Ten lessons on limits, continuity, derivatives and their rules, the chain rule, related rates, approximation, curve sketching, optimization and the Fundamental Theorem, with sliders and worked examples.
A study guide to a first college calculus course, not a full course. Topics follow the standard single-variable sequence used by open textbooks. Practice values in the labs are invented. Work extra problems from a textbook.
['Algebra and trigonometry']
Course outline
Limits: what a function approaches
Estimate and compute limits, including indeterminate forms.
Continuity and the Intermediate Value Theorem
Decide when a function is continuous and use IVT to locate roots.
The derivative as a limit
Define the derivative and read it as slope and rate.
Differentiation rules
Use power, sum, product and quotient rules.
The chain rule
Differentiate compositions.
Implicit differentiation and related rates
Differentiate relations and link rates through time.
Linear approximation and the Mean Value Theorem
Approximate with tangent lines and state MVT.
Curve sketching with the first and second derivative
Read every lesson below. The interactive reader above contains the same explanations, with visual tools and quizzes.
1. Limits: what a function approaches
Learning goal: Estimate and compute limits, including indeterminate forms.
A limit describes what f(x) approaches as x approaches a number, not what happens at that number. The limit of (x squared minus 4) divided by (x minus 2) as x goes to 2 is 4, even though the function is undefined at 2. Factor the top as (x minus 2)(x plus 2), cancel, and substitute.
A limit exists only when the left-hand and right-hand limits agree. A jump or a vertical asymptote means the two sides disagree or grow without bound. Limits at infinity ask how a function behaves for very large x: a ratio of polynomials is decided by the highest powers.
Direct substitution is the first move. If it gives a number, you are done. If it gives 0 over 0, simplify by factoring, rationalizing or finding a common denominator, then try again. Never stop at 0 over 0.
Numerical tables and graphs suggest limits but do not prove them. Use them to guess, and algebra to confirm.
Worked example
Find the limit of (x squared - 4)/(x - 2) as x goes to 2.
Substitute: 0 over 0
Factor (x-2)(x+2)
Cancel x-2
Substitute: 2 + 2 = 4
Practice problem and solution
Find the limit of (x squared - 4)/(x - 2) as x approaches 2. Enter a number.
Factor and cancel: x + 2 gives 4 at x = 2.
Mental model: A limit is about nearby values; simplify 0 over 0.
Common trap: Stopping at 0 over 0 or treating a limit as the value at the point.
2. Continuity and the Intermediate Value Theorem
Learning goal: Decide when a function is continuous and use IVT to locate roots.
A function is continuous at a if three things hold: f(a) exists, the limit as x approaches a exists, and they are equal. Polynomials, exponentials and sine and cosine are continuous everywhere. Rational functions are continuous except where the denominator is 0.
Breaks come in kinds. A removable break (hole) is a limit that exists but does not match the value. A jump has different one-sided limits. An infinite break has a vertical asymptote.
The Intermediate Value Theorem says: if f is continuous on [a, b] and N is between f(a) and f(b), then f(c) = N for some c in (a, b). It guarantees a solution exists but does not say where.
Use IVT to show a root exists: f(x) = x squared minus 5 has f(2) = -1 and f(3) = 4. The sign change proves a root lies between 2 and 3. Bisection repeats the idea to narrow the interval.
Worked example
Show x squared = 5 has a solution between 2 and 3.
f(x) = x^2 - 5 is continuous
f(2) = -1 is negative
f(3) = 4 is positive
IVT gives a root in (2, 3)
Practice problem and solution
f(x) = x squared - 5 changes sign between the integers n and n + 1. What is n? Enter a number.
f(2) = -1 and f(3) = 4.
Mental model: Continuity plus a sign change guarantees a root.
Common trap: Applying IVT to a function with a break.
3. The derivative as a limit
Learning goal: Define the derivative and read it as slope and rate.
The derivative of f at a is the limit of the difference quotient: (f(a+h) minus f(a)) divided by h as h goes to 0. It is the slope of the tangent line and the instantaneous rate of change.
For f(x) = x squared at x = 3 the quotient is ((3+h) squared minus 9) over h = 6 + h. As h goes to 0 this gives 6. So the tangent at (3, 9) has slope 6.
The derivative is itself a function, f prime of x. Read it in context: if s(t) is position, s prime is velocity; if C(q) is cost, C prime is marginal cost. Always attach units: units of f per unit of x.
A function can be continuous but not differentiable: at a corner, a cusp or a vertical tangent. Differentiable implies continuous, not the other way.
