Derivatives of tan, cot, sec and csc

Part 1 · Hook

Why this matters

A surveyor tilts a sight toward a mountaintop, and the height she reads grows with the tangent of the angle. Near flat angles the height barely changes; near straight up it explodes. How fast it changes is the derivative of

\tan x

, and the quotient rule finds it, along with the other three trigonometric derivatives, from sine and cosine alone.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. $\frac{d}{dx}\left( \frac{f(x)}{g(x)} \right)$ is

$\frac{f'(x)g(x) - f(x)g'(x)}{\left( g(x) \right)^2}$
$\frac{f'(x)}{g'(x)}$
$\frac{f(x)g'(x) - f'(x)g(x)}{\left( g(x) \right)^2}$
$f'(x)g(x) + f(x)g'(x)$

Show the answer

Low d-high minus high d-low, over low squared.

Correct: $\frac{f'(x)g(x) - f(x)g'(x)}{\left( g(x) \right)^2}$ :
$\frac{f'(x)}{g'(x)}$ :
$\frac{f(x)g'(x) - f'(x)g(x)}{\left( g(x) \right)^2}$ :
$f'(x)g(x) + f(x)g'(x)$ :

2. $\sin^2 x + \cos^2 x$ equals

$1$
$0$
$2\sin x\cos x$
$\tan^2 x$

Show the answer

The Pythagorean identity, from the unit circle.

Correct: $1$ :
$0$ :
$2\sin x\cos x$ :
$\tan^2 x$ :

3. $\frac{d}{dx}(\cos x)$ is

$-\sin x$
$\sin x$
$\cos x$
$-\cos x$

Show the answer

Cosine falls where sine is positive.

Correct: $-\sin x$ :
$\sin x$ :
$\cos x$ :
$-\cos x$ :

Part 4 · See it

See it first

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. — The graph of $\tan x$ (top) has slope 1 at $x = 0$ and 2 at $x = \frac{\pi}{4}$ . Its derivative, $\sec^2 x$ (bottom), is never below 1. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

$\tan x = \frac{\sin x}{\cos x}$ the quotient rule gives $\frac{\cos x \cos x - \sin x(-\sin x)}{\cos^2 x}$
The numerator is $\cos^2 x + \sin^2 x = 1$ $\frac{d}{dx}(\tan x)$ $= \frac{1}{\cos^2 x}$ $= \sec^2 x$
$\sec x = \frac{1}{\cos x}$ and $\csc x = \frac{1}{\sin x}$ the quotient rule with a constant top gives $\sec x\tan x$ and $-\csc x\cot x$
Each "co-" function is built from cosine where its partner uses sineits derivative carries a minus sign: $-\csc^2 x$ , $-\csc x\cot x$

Part 6 · Key ideas

Key ideas

$\frac{d}{dx}(\tan x) = \sec^2 x$ and $\frac{d}{dx}(\cot x) = -\csc^2 x$ .
$\frac{d}{dx}(\sec x)$ $= \sec x\tan x$ and $\frac{d}{dx}(\csc x)$ $= -\csc x\cot x$ .
Every one comes from writing the function with $\sin x$ and $\cos x$ , using the quotient rule and the identity $\sin^2 x + \cos^2 x = 1$ .
The three "co-" functions (cosine, cotangent, cosecant) have derivatives with a minus sign.
All of these hold in radians, wherever the function is defined.

Part 7 · Misconception

A common mistake

The wrong idea: $\frac{d}{dx}(\sec x) = \tan^2 x$ , or $\frac{d}{dx}(\cot x) = \csc^2 x$ .

What actually happens: Rebuild them instead of guessing: $\sec x = \frac{1}{\cos x}$ , so its derivative is $\frac{0 - 1(-\sin x)}{\cos^2 x}$ $= \frac{\sin x}{\cos^2 x}$ $= \sec x\tan x$ . For $\cot x = \frac{\cos x}{\sin x}$ , the numerator is $-\sin^2 x - \cos^2 x = -1$ , so the derivative is $-\csc^2 x$ , negative like every "co-" derivative.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Graph

The graph of $y = \tan x$ on $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$

1. What is the slope of the graph of $y = \tan x$ at $x = \frac{\pi}{4}$ ?

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

$\sec^2 \frac{\pi}{4}$ $= \frac{1}{\cos^2 \frac{\pi}{4}}$ $= \frac{1}{1/2}$ $= 2$ .

