Derivatives of tangent, secant, cotangent and cosecant

Warm-up

Three quick questions from earlier lessons.

Discover

$\tan x = \dfrac{\sin x}{\cos x}$ . With the quotient rule, the top of $(\tan x)'$ is $\cos x \cdot \cos x - \sin x \cdot (-\sin x)$ . What does it simplify to?

1
$\cos^2 x - \sin^2 x$

Key idea

The quotient rule gives the other four.

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

,

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

,

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

,

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

. The "co" functions get a minus sign.

Worked example

$f(x) = 2\tan x - \csc x$ . Find $f'(x)$ .

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

(tan x)' = sec² x.

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

(csc x)' = −csc x cot x, and − · − = +.

f'(x) = 2\sec^2 x + \csc x\cot x

Combine.

Your turn, guided

Your turn: $f(x) = \sec x + x$ .

f'(x) = \sec x\tan x + 1

sec x tan x, plus 1.

Practice ★

$f(x) = 7\tan x$ . Find $f'(x)$ .
$f(x) = 3\tan x$ . Find $f'(x)$ .
$f(x) = 5\tan x$ . Find $f'(x)$ .

Watch out!

Practice ★★

$f(x) = \sec x + 2\cot x$ . Find $f'(x)$ .
$f(x) = \sec x + 8\cot x$ . Find $f'(x)$ .

Find the slope of the tangent line to $y = \tan x$ at $x = \dfrac{\pi}{ 6 }$ . Give an exact value.
Find the slope of the tangent line to $y = \tan x$ at $x = \dfrac{\pi}{ 4 }$ . Give an exact value.

Exit ticket

Three questions to check you have got it.

$f(x) = 7\tan x$ . Find $f'(x)$ .
$f(x) = 3\tan x$ . Find $f'(x)$ .
$f(x) = 5\tan x$ . Find $f'(x)$ .

Summary card

tan, sec, cot, csc

$(\tan x)' = \sec^2 x$ , $(\sec x)' = \sec x\tan x$ .
$(\cot x)' = -\csc^2 x$ , $(\csc x)' = -\csc x\cot x$ .

Watch out: co-functions take a minus sign. You may also write $\sec^2 x$ as $\dfrac{1}{\cos^2 x}$ .

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