\right).

There is nothing to gain from simplifying further; point-slope form with exact values is a complete answer.

Worked exampleExample 4: an exam-style free-response question (no calculator)

Let $g(x) = 2x - \tan x$ for $-\dfrac{\pi}{2} < x < \dfrac{\pi}{2}$ .

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

(b) Find all $x$ in the interval at which the graph of $g$ has a horizontal tangent line.

Show solution

(a) $g'(x) = 2 - \sec^2 x$ .

(b) Set $g'(x) = 0$ : $\sec^2 x = 2$ , so $\cos^2 x = \frac{1}{2}$ and $\cos x = \frac{\sqrt{2}}{2}$ (cosine is positive on this interval, so the negative root is not possible). Then $x = -\frac{\pi}{4}$ or $x = \frac{\pi}{4}$ .

(c) No. For every $x$ in the interval, $0 < \cos^2 x \le 1$ , so $\sec^2 x = \frac{1}{\cos^2 x} \ge 1$ . Therefore $g'(x) = 2 - \sec^2 x \le 2 - 1 = 1$ . The slope of the graph of $g$ is never greater than 1, so it is never 3.

Part (c) is the kind of question where a bound on the derivative is the whole argument. The justification must say why $\sec^2 x \ge 1$ , not only that $g'(x) = 3$ "has no solution".

Where students lose points

On the exam

These four formulas are expected from memory; no formula sheet gives them, and they appear inside product, quotient and chain-rule problems as well as on their own. On no-calculator questions, evaluating them at $\frac{\pi}{6}$ , $\frac{\pi}{4}$ and $\frac{\pi}{3}$ by hand is part of the task, and exact answers such as $12 - 4\sqrt{3}$ need no decimal approximation.

Check yourself

Question 1
What is the derivative of $\cot x$ at $x = \dfrac{\pi}{4}$ ?
1. A
  $2$
2. B
  $-1$
3. C
  $-2$
4. D
  $-\sqrt{2}$
Show answer
C
$\frac{d}{dx}\cot x = -\csc^2 x$ , and $\csc\frac{\pi}{4} = \sqrt{2}$ , so the value is $-2$ . Choice A drops the minus sign, and choice D forgets to square $\csc x$ .
Question 2
Find $\dfrac{d}{dx}\left(\tan^2 x - \sec^2 x\right)$ .
Show answer
$0$
Since $1 + \tan^2 x = \sec^2 x$ , the expression equals $-1$ wherever it is defined, and a constant has derivative 0. Differentiating each term separately also works, but the identity shows the answer at once.
Question 3
Write an equation for the line tangent to the graph of $y = \csc x$ at $x = \dfrac{\pi}{2}$ .
Show answer
$y = 1$
The point is $\left(\frac{\pi}{2}, \csc\frac{\pi}{2}\right) = \left(\frac{\pi}{2}, 1\right)$ . The slope is $-\csc\frac{\pi}{2}\cot\frac{\pi}{2} = -(1)(0) = 0$ , so the tangent line is horizontal.
Question 4
For how many values of $x$ in $-\dfrac{\pi}{2} < x < \dfrac{\pi}{2}$ is the derivative of $\tan x$ equal to $\dfrac{1}{2}$ ?
1. A
  One
2. B
  None
3. C
  Two
4. D
  Infinitely many
Show answer
B
The derivative is $\sec^2 x = \frac{1}{\cos^2 x}$ , and $\cos^2 x \le 1$ makes it at least 1 everywhere, so it never equals $\frac{1}{2}$ . Choice C comes from solving $\cos^2 x = \frac{1}{2}$ , which is where the derivative equals 2, not $\frac{1}{2}$ .

Go deeper

OpenStax, Calculus Volume 1, 3.5 Derivatives Of Trigonometric Functions