Source: MAT210 Brief Calculus, Arizona State University, Section 2.1 Lecture Notes
Difficulty: Intermediate | Prerequisites: Basic differentiation rules (power rule for xⁿ, derivative of eˣ, derivative of ln x)
Up to this point in MAT210, you have learned to differentiate simple expressions: powers like xⁿ, the exponential eˣ, and the natural logarithm ln x. Those rules break down as soon as the argument of the function is itself a function, for example (x² + 4x + 1)³ or e^(3x²) or ln(x² + 1). These are composite functions, and differentiating them requires a new tool called the chain rule. The chain rule is one of the most frequently used differentiation techniques in calculus, and it will combine with every other rule you learn from here on.
When one function is plugged into another (a composite function), you cannot just apply the basic differentiation rules directly. The chain rule says: differentiate the outer function, leave the inner function alone inside it, then multiply by the derivative of the inner function. It works the same way whether the outer function is a power, an exponential, or a logarithm.
Composite function
A function formed by plugging one function into another, written as f(g(x)) or (f ∘ g)(x). In simple terms, the output of the inner function g(x) becomes the input of the outer function f.
Inner function (inside function)
The function that sits inside the argument of the outer function. In f(g(x)), the inner function is g(x). Think of it as the part you would evaluate first if you were plugging in a number.
Outer function (outside function)
The function that acts on the result of the inner function. In f(g(x)), the outer function is f. Think of it as the last operation you would perform.
Chain rule
The differentiation rule for composite functions. It states that d/dx [f(g(x))] = f'(g(x)) · g'(x). In simple terms, you differentiate the outside, leave the inside unchanged, then multiply by the derivative of the inside.
Function composition notation (f ∘ g)
The symbol ∘ denotes composition: (f ∘ g)(x) means f(g(x)). Read it as "f composed with g" or "f of g of x."
A composite function is built when the output of one function feeds directly into another.
Example: if R(p) = 0.3p³ − p² + 25 gives revenue as a function of patrons, and p(t) = 4t + 20 gives patrons as a function of days, then R(p(t)) = 0.3(4t + 20)³ − (4t + 20)² + 25 gives revenue as a function of days.
You could expand the composite and differentiate term by term, but for anything beyond the simplest cases this is extremely tedious. The chain rule provides the shortcut.
Identify the inner function g(x) and the outer function f.
For y = (x² + 5x − 1)³: the inner function is g(x) = x² + 5x − 1, the outer function is f(u) = u³.
Differentiate the outer function, leaving the inner function in place: f'(g(x)).
f'(u) = 3u², so f'(g(x)) = 3(x² + 5x − 1)².
Multiply by the derivative of the inner function: g'(x).
g'(x) = 2x + 5.
Combine: dy/dx = 3(x² + 5x − 1)² · (2x + 5).
If y = (4x − 3)⁷, then:
Inner: g(x) = 4x − 3, outer: f(u) = u⁷
dy/dx = 7(4x − 3)⁶ · 4 = 28(4x − 3)⁶
If y = 1/(x² + 10) = (x² + 10)⁻¹, then:
Inner: g(x) = x² + 10, outer: f(u) = u⁻¹
dy/dx = −1(x² + 10)⁻² · 2x = −2x / (x² + 10)²
If y = √(z³ + 5z + 1) = (z³ + 5z + 1)^(1/2), then:
Inner: g(z) = z³ + 5z + 1, outer: f(u) = u^(1/2)
dy/dz = (1/2)(z³ + 5z + 1)^(−1/2) · (3z² + 5) = (3z² + 5) / [2√(z³ + 5z + 1)]
If y = e^(0.04x²), then:
Inner: f(x) = 0.04x², outer: eᵘ
dy/dx = e^(0.04x²) · 0.08x = 0.08x · e^(0.04x²)
General pattern: if y = e^(f(x)), then y' = e^(f(x)) · f'(x).
Special case: if y = e^(ax) where a is a constant, then y' = a · e^(ax).
If y = ln(1 + 4t⁷), then:
Inner: f(t) = 1 + 4t⁷, outer: ln(u)
dy/dt = [1/(1 + 4t⁷)] · 28t⁶ = 28t⁶ / (1 + 4t⁷)
General pattern: if y = ln(f(x)), then y' = f'(x) / f(x).
General Chain Rule
\frac{d}{dx}\bigl[f(g(x))\bigr] = f'(g(x)) \cdot g'(x)Power Rule (chain rule form)
\frac{d}{dx}\bigl[f(x)\bigr]^n = n\bigl[f(x)\bigr]^{n-1} \cdot f'(x)Exponential Rule (chain rule form)
\frac{d}{dx}\bigl[e^{f(x)}\bigr] = e^{f(x)} \cdot f'(x)Special case: d/dx [e^(ax)] = a · e^(ax), where a is a constant.
