Premed · Premed · Calculus 1

Lecture 7: Derivatives of Trigonometric Functions

Calculus I — Differential Calculus

Learning Objectives

By the end of this lecture, students will be able to:

1. State the derivatives of all six trigonometric functions
2. Prove the derivative of sin x using the limit definition
3. Apply the chain rule to differentiate compositions involving trig functions
4. Differentiate expressions involving products and quotients of trig functions
5. Solve problems involving rates of change with trigonometric models

Lecture Content

I. Key Limit Results (Prerequisites)

Before deriving the trigonometric derivatives, two fundamental limits must be established: lim_{x -> 0} (sin x)/x = 1 and lim_{x -> 0} (cos x - 1)/x = 0. Both of these are proved using the Squeeze Theorem and geometric arguments involving the unit circle (see Lecture 3). It is essential to note that both limits assume x is measured in radians. If x were measured in degrees, these formulas would not hold.

II. Derivative of sin x

Using the limit definition, we compute d/dx [sin x] = lim_{h -> 0} [sin(x + h) - sin x] / h. Applying the angle addition formula, sin(x + h) = sin x cos h + cos x sin h, so the expression becomes lim_{h -> 0} [sin x cos h + cos x sin h - sin x] / h = lim_{h -> 0} [sin x (cos h - 1) / h + cos x (sin h) / h]. By the two prerequisite limits, this simplifies to sin x 0 + cos x 1 = cos x. The result is d/dx [sin x] = cos x.

III. Derivative of cos x

The derivative of cos x can be found similarly through the limit definition, or more efficiently by writing cos x = sin(pi/2 - x) and applying the chain rule. Either approach yields d/dx [cos x] = -sin x. A useful mnemonic pattern emerges: differentiating sin and cos cycles through the sequence sin -> cos -> -sin -> -cos -> sin -> ..., repeating every four derivatives.

IV. Derivatives of All Six Trig Functions

The derivatives of the six trigonometric functions are: d/dx [sin x] = cos x, d/dx [cos x] = -sin x, d/dx [tan x] = sec^2 x, d/dx [cot x] = -csc^2 x, d/dx [sec x] = sec x tan x, and d/dx [csc x] = -csc x cot x.

The derivative of tan x follows from writing tan x = sin x / cos x and applying the quotient rule: [cos x cos x - sin x (-sin x)] / cos^2 x = [cos^2 x + sin^2 x] / cos^2 x = 1/cos^2 x = sec^2 x. The derivative of sec x can be obtained by writing sec x = (cos x)^{-1} and using the chain rule: -(cos x)^{-2} * (-sin x) = sin x / cos^2 x = sec x tan x. A notable pattern is that the "co-functions" (cos, cot, csc) all carry a negative sign in their derivatives.

<image>A reference table showing all six trigonometric functions and their derivatives in two columns. Left column: f(x) = sin x, cos x, tan x, cot x, sec x, csc x. Right column: f'(x) = cos x, -sin x, sec^2 x, -csc^2 x, sec x tan x, -csc x cot x. A note at the bottom highlights the pattern: "co-functions" (cos, cot, csc) all have a negative sign in their derivatives. Title: "Derivatives of the Six Trigonometric Functions."</image>

V. Chain Rule with Trigonometric Functions

When the argument of a trigonometric function is not simply x, the chain rule must be applied. The general formulas are d/dx [sin(g(x))] = cos(g(x)) g'(x), d/dx [cos(g(x))] = -sin(g(x)) g'(x), d/dx [tan(g(x))] = sec^2(g(x)) * g'(x), and similarly for the other three functions.

For example, d/dx [sin(3x^2)] = cos(3x^2) 6x = 6x cos(3x^2). As another example, d/dx [tan(sqrt(x))] = sec^2(sqrt(x)) 1/(2sqrt(x)). A slightly more subtle case is d/dx [cos^3(x)] = d/dx [(cos x)^3] = 3(cos x)^2 (-sin x) = -3 cos^2(x) sin(x). Here, the notation cos^3(x) means (cos x)^3, so the chain rule applies both to the outer cubic power and to the cosine function itself.

VI. Products and Quotients Involving Trig Functions

Combining trigonometric derivatives with the product and quotient rules is a routine part of calculus. For d/dx [x^2 sin x], the product rule gives 2x sin x + x^2 cos x.

For d/dx [sin x / (1 + cos x)], the quotient rule yields [cos x (1 + cos x) - sin x (-sin x)] / (1 + cos x)^2 = [cos x + cos^2 x + sin^2 x] / (1 + cos x)^2 = [cos x + 1] / (1 + cos x)^2 = 1 / (1 + cos x). Simplifications like this one often rely on trigonometric identities, particularly the Pythagorean identity sin^2 x + cos^2 x = 1 and double-angle formulas.

<image>A unit circle diagram showing the geometric interpretation of sine and cosine derivatives. At an angle theta on the unit circle, the point (cos theta, sin theta) is marked. A tangent vector at that point is drawn, pointing in the direction (-sin theta, cos theta), illustrating that the rate of change of sin theta is cos theta and the rate of change of cos theta is -sin theta. The tangent vector components are labeled. Title: "Why the derivative of sin is cos: motion around the unit circle."</image>

VII. Applications

One of the most important physical applications of trigonometric derivatives is simple harmonic motion. The position of an oscillating object is often modeled as s(t) = A sin(omega t + phi), where A is the amplitude, omega is the angular frequency, and phi is the phase shift. Differentiating gives the velocity v(t) = A omega cos(omega t + phi) and the acceleration a(t) = -A omega^2 sin(omega t + phi) = -omega^2 s(t). The fact that acceleration is proportional to and opposite in sign from the displacement is the hallmark of simple harmonic motion.

These derivatives appear extensively in pendulum problems, wave equations, and alternating current circuits, making fluency with trigonometric differentiation essential for the sciences.

<image>Panel A: A graph of s(t) = sin(t) (position) and v(t) = cos(t) (velocity) plotted on the same axes over the interval [0, 4*pi]. The position curve is in blue, the velocity curve in red. Panel B: Annotations showing that when position is at a maximum (sin t = 1), velocity is zero (cos t = 0), and when position passes through zero, velocity is at its maximum. Vertical dashed lines connect corresponding features. Title: "Position and velocity in simple harmonic motion."</image>

Read this lecture as Markdown