Worked example
Find the derivative of x squared at 3.
Quotient: ((3+h)^2 - 9)/h
Expand: (6h + h^2)/h = 6 + h
Let h go to 0
Slope is 6
Practice problem and solution
f(x) = x squared. What is f prime(3)? Enter a number.
The tangent slope is 6 at x = 3.
Mental model: The derivative is the limit of the difference quotient.
Common trap: Reading the average rate as the instantaneous rate.
4. Differentiation rules
Learning goal: Use power, sum, product and quotient rules.
Power rule: the derivative of x to the n is n times x to the n minus 1. Constants drop out, and sums and constant multiples differentiate term by term.
Product rule: (f g) prime = f prime g + f g prime. Quotient rule: (f over g) prime = (f prime g - f g prime) over g squared. Mnemonic: low d-high minus high d-low, over low squared.
Memorize the basics: sine goes to cosine, cosine to minus sine, e to the x stays e to the x, natural log of x goes to 1 over x. Rewrite roots and reciprocals as powers first: the square root of x is x to the one half.
Check by simplifying before differentiating. If a quotient can be divided out, that is faster and less error-prone than the quotient rule.
Worked example
Differentiate x squared times (x + 1) at x = 2.
f = x^2, g = x + 1
f' = 2x, g' = 1
f'g + fg' = 2x(x+1) + x^2
At x = 2: 4 times 3 + 4 = 16
Practice problem and solution
f(x) = x squared (x + 1). What is f prime(2)? Enter a number.
2x(x+1) + x squared = 4(3) + 4 = 16.
Mental model: Rules turn limits into algebra; simplify first.
Common trap: Writing (fg)' as f'g'.
5. The chain rule
Learning goal: Differentiate compositions.
For a composition f(g(x)) the chain rule gives f prime of g(x) times g prime of x: differentiate the outer function, leave the inside alone, then multiply by the derivative of the inside.
Example: (3x + 1) to the 4th. Outer is u to the 4th, inner is 3x + 1. The derivative is 4 (3x + 1) cubed times 3 = 12 (3x + 1) cubed. At x = 0 that equals 12.
For nested layers work from the outside in: the derivative of sine of (x squared) is cosine of (x squared) times 2x. Combine the chain rule with product and quotient rules as needed.
The most common error is forgetting the inner derivative. Check by asking what the inner function was.
Worked example
Differentiate (3x + 1) to the 4th at x = 0.
Outer: u^4, inner: 3x + 1
4 (3x + 1)^3 times 3
At x = 0: 4 times 1 times 3 = 12
Practice problem and solution
f(x) = (3x + 1) to the 4th. What is f prime(0)? Enter a number.
4(1)^3 times 3 = 12.
Mental model: Outer prime times inner prime.
Common trap: Forgetting the inner derivative.
6. Implicit differentiation and related rates
Learning goal: Differentiate relations and link rates through time.
When y is defined implicitly, such as x squared + y squared = 25, differentiate both sides with respect to x and treat y as a function of x, so each y term gets a factor of dy/dx. Then solve for dy/dx.
Related rates connect rates that change with time. Steps: draw and name the variables, write an equation that holds at all times, differentiate with respect to t, and only then substitute the values at the moment asked about.
Example: a circle's radius grows at 2 cm per second. Area A = pi r squared gives dA/dt = 2 pi r dr/dt. At r = 5, dA/dt = 2 pi times 5 times 2 = 20 pi square centimetres per second.
Never substitute a number for a changing quantity before you differentiate. Keep units with every rate.
Worked example
A circle's radius grows at 2 cm/s. Find dA/dt at r = 5.
A = pi r^2
dA/dt = 2 pi r dr/dt
At r = 5 and dr/dt = 2
20 pi cm^2/s
Practice problem and solution
With A = pi r squared, r = 5 cm and dr/dt = 2 cm/s, dA/dt is how many times pi? Enter a number.
2 times 5 times 2 = 20.
Mental model: Differentiate in time, then plug in.
Common trap: Substituting values before differentiating.
7. Linear approximation and the Mean Value Theorem
Learning goal: Approximate with tangent lines and state MVT.
Near a point a, f(x) is approximately f(a) + f prime(a)(x - a). This tangent-line approximation is accurate close to a. For the square root of 4.1 use a = 4: root 4 + (1/4)(0.1) = 2.025.