Natural Logarithm Rule (chain rule form)
\frac{d}{dx}\bigl[\ln(f(x))\bigr] = \frac{f'(x)}{f(x)}The chain rule appears whenever a quantity depends on another quantity that itself changes over time. In business, revenue might depend on the number of customers, which in turn depends on how many days the shop has been open; the chain rule gives you the rate at which revenue changes with respect to time, without having to expand the entire expression first.
In finance, continuously compounded interest follows the model P(t) = P₀ · e^(rt). Finding how fast the account balance is growing at a particular moment means differentiating with the chain rule: P'(t) = P₀ · r · e^(rt). This is the basis for pricing bonds, valuing options, and modelling any quantity that grows or decays exponentially.
Students often forget to multiply by the derivative of the inner function. Differentiating (x² + 1)⁵ and writing 5(x² + 1)⁴ instead of 5(x² + 1)⁴ · 2x is the single most common chain rule error.
Students sometimes differentiate the inner function and substitute it back into the outer function, rather than leaving the inner function untouched inside the derivative of the outer function. The inner function stays as-is inside f'; you only differentiate it separately as the multiplier.
When the inner function is linear (e.g. 4x − 3), students occasionally skip the chain rule entirely because the derivative of the inner function is just a constant. The constant factor still matters: d/dx [(4x − 3)⁷] = 7(4x − 3)⁶ · 4, not 7(4x − 3)⁶.
Students confuse the chain rule with the product rule. The chain rule applies to nested functions f(g(x)); the product rule applies to functions multiplied together f(x) · g(x). If in doubt, ask: "Is one function inside the other, or are they side by side?"
⚠️ The chain rule is tested in nearly every differentiation problem from this point onward. Expect it combined with the power rule, exponential rule, and logarithm rule on exams.
⚠️ Applied problems (revenue over time, compound interest, stock value) will ask you to find a derivative and then interpret its meaning in context: "the rate at which [quantity] is changing at time t." Always include units in your interpretation.
⚠️ A positive derivative at a given point means the quantity is increasing at that moment; a negative derivative means it is decreasing. Exam questions frequently ask you to state which.
⚠️ Rewriting functions before differentiating is a common exam technique: 1/(x² + 10) becomes (x² + 10)⁻¹, and √(z³ + 5z + 1) becomes (z³ + 5z + 1)^(1/2). Get comfortable converting roots and fractions to negative or fractional exponents.
True or false: The derivative of [f(g(x))] is f'(x) · g'(x). (False. It is f'(g(x)) · g'(x).)
In the function y = (3x + 7)⁴, the inner function is __________. (3x + 7)
True or false: If y = e^(5x), then y' = e^(5x). (False. y' = 5e^(5x). You must multiply by the derivative of the inner function.)
The derivative of ln(f(x)) is __________. (f'(x) / f(x))
True or false: A positive value of the derivative at a point means the original function is decreasing there. (False. Positive derivative means increasing.)
Q: Find dy/dx if y = (2x³ − x + 4)⁵.
A: dy/dx = 5(2x³ − x + 4)⁴ · (6x² − 1)
Q: Find dy/dx if y = e^(3x² + 1).
A: dy/dx = e^(3x² + 1) · 6x = 6x · e^(3x² + 1)
Q: Find dy/dt if y = ln(5t² + 2t).
A: dy/dt = (10t + 2) / (5t² + 2t)
Q: The value of an investment is given by V(t) = 5000e^(0.06t). Find V'(3) and interpret the result.
A: V'(t) = 5000 · 0.06 · e^(0.06t) = 300e^(0.06t). V'(3) = 300e^(0.18) ≈ $359.18. After 3 years, the investment is growing in value at a rate of approximately $359.18 per year.
Q: If y = 1/(3x − 2)⁴, rewrite the function using a negative exponent and find dy/dx.
A: y = (3x − 2)⁻⁴. dy/dx = −4(3x − 2)⁻⁵ · 3 = −12/(3x − 2)⁵.
Q: For R(t) = 0.3(4t + 20)³ − (4t + 20)² + 25, find R'(t) and evaluate R'(10). What does the result mean?
A: R'(t) = 0.3 · 3(4t + 20)² · 4 − 2(4t + 20) · 4 = 3.6(4t + 20)² − 8(4t + 20). R'(10) = 3.6(60)² − 8(60) = 12,960 − 480 = $12,480 per day. On the 10th day, revenue is increasing at a rate of $12,480 per day.
The chain rule connects directly to the product rule and quotient rule covered in the next section (2.2). Most real functions require a combination of these rules, so fluency with the chain rule is a prerequisite for everything that follows.
In later chapters, the chain rule reappears in implicit differentiation, related rates problems, and integration by substitution (where the process runs in reverse). If the chain rule feels mechanical now, that is fine: the pattern becomes second nature once you see it used in these broader contexts.
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