Derivatives of Trigonometric Functions | Tony's Intertwingled Memex

Core Definition

The derivatives of the six fundamental trigonometric functions (with angles in radians) are:

\begin{aligned} > \frac{d}{dx}[\sin x] &= \cos x &\quad \frac{d}{dx}[\cos x] &= -\sin x \\ > \frac{d}{dx}[\tan x] &= \sec^2 x &\quad \frac{d}{dx}[\cot x] &= -\csc^2 x \\ > \frac{d}{dx}[\sec x] &= \sec x \tan x &\quad \frac{d}{dx}[\csc x] &= -\csc x \cot x > \end{aligned}

Mnemonic Pattern: Every "co-" function (

\cos, \cot, \csc

) produces a negative derivative.

Concept Explanation

1. Proofs of Fundamental Trigonometric Derivatives

The derivatives of $\sin x$ and $\cos x$ are established from the limit definition using the addition angle formulas and special trigonometric limits $\lim_{\theta \to 0} \frac{\sin \theta}{\theta} = 1$ and $\lim_{\theta \to 0} \frac{\cos \theta - 1}{\theta} = 0$ :

\frac{d}{dx}[\sin x] = \lim_{h \to 0} \frac{\sin(x+h) - \sin x}{h} = \lim_{h \to 0} \left[ \sin x \left(\frac{\cos h - 1}{h}\right) + \cos x \left(\frac{\sin h}{h}\right) \right] = \cos x

The remaining four trigonometric derivatives are derived directly via the Quotient Rule:

Derivative of $\tan x = \frac{\sin x}{\cos x}$ : $\frac{d}{dx}[\tan x] = \frac{(\cos x)(\cos x) - (\sin x)(-\sin x)}{\cos^2 x} = \frac{\cos^2 x + \sin^2 x}{\cos^2 x} = \frac{1}{\cos^2 x} = \sec^2 x$
Derivative of $\sec x = \frac{1}{\cos x}$ : $\frac{d}{dx}[\sec x] = \frac{(0)(\cos x) - (1)(-\sin x)}{\cos^2 x} = \frac{\sin x}{\cos^2 x} = \frac{1}{\cos x} \cdot \frac{\sin x}{\cos x} = \sec x \tan x$
Derivative of $\cot x = \frac{\cos x}{\sin x}$ : $\frac{d}{dx}[\cot x] = \frac{(-\sin x)(\sin x) - (\cos x)(\cos x)}{\sin^2 x} = \frac{-(\sin^2 x + \cos^2 x)}{\sin^2 x} = -\frac{1}{\sin^2 x} = -\csc^2 x$
Derivative of $\csc x = \frac{1}{\sin x}$ : $\frac{d}{dx}[\csc x] = \frac{(0)(\sin x) - (1)(\cos x)}{\sin^2 x} = -\frac{\cos x}{\sin^2 x} = -\csc x \cot x$

2. Periodicity of Higher-Order Derivatives of Sine and Cosine

Successive derivatives of $\sin(x)$ and $\cos(x)$ cycle in a closed loop of period 4:

\begin{array}{c|c|c} \text{Derivative Order } n & f(x) = \sin x & g(x) = \cos x \\ \hline n = 0 \text{ (Original)} & \sin x & \cos x \\ n = 1 \text{ (1st)} & \cos x & -\sin x \\ n = 2 \text{ (2nd)} & -\sin x & -\cos x \\ n = 3 \text{ (3rd)} & -\cos x & \sin x \\ n = 4 \text{ (4th)} & \sin x & \cos x \\ \end{array}

To evaluate any arbitrarily high $n^{\text{th}}$ derivative, compute $n \bmod 4$ (the remainder after dividing by 4):

f^{(n)}(x) = f^{(n \bmod 4)}(x)

Engineering Applications

Mechanical & Civil Engineering (Harmonic Oscillation): Mass-spring-damper systems and structural vibration modes follow sinusoidal displacements $x(t) = A \sin(\omega t + \phi)$ , where velocity is $v(t) = A \omega \cos(\omega t + \phi)$ and acceleration is $a(t) = -A \omega^2 \sin(\omega t + \phi)$ .
Electrical & Computer Engineering (AC Signal Analysis): Alternating current voltage waveforms $v(t) = V_{\text{peak}} \cos(\omega t)$ are differentiated to evaluate inductor voltages and reactive power dissipation.
Entertainment Engineering (Robotic Stage Motion): Automated stage lifts and rotary actuators utilize sinusoidal velocity profiles to achieve smooth starting and stopping without mechanical shock.

Example Problems

Example 1: Product with Trigonometric Terms

Find the equation of the line tangent to $y = 3x \tan(x)$ at $x = \pi$ .

Solution

Find point of tangency $(x_1, y_1)$ : $y_1 = y(\pi) = 3(\pi)\tan(\pi) = 3\pi(0) = 0 \implies (\pi, 0)$
Find slope $m = y'(\pi)$ using Product Rule: $y' = \frac{d}{dx}[3x]\tan x + 3x \frac{d}{dx}[\tan x] = 3\tan x + 3x \sec^2 x$ Evaluate at $x = \pi$ : $m = y'(\pi) = 3\tan(\pi) + 3\pi \sec^2(\pi) = 3(0) + 3\pi(-1)^2 = 3\pi$
Assemble tangent line equation: $y - 0 = 3\pi(x - \pi) \implies y = 3\pi x - 3\pi^2$

Example 2: Arbitrary Higher-Order Derivative

Find the $97^{\text{th}}$ derivative of $f(x) = \sin(x)$ and the $54^{\text{th}}$ derivative of $g(x) = \cos(x)$ .

Solution

Evaluate $97 \bmod 4$ for $\sin(x)$ : $97 \div 4 = 24 \text{ with remainder } 1 \implies 97 \equiv 1 \pmod 4$ Thus, $f^{(97)}(x) = f^{(1)}(x) = \cos(x)$ .
Evaluate $54 \bmod 4$ for $\cos(x)$ : $54 \div 4 = 13 \text{ with remainder } 2 \implies 54 \equiv 2 \pmod 4$ Thus, $g^{(54)}(x) = g^{(2)}(x) = -\cos(x)$ .

Connections & References

Parent Concepts: Derivatives, Trigonometric Identities
Prerequisites: Special Trigonometric Limits, Solving Right Triangles
Related Rules: Product Rule, Quotient Rule, Higher-Order Derivatives
Course Sessions: Calc-1 Session 07