AP Calculus AB Derivatives: Rules, Applications, and Common Mistakes

24 min readOct 28, 2026

AP Calculus AB derivatives measure the instantaneous rate of change of a function and help students understand how quantities change at a specific point. In AP Calculus AB, derivatives are calculated using rules such as the power rule, product rule, quotient rule, and chain rule. They are also applied to real-world problems involving motion, optimization, related rates, and interpreting functions.

Dark wall covered in glowing purple handwritten math symbols and equations, including a quadratic formula, pi and infinity.

A car’s speedometer does not tell you how far you have travelled or how long the trip has taken. It tells you how fast you are going right now. That instantaneous rate of change is what a derivative measures, and it sits at the centre of AP Calculus AB derivatives.

Graphically, the same idea is the slope of a tangent line: at any point on a curve, the derivative tells you how steeply the graph is rising or falling there. “Rate of change” and “tangent slope” are two descriptions of one number, and moving comfortably between them is half the battle in this part of the course.

College Board: where derivatives sit in the course. AP Calculus AB is described as an introductory college-level course covering differential and integral calculus, organised into eight units. Four of those units are about differentiation. The course page publishes each unit’s weighting for the multiple-choice section:

Unit 2, Differentiation: Definition and Fundamental Properties: 10%–15%
Unit 3, Differentiation: Composite, Implicit, and Inverse Functions: 5%–10%
Unit 4, Contextual Applications of Differentiation: 10%–15%
Unit 5, Analytical Applications of Differentiation: 15%–20%

These ranges say that derivatives matter a great deal. They do not say which questions will appear, in what form, or how often, and nothing in this article predicts that. Derivatives also show up in other units and in the free-response section.

What is official and what is not. Statements labelled “College Board” describe the official course and come from AP Central; the pages are listed at the end. Everything else (the explanations, the worked examples and every practice problem) is original to this article and is not taken from a released exam. This article is independent of the College Board and is not affiliated with or endorsed by it.

Scope. You will work through the core of Unit 2 (the definition, the power, sum, product and quotient rules, and the derivatives of trigonometric, exponential and logarithmic functions), the chain rule from Unit 3, and the most common applications from Units 4 and 5. Implicit differentiation, derivatives of inverse functions, related rates and linearisation are also part of the course, but they need their own articles.

Understand the meaning of a derivative

Take a function f(x). Its derivative, written f'(x), is a second function: for every input x, it outputs the slope of the graph of f at that point, which is also the instantaneous rate of change of f there.

The idea starts with an average. The average rate of change of f from x to x + h is the slope of a secant line through two points on the curve:

(f(x + h) - f(x)) / h

Now shrink h towards 0. The two points slide together, the secant line swings towards the tangent line, and the average rate settles on the instantaneous rate. That limit is the derivative:

f'(x) = lim (h→0) [f(x + h) - f(x)] / h

You will see three notations for the same thing: f'(x), dy/dx and d/dx [f(x)]. College Board: Unit 2 opens with defining the derivative, using derivative notation and estimating derivatives at a point, before it moves on to the rules.

A simple example. Let f(x) = x^2 and look at the point where x = 3, so the point on the graph is (3, 9). The average rate of change from x = 3 to x = 3 + h is:

((3 + h)^2 - 9) / h = (9 + 6h + h^2 - 9) / h = (6h + h^2) / h = 6 + h

Put in smaller and smaller values of h:

h = 1: the secant slope is 7, from (3, 9) to (4, 16)
h = 0.1: the slope is 6.1, because 3.1^2 = 9.61 and (9.61 - 9) / 0.1 = 6.1
h = 0.01: the slope is 6.01

As h approaches 0, 6 + h approaches 6, so f'(3) = 6. That one number has two meanings:

Rate of change: at x = 3, the output of f is growing at 6 units for every 1 unit of input.
Tangent slope: the tangent line to y = x^2 at (3, 9) has slope 6. Its equation is y - 9 = 6(x - 3), which simplifies to y = 6x - 9.

