Difficulty: Intermediate | Prerequisites: Basic differentiation (power rule, exponential and logarithmic derivatives), understanding of critical points.
This topic covers the two main rules for differentiating compound expressions (the product rule and the chain rule) and then puts them to work. The applied side focuses on revenue optimisation with nonlinear demand functions and on computing rates of change for real-world quantities like cost, volume, and investment. If the critical-points material tells you where to look for peaks and valleys, this material gives you the machinery to get there when functions are more complex than simple polynomials.
The product rule handles derivatives of two functions multiplied together. The chain rule handles derivatives of functions nested inside other functions. Together they let you differentiate nearly any expression you will meet in this course. The payoff: you can set up and solve optimisation problems like "find the price that maximises revenue" when the demand function is nonlinear.
Product rule
The derivative of a product of two functions: d/dx [u(x) v(x)] = u'(x) v(x) + u(x) v'(x).
In simple terms, you differentiate the first and multiply by the second, then add the first multiplied by the derivative of the second. Two terms, each with one piece differentiated.
Chain rule
The derivative of a composite function: d/dx [f(g(x))] = f'(g(x)) times g'(x).
Think of it as peeling layers. Differentiate the outer function (leaving the inner function untouched inside it), then multiply by the derivative of the inner function.
Revenue function
R(p) = p times q(p), where p is the price per unit and q(p) is the demand (quantity sold) at price p.
In simple terms, revenue is price times quantity. When demand depends on price, this becomes a function of p alone.
Demand function
q(p) gives the quantity consumers will buy at a given price p. A nonlinear demand function means this relationship is not a straight line (e.g. q(p) = 750 - 10p^2).
Rate of change
The derivative of a quantity with respect to time (or another variable). It tells you how quickly that quantity is growing or shrinking at a particular moment.
Think of it as the speedometer reading for whatever quantity you are tracking.
Use the product rule whenever you need the derivative of two functions multiplied together: u(x) times v(x).
Formula: d/dx [u(x) v(x)] = u'(x) v(x) + u(x) v'(x).
Worked example: f(x) = x e^(2x).
Let u(x) = x and v(x) = e^(2x).
u'(x) = 1.
v'(x) = 2e^(2x) (using the chain rule on e^(2x)).
f'(x) = (1)(e^(2x)) + (x)(2e^(2x)) = e^(2x) + 2x e^(2x) = e^(2x)(1 + 2x).
Use the chain rule whenever one function is nested inside another: f(g(x)).
Formula: d/dx [f(g(x))] = f'(g(x)) times g'(x).
The key step is identifying the outer function and the inner function, then differentiating each in turn.
Worked example: f(x) = ln(2x^5 - 10x + 3).
Outer function: ln(something). Its derivative is 1/(something).
Inner function: 2x^5 - 10x + 3. Its derivative is 10x^4 - 10.
f'(x) = (10x^4 - 10) / (2x^5 - 10x + 3).
Revenue: R(p) = p times q(p), where q(p) is the demand function.
To find the revenue-maximising price:
Write out R(p) = p times q(p) explicitly.
Differentiate R(p) with respect to p. You will often need the product rule here, since both p and q(p) depend on p.
Set R'(p) = 0 and solve for p.
Verify the result is a maximum (use the second derivative or check the sign change of R'(p)).
Worked example: q(p) = 750 - 10p^2.
R(p) = p(750 - 10p^2) = 750p - 10p^3.
R'(p) = 750 - 30p^2.
Set 750 - 30p^2 = 0, so p^2 = 25, giving p = 5 (taking the positive root, since price must be positive).
R''(p) = -60p. At p = 5, R''(5) = -300 < 0, confirming a maximum.
The revenue-maximising price is p = 5.
When a real-world quantity is expressed as a function of time (or another variable), its derivative gives the instantaneous rate of change.
If the quantity is itself a product of two time-dependent functions, you need the product rule.
Example: total cost v(t) = n(t) times c(t), where n(t) is the number of items and c(t) is the cost per item. Then v'(t) = n'(t) c(t) + n(t) c'(t).
If the quantity involves an exponential or a composition, you need the chain rule.
Example: volume V(t) = t^3 e^(0.02t^2). Differentiating requires both the product rule (t^3 times e^(0.02t^2)) and the chain rule (for e^(0.02t^2)).
Product rule
d/dx [u(x) v(x)] = u'(x) v(x) + u(x) v'(x)
Chain rule
d/dx [f(g(x))] = f'(g(x)) times g'(x)
Revenue function
R(p) = p times q(p)
Revenue maximisation procedure
Expand or simplify R(p) = p times q(p).
Differentiate to find R'(p).
Solve R'(p) = 0.
