Difficulty: Intermediate | Prerequisites: Differentiation rules (Ch. 2 Sec. 1–3), basic trig identities.
Section 4 adds the six trig functions to your derivative toolkit and introduces the key limits sin(x)/x → 1 and (cos x - 1)/x → 0. Section 5 presents the chain rule, which handles compositions of functions, the most common pattern in real differentiation problems. Section 6 tackles implicit differentiation, a technique for finding slopes when y is not isolated as a function of x.
The six trig functions each have a derivative you should memorise. The chain rule tells you how to differentiate a "function inside a function" by multiplying the outer derivative by the inner derivative. Implicit differentiation extends the chain rule to equations where y is tangled up with x instead of isolated on one side.
Trigonometric derivatives
The derivatives of the six standard trig functions: d/dx[sin x] = cos x, d/dx[cos x] = -sin x, and the four others derived from these using quotient-rule proofs.
Special trig limits
lim(x→0) sin(x)/x = 1 and lim(x→0) (cos x - 1)/x = 0. These two results are used to prove the trig derivative formulas and to solve a family of limit problems involving trig functions.
Composition of functions
f(g(x)) means "apply g first, then apply f to the result." Also written (f ∘ g)(x). Example: if f(x) = sin x and g(x) = x² + x, then f(g(x)) = sin(x² + x).
Chain rule
If F(x) = f(g(x)), then F'(x) = f'(g(x)) · g'(x). In Leibniz notation: dF/dx = (dF/du) · (du/dx) where u = g(x). In words: differentiate the outer function (leaving the inner function alone), then multiply by the derivative of the inner function.
Implicit differentiation
A method for finding dy/dx when y is defined implicitly by an equation (like x² + y² = 2) rather than explicitly (like y = √(2 - x²)). Differentiate both sides with respect to x, treating y as a function of x (so d/dx[y²] = 2y · dy/dx by the chain rule), then solve for dy/dx.
Memorize these six:
d/dx [sin x] = cos x
d/dx [cos x] = -sin x
d/dx [tan x] = sec²x
d/dx [sec x] = sec x tan x
d/dx [csc x] = -csc x cot x
d/dx [cot x] = -csc²x
The sin and cos derivatives are proved from the limit definition using the special limits. The other four follow from the quotient rule (tan = sin/cos, sec = 1/cos, etc.).
The derivatives of sin follow a four-step cycle: sin → cos → -sin → -cos → sin → ... This means the 83rd derivative of sin(x) is cos(x) (since 83 = 4 · 20 + 3, and the third step in the cycle is cos).
lim(x→0) sin(x)/x = 1. More generally, if lim(x→a) f(x) = 0, then lim(x→a) sin(f(x))/f(x) = 1.
lim(x→0) (cos x - 1)/x = 0.
These let you handle limits like lim(x→0) sin(5x)/(3x) = (5/3) · lim(x→0) sin(5x)/(5x) = 5/3.
For tangent: lim(x→0) tan(2t²)/(4t) requires rewriting as [sin(2t²)/cos(2t²)] / (4t) and using the special limit.
Identifying the composition: Before you can apply the chain rule, you need to see the function as f(g(x)). Look for a "function inside a function": sin(x²+x) has outer function sin(u) and inner function u = x²+x.
The formula: If F(x) = f(g(x)), then F'(x) = f'(g(x)) · g'(x).
In Leibniz notation with u = g(x): dF/dx = (dF/du) · (du/dx). It looks like the du's cancel, which is a useful memory aid (though it is not literally a fraction).
Worked examples:
h(x) = sin(x² + x). Outer: sin(u), derivative cos(u). Inner: u = x² + x, derivative 2x + 1. Result: h'(x) = cos(x² + x) · (2x + 1).
f(x) = (7x³ + 2x² - x + 3)⁵. Outer: u⁵, derivative 5u⁴. Inner: 7x³ + 2x² - x + 3, derivative 21x² + 4x - 1. Result: f'(x) = 5(7x³ + 2x² - x + 3)⁴ · (21x² + 4x - 1).
g(x) = tan(3x² + 5x + 8). Outer: tan(u), derivative sec²(u). Inner: 3x² + 5x + 8, derivative 6x + 5. Result: g'(x) = sec²(3x² + 5x + 8) · (6x + 5).
Using a table of values: If given f, g, f', g' at specific points, find h'(1) where h(x) = f(g(x)) by computing f'(g(1)) · g'(1). Look up g(1) in the table, then use that as the input to f'.
When to use it: Whenever the equation relating x and y cannot be (or has not been) solved for y explicitly. Examples: x² + y² = 2, x³ + y³ = 6xy.
Method:
Differentiate both sides of the equation with respect to x.
Every time you differentiate a term involving y, apply the chain rule: d/dx[yⁿ] = n yⁿ⁻¹ · (dy/dx). The factor dy/dx appears because y is implicitly a function of x.
Collect all terms with dy/dx on one side, everything else on the other.
Factor out dy/dx and solve.
Worked example: x³ + y³ = 6xy.
Differentiate: 3x² + 3y² · (dy/dx) = 6y + 6x · (dy/dx).