Doing the same limit for a general x gives f'(x) = 2x, and 2(3) = 6 agrees. Notice what changed: f(3) = 9 is a height on the graph, while f'(3) = 6 is a steepness. Keeping those two apart is one of the most useful habits in this topic, and it comes back in the mistakes section.

Units help too. If f gives distance in metres and x is time in seconds, then f' is in metres per second. The derivative's units are always the output's units divided by the input's units.

The rules in the next section exist so that you do not have to run this limit every time.

Review the essential derivative rules for AP Calculus

Every rule below can be proved from the limit definition. You do not need the proofs to use them, but you do need to know when each rule applies. A dependable habit is to read an expression from the outside in and ask what the last operation is: a sum, a product, a quotient or a function inside a function. That answer tells you which rule to reach for first.

College Board: Unit 2 lists the power rule; the constant, sum, difference and constant multiple rules; the derivatives of cos x, sin x, e^x and ln x; the product rule; the quotient rule; and the derivatives of tan x, cot x, sec x and csc x. Unit 3 covers composite functions, which is where the chain rule lives.

Constant rule

d/dx [c] = 0

The graph of y = c is a horizontal line, and a horizontal line has slope 0 everywhere. So if f(x) = 7, then f'(x) = 0. Anything that is just a number counts as a constant, including π², e³ and ln 2, so each of those also has derivative 0.

Power rule

d/dx [x^n] = n·x^(n-1)

Bring the exponent down as a multiplier, then lower the exponent by 1. It works for positive, negative and fractional exponents, so rewrite roots and reciprocals as powers before you differentiate.

f(x) = x^5 gives f'(x) = 5x^4.
g(x) = √x = x^(1/2) gives g'(x) = (1/2)·x^(-1/2) = 1/(2√x).
h(x) = 1/x^3 = x^(-3) gives h'(x) = -3·x^(-4) = -3/x^4.

Constant multiple rule

d/dx [c·f(x)] = c·f'(x)

Stretching a graph vertically by a factor of c stretches every slope by the same factor, so the constant simply waits outside while you differentiate. For f(x) = -3x^4, the derivative is -3·4x^3 = -12x^3.

Sum and difference rules

d/dx [f(x) ± g(x)] = f'(x) ± g'(x)

Differentiate term by term. For f(x) = x^4 - 3x^2 + 5x - 9:

f'(x) = 4x^3 - 6x + 5

The 5x term becomes 5 (power rule with exponent 1), and the constant -9 disappears.

Trigonometric, exponential and logarithmic functions

These are facts to memorise, and each has a reason you can picture:

d/dx [sin x] = cos x
d/dx [cos x] = -sin x
d/dx [tan x] = sec^2 x
d/dx [cot x] = -csc^2 x
d/dx [sec x] = sec x·tan x
d/dx [csc x] = -csc x·cot x
d/dx [e^x] = e^x
d/dx [ln x] = 1/x, for x > 0

Why they make sense: the graph of cos x starts to fall just after x = 0, where sin x is rising, which is why the minus sign appears. Every “co-” function (cos, cot, csc) has a negative derivative. The function e^x is the one whose slope at every point equals its height. And ln x rises ever more slowly as x grows, matching a slope of 1/x that shrinks towards 0. These formulas assume x is measured in radians.

Examples:

f(x) = 3 sin x - 2 cos x gives f'(x) = 3 cos x - 2(-sin x) = 3 cos x + 2 sin x.
g(x) = 4e^x + ln x gives g'(x) = 4e^x + 1/x, valid for x > 0.
h(x) = tan x gives h'(x) = sec^2 x.

One warning: the power rule is for a variable base with a constant exponent. In e^x the variable is in the exponent, so d/dx [e^x] is not x·e^(x-1).

Product rule

d/dx [f(x)·g(x)] = f'(x)·g(x) + f(x)·g'(x)

When two changing factors are multiplied, each one contributes: the first factor’s change acts on the second as it currently is, and the second factor’s change acts on the first as it currently is. The derivative of a product is not the product of the derivatives.

