Three rules handle nearly every derivative you will meet in Calculus I: the product rule, the quotient rule and the chain rule. Each tells you how to break a complicated function into simpler pieces, differentiate those pieces, and reassemble. Master the chain rule in particular, because it appears inside the other two whenever the pieces are themselves composite functions.
Difficulty: Intermediate
Prerequisites: Basic differentiation (power rule, trig derivatives, Sec. 2.3/2.4). You should be comfortable differentiating simple polynomials and know the derivatives of sin x and cos x before starting here.
Sections 2.5 and 2.7 of Calculus I extend differentiation to functions built from combinations of simpler ones. Up to this point you could differentiate individual terms; now you need rules for products, quotients and compositions. These rules are the workhorse tools for the rest of the course, turning up in implicit differentiation, related rates and optimisation problems. If you missed the basic derivative rules (power rule, trig derivatives), review those first, because every problem here layers on top of them.
Product rule
The formula for the derivative of two functions multiplied together: (f(x)g(x))' = f'(x)g(x) + f(x)g'(x). Think of it as: "first times derivative of second, plus second times derivative of first."
Quotient rule
The formula for the derivative of one function divided by another: (f(x)/g(x))' = [g(x)f'(x) − f(x)g'(x)] / [g(x)]². In simple terms, this means "low d-high minus high d-low, all over low squared."
Chain rule
The rule for differentiating a composite function: (f(g(x)))' = f'(g(x)) · g'(x). Think of it as: differentiate the outer function, leave the inner function alone inside it, then multiply by the derivative of the inner function.
Composite function
A function formed by plugging one function into another, written f(g(x)) or (f ∘ g)(x). In simple terms, this means the output of g becomes the input of f.
Outer function / inner function
In a composition f(g(x)), f is the outer function (it acts last) and g is the inner function (it acts first). Identifying which is which is the critical first step when applying the chain rule.
Formula
(f(x)g(x))' = f'(x)g(x) + f(x)g'(x)
How to apply it
Identify the two factors, f(x) and g(x)
Differentiate each factor separately
Multiply the first factor's derivative by the second factor (unchanged), then add the first factor (unchanged) times the second factor's derivative
Mnemonic
"First times d-second, plus second times d-first." Some students remember it as: keep one, differentiate the other, then swap roles and add.
Watch out: the product rule does not say (fg)' = f'g'. That is wrong. You must use the full formula with both terms.
Formula
(f(x)/g(x))' = [g(x)f'(x) − f(x)g'(x)] / [g(x)]²
How to apply it
Identify the numerator f(x) and the denominator g(x)
Differentiate each separately
Build the numerator of the result: denominator times derivative of numerator, minus numerator times derivative of denominator
Square the denominator for the result's denominator
Mnemonic
"Low d-high minus high d-low, over the square of what's below." The subtraction order matters: denominator times numerator's derivative comes first.
Worked example from the recitation: g(x) = (x + 4) / ∛(x² + x + 1)
Rewrite the denominator as (x² + x + 1)^(1/3) so you can differentiate it with the power and chain rules. Then apply the quotient rule:
Numerator's derivative: 1 (since d/dx of x + 4 = 1)
Denominator's derivative: (1/3)(x² + x + 1)^(−2/3) · (2x + 1), by the chain rule
Assemble using the quotient formula and simplify
The chain rule is the most frequently used derivative rule in Calculus I. Any time you see a "function inside a function," the chain rule applies.
Formula
(f(g(x)))' = f'(g(x)) · g'(x)
In words: differentiate the outer function, leave the inner function untouched inside it, then multiply by the derivative of the inner function.
Step-by-step process
Identify the outer function f and the inner function g
Differentiate the outer function as if the inner function were a plain variable
Leave g(x) in place where it appears inside f'
Multiply the entire result by g'(x)
Worked example 1: h(x) = cos(∛x) = cos(x^(1/3))
Outer function: cos(u), where u = x^(1/3)
Inner function: x^(1/3)
Derivative of outer: −sin(u) = −sin(x^(1/3))
Derivative of inner: (1/3)x^(−2/3)
Result: h'(x) = −sin(x^(1/3)) · (1/3)x^(−2/3)
Worked example 2: k(x) = (cos x)^(1/3)
Outer function: u^(1/3), where u = cos x
Inner function: cos x
Derivative of outer: (1/3)u^(−2/3) = (1/3)(cos x)^(−2/3)
Derivative of inner: −sin x
Result: k'(x) = (1/3)(cos x)^(−2/3) · (−sin x)
Notice that cos(∛x) and (cos x)^(1/3) look similar but have different outer and inner functions, and therefore different derivatives. Getting the nesting order right is the whole game.
