Derivatives of tan, cot, sec and csc — AP® Calculus AB Notes

You know the derivatives of $\sin x$ and $\cos x$ . The other four trigonometric functions, $\tan x$ , $\cot x$ , $\sec x$ and $\csc x$ , are all fractions built from those two. So the quotient rule, plus one identity, gives every one of their derivatives. This page works them out, shows a pattern that makes the signs easy to remember, and practices them on exact values.

Rewrite with sine and cosine

Function	In terms of sine and cosine
$\tan x$	$\frac{\sin x}{\cos x}$
$\cot x$	$\frac{\cos x}{\sin x}$
$\sec x$	$\frac{1}{\cos x}$
$\csc x$	$\frac{1}{\sin x}$

Every function in the table is a quotient, defined where its denominator is not 0. The one identity you need is the Pythagorean identity, $\sin^2 x + \cos^2 x = 1$ .

The derivative of tangent

Worked example 1. Find $\frac{d}{dx}(\tan x)$ .

High $= \sin x$ , d-high $= \cos x$ . Low $= \cos x$ , d-low $= -\sin x$ .

Quotient rule: $\frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{\cos^2 x}$ $= \frac{\cos^2 x + \sin^2 x}{\cos^2 x}$ .

Identity: the numerator is 1, so the derivative is $\frac{1}{\cos^2 x} = \sec^2 x$ .

Top: y = tan x between its vertical asymptotes at x = minus pi/2 and pi/2 (dashed), with tangent lines at x = 0 (slope 1) and x = pi/4 (slope 2). Bottom, on the same scale: y = sec squared x, which is 1 at x = 0, 2 at x = pi/4, never less than 1, and grows without bound near the asymptotes. — **Figure 1.** $\tan x$ and its derivative $\sec^2 x$ . The slope of tangent is 1 at the origin and grows toward the asymptotes. LevlPrep original diagram.

The picture agrees. The graph of $\tan x$ always rises on $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$ , gently through the origin with slope $\sec^2 0 = 1$ , then more and more steeply near the asymptotes, where $\cos x \to 0$ and $\sec^2 x$ grows without bound. Since $\cos^2 x \le 1$ , $\sec^2 x = \frac{1}{\cos^2 x} \ge 1$ : no slope of $\tan x$ is ever less than 1.

The derivative of cotangent

Worked example 2. Find $\frac{d}{dx}(\cot x)$ .

$\frac{(-\sin x)(\sin x) - (\cos x)(\cos x)}{\sin^2 x}$ $= \frac{-\left( \sin^2 x + \cos^2 x \right)}{\sin^2 x}$ $= \frac{-1}{\sin^2 x}$ $= -\csc^2 x$ .

Secant and cosecant: a constant on top

Worked example 3. Find $\frac{d}{dx}(\sec x)$ .

High $= 1$ , d-high $= 0$ . Low $= \cos x$ , d-low $= -\sin x$ .

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

The same steps with $\csc x = \frac{1}{\sin x}$ give $\frac{0 - \cos x}{\sin^2 x}$ $= -\csc x\cot x$ .

All six, and the "co-" pattern

$f(x)$	$f'(x)$	Partner	Partner's derivative
$\sin x$	$\cos x$	$\cos x$	$-\sin x$
$\tan x$	$\sec^2 x$	$\cot x$	$-\csc^2 x$
$\sec x$	$\sec x\tan x$	$\csc x$	$-\csc x\cot x$

Read across each row. The partner on the right is the "co-" function: cosine, cotangent, cosecant. Its derivative is the left-hand derivative with every function swapped for its co-function and a minus sign in front. So you only need to learn the left column; the right column follows.

Exact values

On no-calculator questions, evaluate these derivatives at the standard angles by first writing each piece in terms of sine and cosine.

Worked example 4. Find the slope of $y = \sec x$ at $x = \frac{\pi}{3}$ .

$\cos \frac{\pi}{3} = \frac{1}{2}$ and $\sin \frac{\pi}{3} = \frac{\sqrt{3}}{2}$ , so $\sec \frac{\pi}{3} = 2$ and $\tan \frac{\pi}{3} = \sqrt{3}$ .

Slope: $\sec x\tan x = 2\sqrt{3}$ .

Worked example 5. Write the line tangent to $y = \tan x$ at $x = \frac{\pi}{4}$ .

Point: $\tan \frac{\pi}{4} = 1$ . Slope: $\sec^2 \frac{\pi}{4}$ $= \frac{1}{\left( \sqrt{2}/2 \right)^2}$ $= 2$ .

Line: $y = 1 + 2\left( x - \frac{\pi}{4} \right)$ .

Combining with the other rules

These derivatives work with every rule so far. For $y = x\tan x$ , the product rule gives $\tan x + x\sec^2 x$ . For $y = \sec x - 2\cot x$ , the difference and constant multiple rules give $\sec x\tan x + 2\csc^2 x$ ; note the double negative.

Common mistakes

Signs: only the "co-" derivatives are negative.
Squares: $(\tan x)' = \sec^2 x$ , not $\sec x$ ; $(\sec x)' = \sec x\tan x$ , not $\tan^2 x$ .
Degrees: all of these formulas are for radians.

That completes the basic derivatives of Unit 2. Unit 3 handles functions placed inside other functions, such as $\sin\left( x^2 \right)$ .