Confirm the critical point is a maximum (second derivative test or sign analysis).
Retailers use revenue maximisation to set prices: if you can model how demand responds to price, calculus tells you the exact price point that brings in the most money. Investment analysts use derivatives to calculate instantaneous growth rates of portfolios under compound interest. Operations managers differentiate cost or volume functions to understand how quickly expenses or output are changing at a given moment.
Students often apply the product rule when they should use the chain rule, or vice versa. The product rule is for two separate functions multiplied together. The chain rule is for one function inside another. If you see f(x) = x times e^(2x), that is a product (x and e^(2x)). If you see f(x) = e^(2x) on its own, that is a composition (e to the power of the inner function 2x).
A frequent mistake in revenue problems is differentiating q(p) alone instead of differentiating R(p) = p times q(p). The revenue function is the product of price and demand, so you typically need the product rule (or expand first, then differentiate).
Students sometimes forget the "times g'(x)" piece of the chain rule. Differentiating ln(2x^5 - 10x + 3) and writing just 1/(2x^5 - 10x + 3) is incomplete. You must multiply by the derivative of the inside: (10x^4 - 10).
When finding the revenue-maximising price, students occasionally accept a negative price as an answer. Always check that your solution makes sense in context.
Product rule and chain rule problems appear on nearly every exam. You should be able to identify which rule applies and execute it cleanly.
Revenue maximisation is a favourite applied problem. Expect a demand function, a setup of R(p), differentiation, and a verification step.
Rate-of-change questions often give you a formula for cost or volume as a function of time and ask for the rate at a specific time. Differentiate, then substitute the given t-value.
Combined problems (product rule inside a chain rule, or vice versa) test whether you can layer the rules. Practice these until they feel routine.
Fill in the blank: The product rule states d/dx [u(x) v(x)] = u'(x) v(x) + ______. (u(x) v'(x).)
True or false: The chain rule is used when two functions are multiplied together. (False. The chain rule is for composite functions, where one function is inside another. Multiplication uses the product rule.)
Fill in the blank: Revenue is defined as R(p) = p times ______. (q(p), the demand function.)
True or false: When differentiating ln(3x^2 + 1), you only need to write 1/(3x^2 + 1). (False. You must also multiply by the derivative of the inner function, which is 6x.)
Fill in the blank: To find the price that maximises revenue, set ______ = 0 and solve for p. (R'(p).)
Q: Find the derivative of f(x) = x^2 e^(3x).
A: Product rule with u = x^2 and v = e^(3x). u' = 2x. v' = 3e^(3x) (chain rule). f'(x) = 2x e^(3x) + x^2 (3e^(3x)) = 2x e^(3x) + 3x^2 e^(3x) = x e^(3x)(2 + 3x).
Q: Find the derivative of f(x) = ln(4x^3 - x).
A: Chain rule. Outer: ln(something), derivative is 1/(something). Inner: 4x^3 - x, derivative is 12x^2 - 1. f'(x) = (12x^2 - 1) / (4x^3 - x).
Q: A product has demand function q(p) = 200 - 5p^2. Find the price that maximises revenue.
A: R(p) = p(200 - 5p^2) = 200p - 5p^3. R'(p) = 200 - 15p^2. Set to zero: 200 - 15p^2 = 0, so p^2 = 200/15 = 40/3, giving p = sqrt(40/3), which is roughly 3.65. Since R''(p) = -30p < 0 for positive p, this is a maximum.
Q: Total cost is v(t) = n(t) times c(t), where n(5) = 100, n'(5) = 3, c(5) = 20, c'(5) = 0.5. Find v'(5).
A: Product rule: v'(t) = n'(t) c(t) + n(t) c'(t). v'(5) = (3)(20) + (100)(0.5) = 60 + 50 = 110. The total cost is increasing at a rate of 110 per unit time at t = 5.
Q: Find the derivative of V(t) = t^3 e^(0.02t^2).
A: Product rule with u = t^3 and v = e^(0.02t^2). u' = 3t^2. v' = e^(0.02t^2) times 0.04t (chain rule on the exponent 0.02t^2). V'(t) = 3t^2 e^(0.02t^2) + t^3 (0.04t) e^(0.02t^2) = t^2 e^(0.02t^2)(3 + 0.04t^2).
This connects directly to critical points and extrema: once you have the derivative (via the product or chain rule), you set it to zero and classify the result using the first or second derivative test.
It also connects to elasticity of demand, which measures how sensitive quantity demanded is to price changes, another derivative-based concept that builds on the demand and revenue framework here.
In later coursework, these rules extend to partial derivatives in multivariable calculus, where you optimise functions of several variables (e.g. profit as a function of both price and advertising spend).
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