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Every derivative involving sine, cosine, or their relatives traces back to six core formulas. Once you know these, differentiating any trigonometric expression becomes a matter of combining them with the product rule, quotient rule, and chain rule.
The remarkable fact: the derivative of $\sin x$ is $\cos x$, and the derivative of $\cos x$ is $-\sin x$. The trig functions differentiate into each other, a beautiful cycle that makes higher derivatives predictable.
| Property | Value |
|---|---|
| Concept | Derivatives |
| Chapter | 2.4 |
| Difficulty | Intermediate |
| Time | ~25 minutes |
| Function | Derivative | Memory Aid |
|---|---|---|
| $\sin x$ | $\cos x$ | Positive pair |
| $\cos x$ | $-\sin x$ | Negative (cosine is "co" = contrary) |
| $\tan x$ | $\sec^2 x$ | Squares of secant |
| $\cot x$ | $-\csc^2 x$ | Negative squares |
| $\sec x$ | $\sec x \tan x$ | Sec stays, brings tan |
| $\csc x$ | $-\csc x \cot x$ | Negative, csc stays |
Pattern: The "co" functions (cos, cot, csc) all have negative derivatives.
$$\boxed{\frac{d}{dx}(\sin x) = \cos x \qquad \frac{d}{dx}(\cos x) = -\sin x}$$
$$\boxed{\frac{d}{dx}(\tan x) = \sec^2 x \qquad \frac{d}{dx}(\cot x) = -\csc^2 x}$$
$$\boxed{\frac{d}{dx}(\sec x) = \sec x \tan x \qquad \frac{d}{dx}(\csc x) = -\csc x \cot x}$$
Using the limit definition of the derivative:
$$\frac{d}{dx}(\sin x) = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h}$$
Apply the angle addition formula $\sin(x+h) = \sin x \cos h + \cos x \sin h$:
$$= \lim_{h \to 0} \frac{\sin x \cos h + \cos x \sin h - \sin x}{h}$$
$$= \lim_{h \to 0} \left(\sin x \cdot \frac{\cos h - 1}{h} + \cos x \cdot \frac{\sin h}{h}\right)$$
Using the special trig limits:
Therefore:
$$\frac{d}{dx}(\sin x) = \sin x \cdot 0 + \cos x \cdot 1 = \cos x$$
Similarly, using $\cos(x+h) = \cos x \cos h - \sin x \sin h$:
$$\frac{d}{dx}(\cos x) = \lim_{h \to 0} \frac{\cos x \cos h - \sin x \sin h - \cos x}{h}$$
$$= \lim_{h \to 0} \left(\cos x \cdot \frac{\cos h - 1}{h} - \sin x \cdot \frac{\sin h}{h}\right)$$
$$= \cos x \cdot 0 - \sin x \cdot 1 = -\sin x$$
Since $\tan x = \frac{\sin x}{\cos x}$, apply the quotient rule:
$$\frac{d}{dx}(\tan x) = \frac{\cos x \cdot \cos x - \sin x \cdot (-\sin x)}{\cos^2 x}$$
$$= \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x$$
The remaining formulas (cot, sec, csc) are derived similarly using the quotient rule.
The derivatives of sine and cosine repeat in a cycle of 4:
| $n$ | $f^{(n)}(x)$ for $f(x) = \sin x$ | $f^{(n)}(x)$ for $f(x) = \cos x$ |
|---|---|---|
| 0 | $\sin x$ | $\cos x$ |
| 1 | $\cos x$ | $-\sin x$ |
| 2 | $-\sin x$ | $-\cos x$ |
| 3 | $-\cos x$ | $\sin x$ |
| 4 | $\sin x$ | $\cos x$ |
| ... | (repeats) | (repeats) |
To find $f^{(n)}(x)$: Compute $n \mod 4$ and look up in the table.
Since $27 = 4 \cdot 6 + 3$, we have $27 \mod 4 = 3$.
From the table: $f^{(3)}(x) = \sin x$ when $f(x) = \cos x$.
Therefore: $\frac{d^{27}}{dx^{27}}(\cos x) = \sin x$
Find the derivative of each function:
Differentiate $f(x) = 3\sin x - 2\cos x$.
For $f(x) = \sec x$, find $f''(x)$.
Find the 100th derivative of $f(x) = \sin x$.
Prove that $\frac{d}{dx}(\sec x) = \sec x \tan x$ using the quotient rule and the known derivative of cosine.
Notice that:
Why does the derivative of cosine have a negative sign but the derivative of sine doesn't?
The trig derivative cycle:
sin x → cos x
↑ ↓
-cos x ← -sin xEvery four derivatives, you return to where you started. The negative signs enter at positions 2 and 3 of the cycle.
For the other four: Remember that "co" functions (those with "co" in the name) always have a negative sign in their derivatives.
Looking back:
Looking ahead:
Real-world connections:
| Previous | Up | Next |
|---|---|---|
| Special Trig Limits | Skills Index | Differentiating Trig Expressions |
Last updated: 2026-01-22