Source: Calculus for Business, Florida International University, Chapter 2
Tags: chain rule, composite function, differentiation, derivative, calculus for business, FIU
Difficulty: Intermediate | Prerequisites: Basic differentiation rules (power rule, constant rule, sum rule), product rule, quotient rule.
The chain rule is one of the most important differentiation techniques you will use in this course and beyond. Up to now, you have learned to differentiate simple functions using the power rule, product rule, and quotient rule. The chain rule extends your toolkit to composite functions, where one function is nested inside another. Nearly every applied calculus problem in business, economics, and the sciences requires it, so if this does not click, the rest of the course will feel significantly harder.
When you have a function inside another function, you differentiate the outer function first (leaving the inner function untouched), then multiply by the derivative of the inner function. In Leibniz notation: dy/dx = (dy/du) × (du/dx). Once you see composite functions as layers, the chain rule is just peeling them one at a time.
Composite function
A function formed by plugging one function into another, written f(g(x)). The output of g becomes the input of f.
Think of it as: a two-step machine where the result of the first step feeds straight into the second.
Chain rule
If y = f(u) and u = g(x), then dy/dx = f'(g(x)) · g'(x). Equivalently, dy/dx = (dy/du) × (du/dx).
In simple terms, this means: differentiate the outside, leave the inside alone, then multiply by the derivative of the inside.
Inner function (u = g(x))
The function that sits inside the other function in a composition. This is the part you differentiate last.
Think of it as: the first operation that happens to x.
Outer function (y = f(u))
The function that wraps around the inner function. This is the part you differentiate first (treating the inner function as a single variable).
Think of it as: the second operation, the one applied to whatever came out of the inner function.
Leibniz notation
The notation dy/dx for derivatives, which makes the chain rule look like fraction cancellation: dy/dx = (dy/du) · (du/dx). The du terms do not literally cancel, but the pattern is a reliable mnemonic.
If y = f(u) is a differentiable function of u, and u = g(x) is a differentiable function of x, then the composite function y = f(g(x)) is differentiable and its derivative is:
dy/dx = f'(g(x)) · g'(x)
Equivalently: dy/dx = (dy/du) · (du/dx)
The pattern in words: "derivative of the outside (evaluated at the inside) times the derivative of the inside."
Identify the outer function f and the inner function g(x).
Differentiate the outer function, leaving the inner function untouched.
Multiply by the derivative of the inner function.
f(x) = (x² + 2)³ – 3(x² + 2)² + 1
Outer: u³ – 3u² + 1 where u = x² + 2
f'(x) = 3(x² + 2)²(2x) – 6(x² + 2)(2x)
The 2x appears each time because d/dx(x² + 2) = 2x
f(x) = (x + ½)²⁰
Outer: u²⁰, inner: u = x + ½
f'(x) = 20(x + ½)¹⁹ · (1 + x⁻²) (note: derivative of ½ with respect to x is included)
Key point: "take 20, everything the same, power drops by 1, derivative of the inside"
f(x) = √(x² + 3x + 2) = (x² + 3x + 2)^(1/2)
Rewrite the radical as a fractional power first
f'(x) = ½(x² + 3x + 2)^(–1/2) · (2x + 3)
f(x) = 1/(2x + 3)⁵ = (2x + 3)⁻⁵
Rewrite as a negative power
f'(x) = –5(2x + 3)⁻⁶ · (2)
f(x) = (2x⁴ – x)³
f'(x) = 3(2x⁴ – x)² · (8x³ – 1)
Inner derivative: d/dx(2x⁴ – x) = 8x³ – 1
The Chain Rule (general form)
If y = f(g(x)), then y' = f'(g(x)) · g'(x)
Leibniz form
dy/dx = (dy/du) · (du/dx)
Generalised Power Rule (most common application)
If f(x) = [g(x)]ⁿ, then f'(x) = n · [g(x)]^(n–1) · g'(x)
This covers positive integer powers, negative powers (rewrite 1/u as u⁻¹), and fractional powers (rewrite √u as u^(1/2)).
Students often forget to multiply by the derivative of the inner function. If you write f'(x) = 3(x² + 2)² and stop there, you have dropped the crucial × 2x factor. The chain rule always has two parts.
Students sometimes differentiate the inner function and substitute it back in the wrong place. The outer function's derivative is evaluated at g(x) (the original inner function, untouched), not at g'(x).
When the inner function is linear (e.g., 2x + 3), the derivative is just a constant (2). Students sometimes skip it because it "feels obvious," but leaving it out changes the answer.
Rewriting before differentiating trips people up. A radical like √(something) must be written as (something)^(1/2) before you apply the generalised power rule. If you try to differentiate the radical sign directly, you will almost certainly get it wrong.
⚠️ The chain rule appears in nearly every subsequent topic: implicit differentiation, related rates, optimisation, and integration by substitution all rely on it.
⚠️ Exam questions frequently present functions that look complicated but are just nested powers. Rewrite radicals and fractions as exponents before differentiating.
⚠️ "Find the rate at which cost is changing with respect to time" is a classic chain rule application (dC/dt = dC/dx · dx/dt). Expect at least one applied chain rule problem on any business calculus exam.
⚠️ Forgetting the inner derivative is the single most common point-losing error on chain rule problems.
True or False: The chain rule states dy/dx = f'(g(x)) + g'(x).
Answer: False. It is a product, not a sum: dy/dx = f'(g(x)) · g'(x).
Fill in the blank: If f(x) = (3x + 1)⁴, then f'(x) = ___(3x + 1)³ · ___.
Answer: 4(3x + 1)³ · 3 = 12(3x + 1)³.
True or False: To differentiate √(x² + 5), you should first rewrite it as (x² + 5)^(1/2).
Answer: True.
Fill in the blank: In Leibniz notation, the chain rule is dy/dx = (dy/) · (/dx).
Answer: du, du.
Q: Find f'(x) for f(x) = (5x² – 3x + 1)⁷.
A: f'(x) = 7(5x² – 3x + 1)⁶ · (10x – 3).
Q: Find f'(x) for f(x) = 1/(4x + 9)³. (Hint: rewrite first.)
A: Rewrite as f(x) = (4x + 9)⁻³. Then f'(x) = –3(4x + 9)⁻⁴ · 4 = –12(4x + 9)⁻⁴.
Q: Find f'(x) for f(x) = √(6x² – x).
A: Rewrite as (6x² – x)^(1/2). Then f'(x) = ½(6x² – x)^(–1/2) · (12x – 1) = (12x – 1) / [2√(6x² – x)].
Q: The cost of producing x units is C(x) = ⅓x² + 4x + 53 dollars, and the production level t hours in is x(t) = 0.2t² + 0.03t units. Write the expression for dC/dt using the chain rule.
A: dC/dt = (dC/dx)(dx/dt) = (⅔x + 4)(0.4t + 0.03). Substitute x = 0.2t² + 0.03t to express everything in terms of t.
The chain rule connects directly to related rates problems (Chapter 3), where you differentiate with respect to time and every variable is itself a function of t. It also underpins integration by substitution (u-substitution), which is essentially the chain rule run in reverse. If you are comfortable identifying the inner and outer functions here, substitution in integration will feel far more natural.
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