Example: differentiate y = (x^2 + 1)(x^3 - 2x).

Let f = x^2 + 1 and g = x^3 - 2x, so f' = 2x and g' = 3x^2 - 2.
y' = 2x(x^3 - 2x) + (x^2 + 1)(3x^2 - 2)
y' = (2x^4 - 4x^2) + (3x^4 + x^2 - 2) = 5x^4 - 3x^2 - 2

You can confirm this by expanding first: y = x^5 - x^3 - 2x, and the power rule gives y' = 5x^4 - 3x^2 - 2. Same answer, which is a useful check whenever the expansion is short.

Quotient rule

d/dx [f(x)/g(x)] = (f'(x)·g(x) - f(x)·g'(x)) / (g(x))^2, where g(x) ≠ 0

In words: bottom times the derivative of the top, minus top times the derivative of the bottom, all over the bottom squared. The order of the two terms in the numerator matters, because swapping them flips the sign of the answer.

Example: differentiate y = x^2 / (x + 1).

Let f = x^2 and g = x + 1, so f' = 2x and g' = 1.
y' = (2x(x + 1) - x^2·1) / (x + 1)^2
y' = (2x^2 + 2x - x^2) / (x + 1)^2 = (x^2 + 2x) / (x + 1)^2 = x(x + 2) / (x + 1)^2

This holds for x ≠ -1, because the original function is undefined there.

Chain rule

d/dx [f(g(x))] = f'(g(x))·g'(x)

A composite function is a function inside a function, and its rate of change is the rate of the outer layer, evaluated at the inner layer, multiplied by the rate of the inner layer. In practice: spot the outer and inner functions, differentiate the outer one while leaving the inner one untouched, then multiply by the derivative of the inner one.

y = (3x^2 - 1)^4: the outer function is something to the fourth power, and the inner function is 3x^2 - 1. So y' = 4(3x^2 - 1)^3·6x = 24x(3x^2 - 1)^3.
y = e^(5x): the inner function is 5x, so y' = e^(5x)·5 = 5e^(5x).
y = sin(3x): the inner function is 3x, so y' = cos(3x)·3 = 3cos(3x).
y = ln(x^2 + 1): the inner function is x^2 + 1, so y' = (1/(x^2 + 1))·2x = 2x/(x^2 + 1).

Real problems usually blend rules. y = x^2·e^(3x) needs the product rule on the outside and the chain rule inside the exponential, which you will see in the practice questions. College Board: Unit 3 includes a topic on selecting procedures for calculating derivatives, which is exactly this skill of choosing the right rule.

Rules at a glance

Copy this block into your notes:

Constant:           d/dx [c] = 0
Power:              d/dx [x^n] = n·x^(n-1)
Constant multiple:  d/dx [c·f] = c·f'
Sum / difference:   d/dx [f ± g] = f' ± g'
Product:            d/dx [f·g] = f'·g + f·g'
Quotient:           d/dx [f/g] = (f'·g - f·g') / g^2
Chain:              d/dx [f(g(x))] = f'(g(x))·g'(x)

sin x → cos x          cos x → -sin x
tan x → sec^2 x        cot x → -csc^2 x
sec x → sec x·tan x    csc x → -csc x·cot x
e^x → e^x              ln x → 1/x  (x > 0)

Understand derivatives from graphs and tables

Not every derivative question hands you a formula. College Board: the course framework describes problems represented graphically, numerically, analytically and verbally, and it lists translating between representations as one of its mathematical practices. So it pays to read derivative information from pictures and tables as well as from algebra.

What the sign of f’ tells you about f

If f'(x) > 0 on an interval, f is increasing there.
If f'(x) < 0 on an interval, f is decreasing there.
If f'(x) = 0, the tangent line is horizontal. That makes x a candidate for a peak or a valley, but only a candidate.
If f' changes from positive to negative at x = c, f has a local maximum there. If it changes from negative to positive, f has a local minimum. If the sign does not change, there is no local extremum at c.

