Course: Calculus for Business (FIU) | Section 3.4 | Warmup 3
Difficulty: Intermediate | Prerequisites: Basic derivative rules (power rule, product rule, quotient rule, exponential and logarithmic derivatives)
The chain rule is the tool you reach for whenever you need to differentiate a composite function, which is a function nested inside another function. Most real-world calculus problems involve compositions (think: the derivative of "something raised to a power" or "e to the something"), so this rule comes up constantly. You should already be comfortable with the power rule, product rule, quotient rule, and the basic derivatives of eˣ and ln(x). If those feel shaky, revisit them first, because the chain rule layers on top of all of them.
The chain rule tells you how to take the derivative of a function inside another function. Differentiate the outer function first (leaving the inner one alone), then multiply by the derivative of the inner function. Once you see it, you will use it in nearly every derivative problem from here on.
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 function inside a function, like peeling layers of an onion.
Outer function
The "wrapper" function applied last. In f(g(x)), the outer function is f.
In simple terms, this means: the operation you see on the outside. If you have (3x² – 8x)³, the outer function is "something cubed."
Inner function
The function that gets plugged in first. In f(g(x)), the inner function is g(x).
In simple terms, this means: the expression inside the parentheses, exponent, or argument. In (3x² – 8x)³, the inner function is 3x² – 8x.
Chain rule (power rule form)
d/dx [f(x)]ⁿ = n[f(x)]ⁿ⁻¹ · f'(x). Bring the power down, reduce by one, multiply by the derivative of the inside.
Think of it as: the regular power rule, but with a "times the derivative of the inside" tacked on.
Chain rule (composition form)
d/dx f(g(x)) = f'(g(x)) · g'(x). Differentiate the outer function evaluated at the inner, then multiply by the inner function's derivative.
Think of it as: "derivative of the outside, times derivative of the inside."
The chain rule can be written two ways. They say the same thing; which one you reach for depends on the problem.
Power rule form (for [f(x)]ⁿ)
dy/dx = n[f(x)]ⁿ⁻¹ · f'(x)
Use when the outer function is a power: something raised to the nth power.
Composition form (for f(g(x)))
dy/dx = f'(g(x)) · g'(x)
Use for exponentials, logarithms, or any non-power outer function.
Both forms follow the same logic: differentiate the outer function (leaving the inner function untouched inside), then multiply by the derivative of the inner function.
Identify outer and inner: outer = ( )³, inner = 3x² – 8x
Apply the power rule form: bring the 3 down, reduce the exponent by 1
f'(x) = 3(3x² – 8x)² · d/dx(3x² – 8x)
Differentiate the inner: d/dx(3x² – 8x) = 6x – 8
Combine: f'(x) = 3(3x² – 8x)²(6x – 8)
Expanded form: 162x⁵ – 1080x⁴ + 2304x³ – 1536x²
Rewrite first: Q(t) = 2(2t² + 5)⁻²
Outer = ( )⁻², inner = 2t² + 5
Bring the exponent down: Q'(t) = 2 · (–2)(2t² + 5)⁻³ · d/dt(2t² + 5)
Differentiate the inner: d/dt(2t² + 5) = 4t
Combine: Q'(t) = –4(2t² + 5)⁻³ · 4t = –16t / (2t² + 5)³
Key move here: rewriting a fraction as a negative exponent before applying the chain rule.
This function is a sum of two terms, each requiring the chain rule separately.
Term 1: e^(x² – 3x + 4)
Outer = eˢᵒᵐᵉᵗʰⁱⁿᵍ, inner = x² – 3x + 4
The derivative of eᵘ is eᵘ · u'
d/dx = e^(x² – 3x + 4) · (2x – 3)
Term 2: (3x² – 5)²
Outer = ( )², inner = 3x² – 5
d/dx = 2(3x² – 5) · 6x = 12x(3x² – 5)
Expanded: 2(3x² – 5)(6x) = 36x³ – 60x
Combined result:
f'(x) = (2x – 3)e^(x² – 3x + 4) + 36x³ – 60x
Or equivalently: f'(x) = 2xe^(x² – 3x + 4) – 3e^(x² – 3x + 4) + 36x³ – 60x
Two terms again, each with the chain rule.
Term 1: ln(3x² – 5)
Outer = ln( ), inner = 3x² – 5
The derivative of ln(u) is (1/u) · u'
d/dx = [1/(3x² – 5)] · 6x = 6x / (3x² – 5)
Term 2: xe^(–x²)
This requires both the product rule and the chain rule.
