Calculus 1: practice problems and exam questions

Derivatives and integrals: the rules of differentiation, applications such as tangent lines and optimisation, and the main techniques of integration.

Cram pack (PDF)Formula sheets, method steps and practice problems with full solutions, ready to print.

Topics in this course

  1. Limitsdirect substitution and factoring · limits at infinity · L'Hôpital's rulecalculations
  2. Derivativespower rule · chain and product rules · tangents and extremacalculations
  3. Optimizationmaximum area or volume · minimum cost or material · profit and revenuecalculations
  4. Integralspolynomials · exponentials, logs and substitution · areas between curvescalculations

Exam tips for Calculus 1

  • Write the chain rule out in full: the derivative of the outside times the derivative of the inside.
  • Check an integral by differentiating your answer.
  • For a definite integral, a substitution also changes the limits.

Worked examples

LimitsEasy
xyx = 1

Find the limit L = limx→1 (x4 − 1)/(x − 1).

f(x) = (x4 − 1)/(x − 1)x → 1

Find L

Show hints
  1. Substitution gives 0/0. Factor the numerator.
  2. xⁿ − aⁿ = (x − a)(xⁿ⁻¹ + a·xⁿ⁻² + … + aⁿ⁻¹).
  3. x4 − 1 = (x − 1)(x3 + x2 + x + 1).
Show the answer and the step-by-step solution

Limits: first try direct substitution. If it gives 0/0, simplify first: factor and cancel the common factor, or multiply by the conjugate. As x → ∞, divide numerator and denominator by the highest power of x. L'Hôpital's rule: if f/g gives 0/0 or ∞/∞, then lim f/g = lim f′/g′, differentiating the numerator and the denominator separately (not with the quotient rule).

Try direct substitution

numerator → 0 and denominator → 0, so the form is 0/0.

Factor and cancel

(x4 − 1)/(x − 1) = x3 + x2 + x + 1

Substitute

L = 1·13 + 1·12 + 1·1 + 1·1 = 4

Answer

L = 4

This is the derivative of xⁿ at x = a: n·aⁿ⁻¹.

DerivativesMedium
xyx = 1

Let f(x) = x3·e−x. Find f′(1).

f(x) = x3·e−xx = 1

Find f′(1)

Show hints
  1. Product rule with u = x3 and v = e−x.
  2. u′ = 3x2, v′ = −1·e−x.
  3. f′(x) = e−x·(3x2 + (−1)·x3).
Show the answer and the step-by-step solution

Differentiation: the derivative f′(x) is the slope of the graph at x. Power rule: (xⁿ)′ = n·xⁿ⁻¹. Product rule: (uv)′ = u′v + uv′. Chain rule: (f(g(x)))′ = f′(g(x))·g′(x). Where f′(x) = 0 the graph has a horizontal tangent: a maximum, minimum or saddle point.

Product rule

f′(x) = u′v + uv′ = e−x·(3x2 + (−1)·x3)

Substitute

f′(1) = 0.3679·(3 + (−1)) = 0.7358

Answer

f′(1) = 0.7358

How practising works

  • “Practise this course” mixes problems from all its topics, like a real exam.
  • Every problem has new values, so you learn the method instead of an answer.
  • Stuck? Open a small hint first, and the full explanation only when you need it.
  • Mock exams are timed and have no hints, and you get your score and a full review at the end.

Calculators for this course

Related courses