Trigonometric functions appear everywhere in calculus — from oscillating motion to waves to periodic phenomena. Their derivatives follow a beautiful (and memorable) pattern. You absolutely must memorize these six formulas.
🎯 In this section you will learn
- Derivatives of $\sin x$, $\cos x$, and $\tan x$
- Derivatives of $\sec x$, $\csc x$, and $\cot x$
- How to apply the Chain Rule with trig functions
- Common patterns and memory tricks
📌 The Six Trigonometric Derivatives (Reference Table)
| Function | Derivative | Memory Trick |
|---|
| $\sin x$ | $\cos x$ | "sine goes to cosine" |
| $\cos x$ | $-\sin x$ | "cosine goes to negative sine" |
| $\tan x$ | $\sec^2 x$ | "tangent squared" |
| $\cot x$ | $-\csc^2 x$ | "negative cosecant squared" |
| $\sec x$ | $\sec x \tan x$ | "secant times tangent" |
| $\csc x$ | $-\csc x \cot x$ | "negative cosecant times cotangent" |
💡 Pattern Recognition
- Derivative of $\sin x$ is $\cos x$ (starts with "c")
- Derivative of $\cos x$ is $-\sin x$ (starts with "s", negative)
- Derivatives of $\sec x$ and $\tan x$ also contain $\sec x$ and $\tan x$
- Derivatives of $\csc x$ and $\cot x$ are negative versions of the above
📌 Derivative of $\sin x$ and $\cos x$
- $\frac{d}{dx}\sin x = \cos x$
- $\frac{d}{dx}\cos x = -\sin x$
- $\frac{d}{dx}(5\sin x) = 5\cos x$
- $\frac{d}{dx}(\sin x + \cos x) = \cos x - \sin x$
Find $\frac{d}{dx}\sin(4x)$.
①
Derivative of outside ($\sin u$)
$\cos(4x)$
②
Multiply by derivative of inside ($4x$)
$4$
③
Result
$\frac{d}{dx}\sin(4x) = 4\cos(4x)$
Find $\frac{d}{dx}\cos(3x)$.
- Derivative of $\cos u$ is $-\sin u$
- $\frac{d}{dx}\cos(3x) = -\sin(3x) \cdot 3 = -3\sin(3x)$
📌 Derivative of $\tan x$ and $\sec x$
- $\frac{d}{dx}\tan x = \sec^2 x$
- $\frac{d}{dx}\sec x = \sec x \tan x$
- $\frac{d}{dx}(2\tan x) = 2\sec^2 x$
- $\frac{d}{dx}(\sec x - \tan x) = \sec x \tan x - \sec^2 x$
Find $\frac{d}{dx}\tan(2x)$.
- Derivative of $\tan u$ is $\sec^2 u$
- $\frac{d}{dx}\tan(2x) = \sec^2(2x) \cdot 2 = 2\sec^2(2x)$
Find $\frac{d}{dx}\sec(5x)$.
- Derivative of $\sec u$ is $\sec u \tan u$
- $\frac{d}{dx}\sec(5x) = \sec(5x) \tan(5x) \cdot 5 = 5\sec(5x) \tan(5x)$
📌 Derivative of $\cot x$ and $\csc x$
- $\frac{d}{dx}\cot x = -\csc^2 x$
- $\frac{d}{dx}\csc x = -\csc x \cot x$
Find $\frac{d}{dx}\cot(x^2)$.
- Derivative of $\cot u$ is $-\csc^2 u$
- $\frac{d}{dx}\cot(x^2) = -\csc^2(x^2) \cdot 2x = -2x \csc^2(x^2)$
Find $\frac{d}{dx}\csc(3x)$.
- Derivative of $\csc u$ is $-\csc u \cot u$
- $\frac{d}{dx}\csc(3x) = -\csc(3x) \cot(3x) \cdot 3 = -3\csc(3x) \cot(3x)$
🔄 Combining Trig Derivatives with Other Rules
Find $\frac{d}{dx}(x^2 \sin x)$.
- Product Rule: $f'g + fg'$
- $f(x) = x^2$, $g(x) = \sin x$
- $f'(x) = 2x$, $g'(x) = \cos x$
- $\frac{d}{dx}(x^2 \sin x) = 2x \sin x + x^2 \cos x$
Find $\frac{d}{dx}\left(\frac{\sin x}{x}\right)$.
- Quotient Rule: $\frac{f'g - fg'}{g^2}$
- $f(x) = \sin x$, $g(x) = x$
- $f'(x) = \cos x$, $g'(x) = 1$
- $\frac{d}{dx}\left(\frac{\sin x}{x}\right) = \frac{(\cos x)(x) - (\sin x)(1)}{x^2} = \frac{x\cos x - \sin x}{x^2}$
⚠️ Common Mistakes
- Sign errors: $\frac{d}{dx}\cos x = -\sin x$ (many forget the negative sign!)
- Confusing $\sec x$ and $\csc x$ derivatives: $\frac{d}{dx}\sec x = \sec x \tan x$, $\frac{d}{dx}\csc x = -\csc x \cot x$
- Forgetting the Chain Rule: $\frac{d}{dx}\sin(3x) \neq \cos(3x)$ — missing the $3$ factor!
- Mixing up $\tan^2 x$ and $\tan(x^2)$: They are different functions and require different applications of the chain rule.
🔍 Key Takeaways
- Memorize the six trig derivatives — they appear constantly on the AP exam.
- Pattern: sine ↔ cosine, secant ↔ tangent, cosecant ↔ cotangent (with signs).
- Always apply the Chain Rule when the argument is not simply $x$ (e.g., $\sin(2x)$, $\cos(5x)$, $\tan(x^2)$).
- Combine trig derivatives with product, quotient, and chain rules as needed.
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