ECON5005 · Quantitative Tools for Economics
Differentiation
Differentiation is the calculus core of ECON5005 Quantitative Tools for Economics at the University of Sydney: the derivative measures how fast a function changes, and in economics that instantaneous rate of change is the marginal quantity — marginal revenue, marginal cost, the marginal propensity to consume, and the elasticity of demand. This chapter builds the five differentiation rules (power, constant-multiple, sum, product, quotient, chain), then higher derivatives and the second-derivative test for curvature. It is the engine every later topic — optimisation, level curves, integration and difference equations — runs on, and it is examined in both the mid-semester test and the final.
What this chapter covers
- 01Read the derivative as the slope of the tangent — the instantaneous rate of change
- 02Apply the base rules: constant, power (x^n → n·x^(n-1)), exponential and natural-log derivatives
- 03Use the product rule (fg)' = f'g + fg' and know it has two terms, not one
- 04Use the quotient rule with the numerator in the right order: (gf' − fg')/g²
- 05Use the chain rule for compositions: differentiate outside, then multiply by the inner derivative
- 06Turn totals into marginal functions: MR = d(TR)/dQ, MC = d(TC)/dQ, MPC = dC/dY
- 07Compute point elasticity of demand ε = (dQ/dP)(P/Q) and classify elastic / unit / inelastic
- 08Take higher derivatives and read curvature: f'' > 0 convex, f'' < 0 concave, sign change = inflection
- 09Connect the second derivative to the max/min test used in the optimisation chapter
Marginal revenue and point elasticity for a monopolist
- +1Total revenue is price times quantity: TR = P·Q = (120 − 3Q)Q = 120Q − 3Q².
- +1Marginal revenue is the derivative of TR: MR = d(TR)/dQ = 120 − 6Q. At Q = 10, MR = 120 − 60 = 60.
- +1For elasticity we need dQ/dP. From the inverse demand dP/dQ = −3, so dQ/dP = 1/(dP/dQ) = −1/3.
- +1Find the price at Q = 10: P = 120 − 3(10) = 90. Then ε = (dQ/dP)·(P/Q) = (−1/3)·(90/10) = (−1/3)(9) = −3.
- +1Interpret: |ε| = 3 > 1, so demand is elastic at this point — a 1% price rise cuts quantity demanded by about 3%.
Key terms
- Derivative
- The instantaneous rate of change of a function, equal to the slope of the tangent at a point; written f'(x) or dy/dx. Positive means the function is rising, negative means falling.
- Power rule
- d/dx(x^n) = n·x^(n-1). Rewrite roots and reciprocals as powers first (√x = x^(1/2), 1/x² = x^(-2)) so the rule applies directly.
- Product rule
- For a product of two functions, (fg)' = f'g + fg'. It always yields two terms; writing only f'g' is a common error.
- Quotient rule
- For a ratio, (f/g)' = (gf' − fg')/g², with the numerator's derivative taken first. The order gf' − fg' fixes the sign of the answer.
- Chain rule
- For a composition y = f(g(x)), dy/dx = f'(g(x))·g'(x): differentiate the outer function, then multiply by the derivative of the inner function.
- Marginal function
- The derivative of a total with respect to output or income: marginal revenue MR = d(TR)/dQ, marginal cost MC = d(TC)/dQ, marginal propensity to consume MPC = dC/dY — the effect of one more unit.
- Point elasticity of demand
- ε = (dQ/dP)·(P/Q), the percentage change in quantity for a 1% change in price. Demand is elastic if |ε| > 1, unit-elastic if |ε| = 1, inelastic if |ε| < 1; ε is normally negative.
- Second derivative
- The derivative of the derivative, f''(x). Its sign gives curvature: f'' > 0 is convex (concave up), f'' < 0 is concave (concave down); a sign change marks an inflection point and it underpins the maximum/minimum test.
Differentiation FAQ
Is differentiation examined in ECON5005, and where?
Yes. Differentiation is Week 4 material and one of the most load-bearing skills in the unit: it appears directly in the mid-semester test (algebra through differentiation and curve analysis) and again in the final, and it also underpins later questions on optimisation, level curves and integration. Expect to differentiate with the power, product, quotient and chain rules, form marginal functions, and use the second derivative to describe curvature.
What is the difference between the product rule and the chain rule, and how do I know which to use?
Use the product rule when two functions are multiplied, e.g. x·e^x, giving (fg)' = f'g + fg'. Use the chain rule when one function sits inside another, e.g. (3x² + 1)^5 or e^(3x²), giving f'(g(x))·g'(x) — differentiate the outside, then multiply by the inner derivative. Many terms need both: x·e^(2x) is a product whose second factor also needs the chain rule. Name the structure first, then pick the rule.
Can AI help me with differentiation in ECON5005?
Yes, as a study aid. Sia, the AskSia AI tutor, can explain a differentiation rule step by step, work through a similar practice problem so you can see the method, check where a sign or an inner derivative went wrong, and quiz you on marginal functions and elasticity. It is there to help you understand and practise the technique — not to hand in answers for you. Use it to build the skill, then do the actual assessed quiz, test and exam questions yourself, in line with the University of Sydney's academic-integrity rules.
Studying with AI? Sia — free AI economics tutor works through ECON5005 step by step.
Exam move
Treat the five rules as reflexes rather than reference material: power, constant-multiple, sum, product, quotient and chain. Drill a handful of mixed derivatives daily until you can name the structure (product? quotient? composition?) at a glance and never drop an inner derivative or flip the quotient numerator. Then practise the economic translations — differentiate a total-revenue or total-cost function to get its margin, and compute a point elasticity from an inverse demand, keeping the sign and classifying by magnitude. Finish each session with one higher-derivative problem so the convex/concave (f'' > 0 / f'' < 0) reading becomes automatic, because the optimisation chapter reuses it as the second-order test. Because most marks are method marks, always write the rule you are using and keep every factor visible.
