Finding the Derivatives of Tangent, Cotangent, Secant, and Cosecant Functions
| English | Chinese | Pinyin |
|---|---|---|
| trigonometric | 三角函数 | sān jiǎo hán shù |
The other four trig functions
- We know $\sin'=\cos$ and $\cos'=-\sin$. The other four trig functions are all quotients of these.
- So the Quotient Rule derives their derivatives — no new magic, just careful algebra.
- $\tan x=\dfrac{\sin x}{\cos x}$, $\cot x=\dfrac{\cos x}{\sin x}$, $\sec x=\dfrac1{\cos x}$, $\csc x=\dfrac1{\sin x}$.
- Learn the four results, but know they come from the Quotient Rule.
These four derivatives are all obtained using the...
Each of $\tan,\cot,\sec,\csc$ is a quotient of $\sin$/$\cos$.
Deriving $\tan x$
- $\dfrac{d}{dx}[\tan x]=\dfrac{d}{dx}\!\left[\dfrac{\sin x}{\cos x}\right]$.
- Quotient Rule: $\dfrac{\cos x\cos x-\sin x(-\sin x)}{\cos^2 x}=\dfrac{\cos^2 x+\sin^2 x}{\cos^2 x}$.
- Since $\cos^2 x+\sin^2 x=1$: $=\dfrac{1}{\cos^2 x}=\sec^2 x$.
- So $\boxed{\dfrac{d}{dx}[\tan x]=\sec^2 x}$ and, similarly, $\dfrac{d}{dx}[\cot x]=-\csc^2 x$.
The sine wave whose ratio makes tan
$\tan x=\sin x/\cos x$; its derivative $\sec^2 x$ is always positive — tan is increasing wherever it is defined.
What is $\dfrac{d}{dx}[\tan x]$?
From the Quotient Rule on $\sin/\cos$: $\sec^2 x$.
$\dfrac{d}{dx}[\cot x]=$ ____ $\csc^2 x$.
It is $-\csc^2 x$.
The secant and cosecant pair
- $\dfrac{d}{dx}[\sec x]=\sec x\tan x$.
- $\dfrac{d}{dx}[\csc x]=-\csc x\cot x$.
- Pattern: the "co-" functions ($\cos,\cot,\csc$) all carry a minus sign in their derivative.
- Each of these four also follows from the Quotient Rule applied to a $\tfrac1{\cos}$ or $\tfrac1{\sin}$.
What is $\dfrac{d}{dx}[\sec x]$?
$\frac{d}{dx}[\sec x]=\sec x\tan x$.
The derivatives of $\cot x$ and $\csc x$ both carry a minus sign.
$-\csc^2 x$ and $-\csc x\cot x$ — the "co-" functions carry a minus.
Using them with other rules
- These trigonometric derivatives combine with the product and quotient rules like any others.
- $\dfrac{d}{dx}[x\tan x]=\tan x+x\sec^2 x$ (Product Rule).
- $\dfrac{d}{dx}[3\sec x-\cot x]=3\sec x\tan x+\csc^2 x$.
- Keep the four boxed results handy; they turn up throughout the rest of the course.
Select all correct derivatives.
The last is wrong: $\frac{d}{dx}[\sec x]=\sec x\tan x$, not $\sec^2 x$.
Watch the minus signs on the "co-" functions: $\frac{d}{dx}[\cot x]=-\csc^2 x$ and $\frac{d}{dx}[\csc x]=-\csc x\cot x$ both carry a minus, matching $\cos'=-\sin$. Mixing up $\sec^2 x$ (from $\tan$) with $\csc^2 x$ (from $\cot$) is another frequent slip — pair each derivative with its source.
Differentiate $y=x^2\sec x$.
- Product Rule with $f=x^2$, $g=\sec x$: $f'=2x$, $g'=\sec x\tan x$.
- $y'=f'g+fg'=2x\sec x+x^2\sec x\tan x$.
- Factor: $y'=x\sec x\,(2+x\tan x)$.
The four extra trigonometric derivatives all follow from the Quotient Rule: $\frac{d}{dx}[\tan x]=\sec^2 x$, $\frac{d}{dx}[\cot x]=-\csc^2 x$, $\frac{d}{dx}[\sec x]=\sec x\tan x$, $\frac{d}{dx}[\csc x]=-\csc x\cot x$. The "co-" functions carry a minus. Combine them freely with the product and quotient rules.