Product rule: d/dx[x · e^(–x²)] = (1) · e^(–x²) + x · d/dx[e^(–x²)]
Chain rule on e^(–x²): derivative = e^(–x²) · (–2x)
So: e^(–x²) + x · e^(–x²) · (–2x) = e^(–x²) – 2x²e^(–x²) = e^(–x²)(1 – 2x²)
Combined result:
g'(x) = 6x / (3x² – 5) + (1 – 2x²) / e^(x²)
Chain rule (power form):
\frac{d}{dx}[f(x)]^n = n[f(x)]^{n-1} \cdot f'(x)Chain rule (composition form):
\frac{d}{dx} f(g(x)) = f'(g(x)) \cdot g'(x)Chain rule with exponentials:
\frac{d}{dx} e^{g(x)} = e^{g(x)} \cdot g'(x)Chain rule with logarithms:
\frac{d}{dx} \ln(g(x)) = \frac{g'(x)}{g(x)}In business calculus, the chain rule shows up whenever a quantity you care about depends on an intermediate variable. For example, if revenue depends on price and price depends on time, the chain rule tells you how revenue changes with respect to time. The same logic applies in cost modelling, compound interest calculations, and any situation where you are differentiating a function of a function, which is most of them.
Students often forget to multiply by the derivative of the inner function. They apply the power rule to the outer function and stop. The chain rule is not optional when there is a composite function; leaving off the inner derivative is the single most common error.
Students sometimes differentiate the inner function and replace it everywhere, rather than leaving the inner function intact inside the outer derivative. In 3(3x² – 8x)², the (3x² – 8x)² stays as is; you do not substitute 6x – 8 into it.
When the chain rule combines with the product rule (as in xe^(–x²)), students frequently apply one rule but forget the other. If a term is a product of two functions AND one of those functions is composite, both rules are needed.
Students sometimes confuse the chain rule with the product rule. The chain rule applies to functions nested inside each other, f(g(x)). The product rule applies to functions multiplied together, f(x) · g(x). A useful test: if you can identify an "inside" function and an "outside" function, it is the chain rule.
⚠️ The chain rule is tested in nearly every exam from this point forward. Expect it to appear in combination with power, product, quotient, exponential, and logarithmic rules.
⚠️ Problems that combine the chain rule with the product rule (Example D, Term 2) are a favourite exam format. Practise these until the two-rule combination feels routine.
⚠️ Rewriting fractions as negative exponents before differentiating (Example B) is a common setup. If you see a function in the denominator, rewrite it with a negative power first.
⚠️ Exam questions often ask for the derivative in both factored and expanded form. Be comfortable simplifying after applying the chain rule.
True or False: The chain rule applies whenever you differentiate a function raised to a power where the base is itself a function of x.
Answer: True.
Fill in the blank: d/dx [f(x)]ⁿ = n[f(x)]ⁿ⁻¹ · ______
Answer: f'(x)
True or False: When differentiating e^(g(x)), the result is simply e^(g(x)).
Answer: False. You must multiply by g'(x).
Fill in the blank: d/dx ln(g(x)) = ______ / g(x)
Answer: g'(x)
True or False: If a term requires both the product rule and the chain rule, you can choose to apply only one.
Answer: False. Both rules must be applied when both conditions are present.
Q: Find the derivative of f(x) = (5x³ + 2x)⁴.
A: f'(x) = 4(5x³ + 2x)³ · (15x² + 2). Outer = ( )⁴, inner = 5x³ + 2x, inner derivative = 15x² + 2.
Q: Find the derivative of y = e^(4x² + 1).
A: y' = e^(4x² + 1) · 8x. The outer function is eˢᵒᵐᵉᵗʰⁱⁿᵍ (whose derivative is itself), multiplied by the derivative of the inner (8x).
Q: Find the derivative of h(x) = ln(x² + 3x).
A: h'(x) = (2x + 3) / (x² + 3x). The derivative of ln(u) is u'/u.
Q: Find the derivative of g(x) = x² · e^(3x).
A: g'(x) = 2x · e^(3x) + x² · 3e^(3x) = e^(3x)(2x + 3x²). This requires the product rule first, then the chain rule on e^(3x).
Q: Rewrite 5/(2x + 1)³ and then differentiate.
A: Rewrite as 5(2x + 1)⁻³. Then f'(x) = 5 · (–3)(2x + 1)⁻⁴ · 2 = –30(2x + 1)⁻⁴ = –30 / (2x + 1)⁴.
The chain rule connects directly to implicit differentiation (Section 3.5 in most texts), where you treat y as a function of x and apply the chain rule every time you differentiate a y term. It also underpins the reverse process in integration: u-substitution is the chain rule run backwards. If you are heading into optimisation or related rates, both depend heavily on chaining together rates of change, which is exactly what this rule formalises.
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