Derivatives of All Six Trigonometric Functions
Derive d/dx[sin x] = cos x and d/dx[cos x] = −sin x from the special limits. Derive the remaining four trig derivatives from these using the quotient rule. Apply in combination with product and chain rules.
A lesson of AP Calculus AB, in Differentiation: Definition and Fundamental Properties, about 10 minutes.
Trigonometric Derivatives
By the end of this lesson, you will be able to differentiate all six trigonometric functions from memory and apply them in combination with other rules.
Derivatives of the Six Trig Functions
These must be memorized:
| Function | Derivative |
|---|---|
| $\sin x$ | $\cos x$ |
| $\cos x$ | $-\sin x$ |
| $\tan x$ | $\sec^2 x$ |
| $\csc x$ | $-\csc x \cot x$ |
| $\sec x$ | $\sec x \tan x$ |
| $\cot x$ | $-\csc^2 x$ |
Memory patterns:
- The "co-" functions (cosine, cosecant, cotangent) all have a negative sign in their derivative
- $\tan x$ → $\sec^2 x$ and $\cot x$ → $-\csc^2 x$ are "squared" forms
- $\sec x$ and $\csc x$ keep themselves as a factor in their own derivative
Deriving sin x and cos x from the Limit Definition
$$\frac{d}{dx}[\sin x] = \lim_{h\to 0}\frac{\sin(x+h)-\sin x}{h}$$
Using the sum identity $\sin(x+h) = \sin x\cos h + \cos x\sin h$:
$$= \lim_{h\to 0}\frac{\sin x\cos h + \cos x\sin h - \sin x}{h}$$
$$= \sin x \cdot \underbrace{\lim_{h\to 0}\frac{\cos h - 1}{h}}{= 0} + \cos x \cdot \underbrace{\lim{h\to 0}\frac{\sin h}{h}}_{= 1} = \cos x$$
The key special limits: $\lim_{h\to 0}\frac{\sin h}{h} = 1$ and $\lim_{h\to 0}\frac{\cos h - 1}{h} = 0$.
Deriving the Other Trig Derivatives
$\tan x = \frac{\sin x}{\cos x}$ → apply the quotient rule:
$$\frac{d}{dx}[\tan x] = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x$$
Examples
Example 1: Differentiate $f(x) = 3\sin x - 2\cos x + \tan x$
$f'(x) = 3\cos x + 2\sin x + \sec^2 x$
Example 2: Differentiate $g(x) = x^2\sin x$
Product rule: $g'(x) = 2x\sin x + x^2\cos x$
Example 3: Differentiate $h(x) = \dfrac{\tan x}{x}$
Quotient rule: $h'(x) = \dfrac{\sec^2 x \cdot x - \tan x \cdot 1}{x^2} = \dfrac{x\sec^2 x - \tan x}{x^2}$