The Mean Value Theorem: if f is continuous on [a, b] and differentiable on (a, b), there is a c in (a, b) with f prime(c) = (f(b) - f(a))/(b - a). Somewhere the instantaneous rate equals the average rate. Rolle's theorem is the case f(a) = f(b), giving f prime(c) = 0.
Example: f(x) = x squared on [1, 3] has average slope (9 - 1)/2 = 4. Solve 2c = 4 to get c = 2, which is inside the interval.
MVT is an existence theorem. It proves results such as: if f prime is 0 on an interval then f is constant.
Worked example
Find c for x squared on [1, 3].
Average slope (9-1)/(3-1) = 4
f'(x) = 2x
Solve 2c = 4
c = 2
Practice problem and solution
For f(x) = x squared on [1, 3], MVT gives f prime(c) = 4. What is c? Enter a number.
2c = 4, so c = 2.
Mental model: Tangent lines approximate; MVT matches average to instant.
Common trap: Using MVT without checking continuity and differentiability.
8. Curve sketching with the first and second derivative
Learning goal: Find extrema, concavity and inflection points.
Critical points occur where f prime is 0 or undefined. The first derivative test: f prime changes from positive to negative at a local maximum, and from negative to positive at a local minimum.
The second derivative tells concavity: f double prime positive means concave up, negative means concave down. A point where concavity changes is an inflection point. The second derivative test at a critical point: positive gives a minimum, negative gives a maximum.
Example: f(x) = x cubed - 3x. f prime = 3x squared - 3 = 0 at x = -1 and 1. f double prime = 6x, so x = 1 is a local minimum with value 1 - 3 = -2, and x = -1 a local maximum with value 2. The inflection is at x = 0.
For a closed interval check the endpoints too: the absolute maximum or minimum may be at an end.
Worked example
Find the local minimum value of x cubed - 3x.
f' = 3x^2 - 3 = 0, so x = -1, 1
f'' = 6x, so f''(1) = 6 is positive
Local minimum at x = 1
f(1) = 1 - 3 = -2
Practice problem and solution
f(x) = x cubed - 3x has a local minimum at x = 1. What is its value there? Enter a number.
f(1) = 1 - 3 = -2.
Mental model: Critical points, then signs of f' and f''.
Common trap: Forgetting to check endpoints on a closed interval.
9. Optimization
Learning goal: Set up and solve max and min problems.
Optimization turns a story into a function of one variable. Name the quantity to maximize or minimize (the objective), write the constraint that links the variables, use it to eliminate one variable, then differentiate and find critical points.
Example: 100 metres of fence for a rectangle. Perimeter gives 2x + 2y = 100, so y = 50 - x. Area A = x (50 - x). A prime = 50 - 2x = 0 at x = 25. Then y = 25 and area = 625 square metres.
Always verify the type of the extremum: use the second derivative or compare with the endpoints. Include the domain: here x must lie between 0 and 50.
State the answer with units and answer the question asked: it may be the dimensions, not the maximum area.
Worked example
Maximize the area with 100 m of fence.
A = xy, 2x + 2y = 100
y = 50 - x
A = 50x - x^2
A' = 50 - 2x = 0 at x = 25
Area 625
Practice problem and solution
A rectangle has perimeter 100 m. What is its maximum area in square metres? Enter a number.
25 times 25 = 625.
Mental model: Constraint reduces to one variable; verify the extremum.
Common trap: Skipping the check that the critical point is a maximum.
10. Integrals and the Fundamental Theorem
Learning goal: Link area, antiderivatives and the net change.
A definite integral is the limit of Riemann sums: slice the interval, multiply height by width, add. It represents signed area: area above the axis counts positive, below counts negative.
An antiderivative F of f satisfies F prime = f. The Fundamental Theorem of Calculus, part 2: the integral from a to b of f equals F(b) - F(a). Part 1: the derivative of the integral from a to x of f(t) dt is f(x). Differentiation and integration undo each other.
Example: the integral from 0 to 3 of 2x dx. An antiderivative is x squared. The value is 9 - 0 = 9. A left Riemann sum with n rectangles gives 9 - 9/n, approaching 9.
Integrals of a rate give net change: the integral of velocity is displacement, and of marginal cost is total cost change.
Worked example
Compute the integral from 0 to 3 of 2x dx.
Antiderivative: x^2
F(3) = 9
F(0) = 0
9 - 0 = 9
Practice problem and solution
What is the integral from 0 to 3 of 2x dx? Enter a number.
x squared from 0 to 3 is 9.
Mental model: Integral is signed area; FTC computes it by antiderivatives.
Common trap: Forgetting that area below the axis is negative.