Answer: 2

2. Which is an equation of the line tangent to $y = \tan x$ at $x = \frac{\pi}{4}$ ?

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

Show the answer

Point $\left( \frac{\pi}{4}, 1 \right)$ , slope $\sec^2 \frac{\pi}{4} = 2$ .

Correct: $y = 1 + 2\left( x - \frac{\pi}{4} \right)$ : Height 1, slope 2.
$y = 2 + \left( x - \frac{\pi}{4} \right)$ : The height and the slope are swapped.
$y = 1 + \sqrt{2}\left( x - \frac{\pi}{4} \right)$ : Slope $\sqrt{2}$ is $\sec \frac{\pi}{4}$ , not squared.
$y = 1 + 2\left( x + \frac{\pi}{4} \right)$ : Point-slope form uses $x - \frac{\pi}{4}$ .

3. Why is every slope of the graph of $y = \tan x$ on $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$ at least 1?

The slope is $\sec^2 x = \frac{1}{\cos^2 x}$ , and $0 < \cos^2 x \le 1$ there.
The slope is $\tan x$ , which is at least 1.
The graph passes through the origin with slope 1, and slopes stay the same.
The slope is $\sec x$ , which is at least 1 on this interval.

Show the answer

Dividing 1 by a number between 0 and 1 gives a number of at least 1, with equality at $x = 0$ .

Correct: The slope is $\sec^2 x = \frac{1}{\cos^2 x}$ , and $0 < \cos^2 x \le 1$ there.: $\cos^2 x$ is at most 1, so its reciprocal is at least 1.
The slope is $\tan x$ , which is at least 1.: $\tan x$ is the height, and it is negative for $x < 0$ .
The graph passes through the origin with slope 1, and slopes stay the same.: The slope changes: it is 1 at 0 and 2 at $\frac{\pi}{4}$ .
The slope is $\sec x$ , which is at least 1 on this interval.: The derivative is $\sec^2 x$ , not $\sec x$ .

4. Which line is tangent to $y = \tan x$ at the origin?

$y = x$
$y = 0$
$y = 1 + x$
$x = 0$

Show the answer

Slope $\sec^2 0 = 1$ through $(0, 0)$ .

Correct: $y = x$ : Slope 1 through the origin.
$y = 0$ : The graph rises through the origin; it is not level there.
$y = 1 + x$ : The line must pass through $(0, \tan 0) = (0, 0)$ .
$x = 0$ : A vertical line would need an infinite slope; $\sec^2 0 = 1$ .

5. Use the quotient rule to find $\frac{d}{dx}\left( \frac{\sin x}{\cos x} \right)$ .

Type an expression for $\frac{d}{dx}\left( \frac{\sin x}{\cos x} \right) =$ .

Show the answer

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

Answer: $\frac{d}{dx}\left( \frac{\sin x}{\cos x} \right) =$ $(\sec(x))^{2}$

6. Deriving $\frac{d}{dx}(\tan x) = \sec^2 x$ with the quotient rule, which identity simplifies the numerator?

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

Show the answer

The numerator is $\cos^2 x + \sin^2 x$ , which is 1.

Correct: $\sin^2 x + \cos^2 x = 1$ : The Pythagorean identity collapses the numerator to 1.
$\tan x = \frac{\sin x}{\cos x}$ : That identity sets up the quotient; it does not simplify the numerator.
$\sin(2x) = 2\sin x\cos x$ : No $2\sin x\cos x$ appears in the numerator.
$\sec x = \frac{1}{\cos x}$ : That identity rewrites the final answer; the numerator needs the Pythagorean identity.

Part 9 · Summary

Summary

Write tangent, cotangent, secant and cosecant in terms of sine and cosine, apply the quotient rule, and simplify with

\sin^2 x + \cos^2 x = 1

. The results:

(\tan x)' = \sec^2 x

,

(\cot x)' = -\csc^2 x

,

(\sec x)' = \sec x\tan x

,

(\csc x)' = -\csc x\cot x

. The "co-" functions get the minus signs.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

The graph of y=tan⁡xy = \tan x on (−π2,π2)\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)

Summary

What comes next

Connections

The graph of $y = \tan x$ on $\left( -\frac{\pi}{2}, \frac{\pi}{2} \right)$