Worked example 3: h(x) = cos(x² + 5x)
Outer: cos(u), inner: u = x² + 5x
h'(x) = −sin(x² + 5x) · (2x + 5)
Worked example 4: f(x) = [1 + sin(2x)] / [1 − sin(2x)]
This requires the quotient rule with chain rule nested inside. Both numerator and denominator contain sin(2x), whose derivative is 2cos(2x) by the chain rule.
f'(x) = 4cos(2x) / [1 − sin(2x)]²
Worked example 5: p(x) = sin(cos(tan x)), triple composition
Outermost: sin(u₁), where u₁ = cos(tan x)
Middle: cos(u₂), where u₂ = tan x
Innermost: tan x
Apply the chain rule from outside in: p'(x) = cos(cos(tan x)) · (−sin(tan x)) · sec²x
Each layer peels off one derivative and multiplies.
Rule | Formula |
|---|---|
Product rule | (fg)' = f'g + fg' |
Quotient rule | (f/g)' = [gf' − fg'] / g² |
Chain rule | (f(g(x)))' = f'(g(x)) · g'(x) |
Students often think (fg)' = f'g'. It does not. You must use the product rule, which has two terms added together.
Students often subtract in the wrong order in the quotient rule. The numerator is g·f' minus f·g', not the other way round. Swapping the order flips the sign of the entire derivative.
Students frequently forget to multiply by the inner derivative when using the chain rule. Differentiating cos(x²) as −sin(x²) is incomplete; the correct answer is −sin(x²) · 2x.
When a function has multiple layers of composition, students sometimes apply the chain rule to the outermost layer and stop. Every nested layer needs its own chain-rule factor. For sin(cos(tan x)), there are three layers and therefore three factors multiplied together.
⚠️ The chain rule is the single most tested derivative technique on MTH 132 exams. Expect at least one problem requiring it on every assessment.
⚠️ Quotient rule problems love to test whether you get the subtraction order right. Write g·f' first, then subtract f·g'.
⚠️ Triple-composition problems (like sin(cos(tan x))) appear as "show you really understand the chain rule" exam questions. Practice peeling off layers systematically.
⚠️ Tangent-line problems (see the companion notes) combine derivative calculation with point-slope form, so you need both the mechanics of the rules and the ability to evaluate the derivative at a specific point.
True or False: (f(x)g(x))' = f'(x) · g'(x).
In the quotient rule, the numerator of the result is g·f' ______ f·g'. (Fill in: + or −)
True or False: The derivative of cos(x³) is −sin(3x²).
When differentiating sin(cos(x)), the outermost derivative you write down first is ________.
True or False: The chain rule only applies when there are exactly two functions composed.
Answers: 1. False (product rule has two added terms). 2. − (minus). 3. False (correct answer is −sin(x³) · 3x²; you must keep x³ inside and multiply by the inner derivative). 4. cos(cos(x)), the derivative of sin evaluated at cos(x). 5. False (the chain rule extends to any number of nested compositions; you apply it once per layer).
Q: Let f(x) = cos x and g(x) = ∛x. Find h(x) = f(g(x)) and compute h'(x).
A: h(x) = cos(x^(1/3)). By the chain rule, h'(x) = −sin(x^(1/3)) · (1/3)x^(−2/3).
Q: Using the same f and g, find k(x) = g(f(x)) and compute k'(x).
A: k(x) = (cos x)^(1/3). By the chain rule, k'(x) = (1/3)(cos x)^(−2/3) · (−sin x).
Q: Differentiate h(x) = cos(x² + 5x).
A: h'(x) = −sin(x² + 5x) · (2x + 5).
Q: Differentiate f(x) = [1 + sin(2x)] / [1 − sin(2x)].
A: Apply the quotient rule. The derivative of sin(2x) is 2cos(2x) by the chain rule. After simplifying, f'(x) = 4cos(2x) / [1 − sin(2x)]².
Q: Differentiate p(x) = sin(cos(tan x)).
A: Apply the chain rule three times, peeling from outside in. p'(x) = cos(cos(tan x)) · (−sin(tan x)) · sec²x.
Q: Find the equation of the tangent line to y = 8/√(4 + 3x) at the point (4, 2).
A: Rewrite as y = 8(4 + 3x)^(−1/2). Differentiate using the chain rule: y' = −4/(√(4 + 3x) · (4 + 3x)) = −4/[(4 + 3x)^(3/2)]. At x = 4, y' = −4/64 = −1/16. The tangent line is y = −(1/16)x + 9/4.
These derivative rules feed directly into implicit differentiation (Sec. 2.6), where every term you differentiate with respect to x requires the chain rule because y is a function of x. They also underpin related rates problems, where you differentiate an equation involving multiple changing quantities and the chain rule links their rates together. Later, in optimisation (Chapter 3), you will use these rules to find critical points of more complex functions.
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