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Calculus AB · Unit 2 · 2.10
AP Calculus · Differentiation: Definition and Fundamental Properties
Derivatives of $\tan$, $\cot$, $\sec$ and $\csc$
Two results to memorise; the other two are their mirrors with a minus sign.
key idea$\text{Every \emph{co-} function carries a MINUS sign}$
The rule
Four more derivatives, and they come in pairs. Learn $\tan$ and $\sec$; then $\cot$ and $\csc$ are their mirror images with a minus sign attached. Every function beginning with co- differentiates to something negative.
| $f(x)$ | $f'(x)$ | Mirror partner | Sign |
|---|
| $\tan x$ | $\sec^{2}x$ | $\cot x\to-\csc^{2}x$ | + |
| $\sec x$ | $\sec x\tan x$ | $\csc x\to-\csc x\cot x$ | + |
| $\cot x$ | $-\csc^{2}x$ | — | − |
| $\csc x$ | $-\csc x\cot x$ | — | − |
Two facts to memorise, not four. Each co- row is its partner with every function co-ed and a minus added.
The pattern$\tan\!\to\!\cot$, $\sec\!\to\!\csc$, then prepend a minus
Domain$\tan,\sec$ break where $\cos x=0$; $\cot,\csc$ where $\sin x=0$
Where they come fromall four follow from the quotient rule on $\sin$ and $\cos$
Pattern recognition
| If you see… | Recognise | Because |
|---|
| $\dfrac{\sin x}{\cos x}$ | $\tan x$ → $\sec^{2}x$ | no quotient rule needed |
| $\dfrac{1}{\cos x}$ | $\sec x$ → $\sec x\tan x$ | reciprocal identity |
| $\dfrac{1}{\sin^{2}x}$ | $\csc^{2}x$ | the square of a reciprocal, not $\csc$ of a square |
| $\sec^{2}x-1$ | $\tan^{2}x$ | Pythagorean identity — simplifies many answers |
| $\tan x\sec x$ | already a derivative | it is $\dfrac{d}{dx}[\sec x]$ |
Rewriting into $\tan$, $\sec$, $\cot$ or $\csc$ before differentiating usually saves a full quotient-rule expansion.
Worked examples
EX 1
Derive one from the quotient rule
$\sec x$Show that $\dfrac{d}{dx}[\sec x]=\sec x\tan x$.
1
Write as a reciprocal
$\sec x=\dfrac{1}{\cos x}$, so $f=1$ and $g=\cos x$.
2
Apply the quotient rule
$\dfrac{0\cdot\cos x-1\cdot(-\sin x)}{\cos^{2}x}=\dfrac{\sin x}{\cos^{2}x}$.
3
Split the fraction
$\dfrac{\sin x}{\cos^{2}x}=\dfrac{1}{\cos x}\cdot\dfrac{\sin x}{\cos x}$.
4
Name the pieces
$=\sec x\tan x$.
5
Note the source of the sign
The $-\sin x$ from differentiating $\cos$ met the minus in the quotient rule and became positive.
EX 2
Use the mirror
$\csc x$Write down $\dfrac{d}{dx}[\csc x]$ using the pattern, then check it.
1
Start from the partner
$\dfrac{d}{dx}[\sec x]=\sec x\tan x$.
2
Co- every function
$\sec\to\csc$ and $\tan\to\cot$, giving $\csc x\cot x$.
3
Attach the minus
$\dfrac{d}{dx}[\csc x]=-\csc x\cot x$.
4
Verify by quotient rule
$\csc x=\dfrac{1}{\sin x}$, so the derivative is $\dfrac{-\cos x}{\sin^{2}x}=-\csc x\cot x$. ✓
5
Why the pattern works
The co- functions are built from $\sin$ where the originals use $\cos$, and $\cos$ is the one that differentiates to a negative.
EX 3
Combined with a product
$x\tan x$Differentiate $y=x\tan x$.
1
Identify the structure
A product: $f=x$, $g=\tan x$.
2
Differentiate each
$f'=1$ and $g'=\sec^{2}x$.
3
Apply the product rule
$y'=1\cdot\tan x+x\sec^{2}x$.
4
Present it
$y'=\tan x+x\sec^{2}x$.
5
State the domain
Both $\tan$ and $\sec$ break where $\cos x=0$, so $y'$ is undefined at $x=\tfrac{\pi}{2}+k\pi$.
EX 4
Simplify with an identity
$\tan^{2}x$Differentiate $y=\tan^{2}x$ and simplify.
1
Recognise a composition
$y=\left(\tan x\right)^{2}$, so the chain rule applies with outer $u^{2}$ and inner $\tan x$.
2
Differentiate
$y'=2\tan x\cdot\sec^{2}x$.
3
Consider rewriting
Using $\sec^{2}x=1+\tan^{2}x$: $y'=2\tan x\left(1+\tan^{2}x\right)$.
4
Choose a form
Both are correct. Keep $2\tan x\sec^{2}x$ unless a later step needs everything in one function.
5
Guard the notation
$\tan^{2}x$ means $(\tan x)^{2}$, not $\tan(x^{2})$. The second would need a different inner derivative.
Watch out
Every co- function gets a minus
$\cot$, $\csc$ — and $\cos$ from the previous topic. If your answer for a co- function is positive, check the sign.
$\sec^{2}x$ is not $\sec\left(x^{2}\right)$
The exponent applies to the whole function value. Same for $\tan^{2}x$ and $\csc^{2}x$.
$\tan$ and $\sec$ share a domain; $\cot$ and $\csc$ share the other
$\tan,\sec$ break where $\cos x=0$; $\cot,\csc$ break where $\sin x=0$. State the exclusions when asked for a domain.
Recognise before you expand
$\dfrac{\sin x}{\cos x}$ is $\tan x$ with a one-step derivative. Running the quotient rule on it works but wastes time and invites slips.