Rearrange: 3y² · (dy/dx) - 6x · (dy/dx) = 6y - 3x².
Factor: (dy/dx)(3y² - 6x) = 6y - 3x².
Solve: dy/dx = (6y - 3x²) / (3y² - 6x) = (2y - x²) / (y² - 2x).
At the point (3, 3): dy/dx = (6 - 9) / (9 - 6) = -3/3 = -1. Tangent line: y - 3 = -1(x - 3), or y = -x + 6.
Horizontal tangent lines occur where the numerator of dy/dx equals 0 (and the denominator does not). Vertical tangent lines occur where the denominator equals 0 (and the numerator does not).
\frac{d}{dx}[\sin x] = \cos x \qquad \frac{d}{dx}[\cos x] = -\sin x \qquad \frac{d}{dx}[\tan x] = \sec^2 x\frac{d}{dx}[\sec x] = \sec x\tan x \qquad \frac{d}{dx}[\csc x] = -\csc x\cot x \qquad \frac{d}{dx}[\cot x] = -\csc^2 x\lim_{x \to 0} \frac{\sin x}{x} = 1 \qquad\qquad \lim_{x \to 0} \frac{\cos x - 1}{x} = 0\textbf{Chain Rule:}\quad [f(g(x))]' = f'(g(x))\cdot g'(x) \quad\text{or}\quad \frac{dF}{dx} = \frac{dF}{du}\cdot\frac{du}{dx}Implicit differentiation: differentiate both sides with respect to x, attach dy/dx to every y-term via the chain rule, solve for dy/dx.
Students often forget the negative sign in d/dx[cos x] = -sin x. A mnemonic: the "co-" functions (cos, csc, cot) all pick up a negative sign in their derivatives.
Forgetting to multiply by the inner derivative when using the chain rule. d/dx[sin(3x)] is cos(3x) · 3, not just cos(3x).
In implicit differentiation, forgetting to attach dy/dx when differentiating y-terms. d/dx[y²] is 2y · (dy/dx), not just 2y.
Confusing when to use the product rule versus the chain rule. The product rule handles f(x) · g(x); the chain rule handles f(g(x)). sin(x) · cos(x) is a product. sin(cos(x)) is a composition.
⚠️ All six trig derivatives are fair game on every exam from this point on. Memorise them.
⚠️ The chain rule is the single most-used differentiation technique in the course. Most functions you encounter in practice are compositions, so the chain rule appears in nearly every problem.
⚠️ Implicit differentiation problems are an exam staple. Expect to find dy/dx and then the equation of a tangent line at a given point.
⚠️ Special trig limit problems (sin(5x)/3x style) appear on Exam 1 and the final. The technique is to manipulate the expression so that sin(something)/something appears, then use the fact that the limit is 1.
Fill in the blank: d/dx[cos x] = ___.
True or False: d/dx[sin(x²)] = cos(x²).
Fill in the blank: lim(x→0) sin(7x)/(7x) = ___.
True or False: In implicit differentiation, d/dx[xy] = x · (dy/dx) (with no other terms).
Fill in the blank: The 84th derivative of sin(x) is ___.
Answers: 1. -sin x. 2. False (it is cos(x²) · 2x by the chain rule). 3. 1. 4. False (product rule gives y + x · dy/dx). 5. sin(x) (since 84 = 4 · 21, completing 21 full cycles back to sin).
Q: Find the derivative of f(x) = cos⁶(x).
A: Rewrite as (cos x)⁶. Chain rule: 6(cos x)⁵ · (-sin x) = -6 cos⁵(x) sin(x).
Q: Evaluate lim(x→0) sin(5x)/(3x).
A: Rewrite as (5/3) · sin(5x)/(5x). As x → 0, sin(5x)/(5x) → 1. So the limit is 5/3.
Q: Find dy/dx for x² + y² = 2.
A: Differentiate: 2x + 2y(dy/dx) = 0. Solve: dy/dx = -x/y.
Q: Find the tangent line to x³ + y³ = 6xy at (3, 3).
A: dy/dx = (2y - x²)/(y² - 2x). At (3,3): (6 - 9)/(9 - 6) = -1. Tangent: y - 3 = -1(x - 3), so y = -x + 6.
Q: Given h(x) = f(g(x)) with f(2) = 3, g(1) = 2, f'(2) = 4, g'(1) = 6, find h'(1).
A: h'(1) = f'(g(1)) · g'(1) = f'(2) · 6 = 4 · 6 = 24.
The chain rule is essential for related rates (Section 2.8), since related-rates problems differentiate equations involving time-dependent quantities, producing chain-rule terms like dr/dt. Implicit differentiation is the technique behind related rates and also returns in later courses (multivariable calculus, differential equations). The trig limits and derivatives connect back to the Squeeze Theorem from Chapter 1.
trig derivatives, derivative of sin, derivative of cos, derivative of tan, derivative of sec, derivative of csc, derivative of cot, sinx over x limit, special trig limits, chain rule, composition of functions, inner function, outer function, Leibniz notation chain rule, implicit differentiation, dy/dx, tangent line implicit, horizontal tangent implicit, MTH 132, Calculus I, Chapter 2, Sections 2.4, 2.5, 2.6, Stewart Calculus