Differentiation
Differentiation transforms A-level maths students from passive observers to active problem-solvers, yet many struggle with the transition from GCSE algebra to calculus concepts. Year 12 students often find the power rule intuitive, but applying derivatives to real-world optimisation problems requires substantial practice across varying difficulty levels.
Why it matters
Differentiation underpins countless real-world applications that A-level students encounter daily. Engineers use derivatives to optimise fuel efficiency in cars, calculating that reducing speed from 70 mph to 60 mph can improve consumption by 15%. Economics students apply differentiation to find maximum profit points — if a company's profit function is P(x) = -2x² + 120x - 800, differentiation reveals the optimal production level at 30 units. Medical researchers use rates of change to model drug concentration in bloodstreams, where a derivative of -0.3t indicates medication levels drop by 30% per hour. Physics students calculate acceleration as the derivative of velocity, essential for understanding motion in Year 13 mechanics modules. Even everyday scenarios like finding the most economical mobile phone tariff involve optimisation through differentiation techniques.
How to solve differentiation
Differentiation
- Power rule: d/dx [xⁿ] = nxⁿ⁻¹.
- Chain rule: d/dx [f(g(x))] = f'(g(x)) × g'(x).
- Product rule: d/dx [uv] = u'v + uv'.
- Derivative = gradient of the tangent = instantaneous rate of change.
Example: f(x) = 3x⁴ → f'(x) = 12x³. At x=2: f'(2) = 96.
Worked examples
Differentiate: f(x) = 2 x3
Answer: f'(x) = 6 x2
- Apply the power rule: d/dx[ax^n] = nax^(n-1) → f'(x) = 3·2x^2 = 6 x^2 — Multiply the exponent 3 by the coefficient 2, then reduce the exponent by 1.
Differentiate: f(x) = 3 x3 - x2 + 5 x + 3
Answer: f'(x) = 9 x2 - 2 x + 5
- Write out the rule → d/dx[x^n] = n·x^(n-1) — The power rule: multiply by the exponent, then reduce the exponent by 1.
- Differentiate 3 x^3 → 3·3x^2 = 9 x^2 — Exponent 3 comes down as a factor, exponent becomes 3−1 = 2.
- Differentiate - x^2 → 2·-1x = - 2 x — Exponent 2 comes down, exponent becomes 2−1 = 1.
- Differentiate 5x → 5 — The derivative of kx is just k. The constant d vanishes.
- Combine all terms → f'(x) = 9 x^2 - 2 x + 5 — Write the derivative as one expression.
Differentiate: f(x) = 4 ex
Answer: f'(x) = 4 ex
- Apply the rule: d/dx[e^x] = e^x → f'(x) = 4 e^x — The constant 4 is carried through.
Common mistakes
- Students frequently forget to reduce the power by 1, writing d/dx[3x⁴] = 12x⁴ instead of 12x³, particularly when working under exam pressure.
- When differentiating constants, many write d/dx[7] = 7 rather than 0, not recognising that constant terms vanish completely in derivatives.
- Chain rule errors are common with composite functions — students often write d/dx[(2x+3)³] = 3(2x+3)² instead of 6(2x+3)², missing the inner function's derivative of 2.
- Sign errors plague polynomial differentiation, with students writing d/dx[x³ - 2x²] = 3x² + 4x instead of 3x² - 4x, particularly when negative coefficients appear.