The last two bullets are the first derivative test. The graph of f' is a map of slopes: its height at any x is the slope of f at that x. Where the f' graph is above the x-axis, f climbs. Where it is below, f falls. Where it crosses the axis, f turns around, or at least pauses.

Worked example. Let f(x) = x^3 - 3x. Then f'(x) = 3x^2 - 3 = 3(x - 1)(x + 1), which is zero at x = -1 and x = 1. Test one point in each region:

Left of -1, try x = -2: f'(-2) = 9, which is positive, so f is increasing.
Between -1 and 1, try x = 0: f'(0) = -3, which is negative, so f is decreasing.
Right of 1, try x = 2: f'(2) = 9, which is positive, so f is increasing.

So f has a local maximum at x = -1, where f(-1) = 2, and a local minimum at x = 1, where f(1) = -2. If you had been given only the graph of f', crossing the axis downwards at -1 and upwards at 1, you would reach the same conclusions.

Estimating a derivative from a table

Suppose a differentiable function g has these values:

g(0) = 3
g(2) = 7
g(5) = 19
g(9) = 31

To estimate g'(3), use the closest listed values on either side of 3, which are x = 2 and x = 5:

g'(3) ≈ (g(5) - g(2)) / (5 - 2) = (19 - 7) / 3 = 4

Treat that as an estimate, never as an exact value: a table only gives average rates over intervals. The average rates between the listed points are 2, 4 and 3, all positive, which suggests g is increasing. A table cannot rule out wiggles between the points it lists, so word your conclusion as "suggests" rather than "proves" unless the problem gives you more.

Where derivative information needs care

Corners. f(x) = |x| is continuous at 0, but its slope is -1 just to the left and +1 just to the right. There is no single tangent line, so f'(0) does not exist.
Vertical tangents. For f(x) = x^(1/3), the derivative is f'(x) = (1/3)·x^(-2/3), which is undefined at 0. The graph is continuous there, but the tangent line is vertical, so there is no finite slope.
Breaks in the graph. If a function has a jump or a hole at a point, it has no derivative there. Differentiable at a point implies continuous at that point, but not the other way round, as |x| shows.
Zero is not enough. For f(x) = x^3, f'(0) = 0, yet f increases on both sides of 0, so there is no extremum at 0. Always check for a sign change.

College Board: Unit 5 covers these ideas in more depth, including the Extreme Value Theorem, critical points, the first and second derivative tests, concavity, and connecting a function with its first and second derivatives.

Apply derivatives to real mathematical problems

College Board: Unit 4 is titled Contextual Applications of Differentiation and Unit 5 is titled Analytical Applications of Differentiation. The examples below sit in that territory. They are my own, written to practise the skills, not copies of released questions.

Tangent-line equations

A tangent line needs a point and a slope. The point is (a, f(a)), the slope is f'(a), and the point-slope form y - f(a) = f'(a)(x - a) puts them together.

Example: find the tangent line to f(x) = x^3 - 2x + 1 at x = 2.

Point: f(2) = 8 - 4 + 1 = 5, so the point is (2, 5).
Slope: f'(x) = 3x^2 - 2, so f'(2) = 3(4) - 2 = 10.
Line: y - 5 = 10(x - 2), which simplifies to y = 10x - 15.

Check: at x = 2 the line gives 10(2) - 15 = 5, so it does pass through the point.

Interpreting rates of change

A good interpretation names the quantity, the moment, the direction and the units: “At time t = a, the quantity is increasing (or decreasing) at |f’(a)| units of output per unit of input.”

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Example: the volume of water in a tank is V(t) = 200 - 3t^2 litres, t minutes after draining starts, for 0 ≤ t ≤ 8.

V'(t) = -6t, so V'(5) = -30.
Interpretation: at t = 5 minutes, the volume is decreasing at 30 litres per minute.

Keep the two numbers apart. V(5) = 125 is how much water is in the tank, and V'(5) = -30 is how fast it is changing. The negative sign is what tells you "decreasing", so write "decreasing at 30" and not "decreasing at -30".

Critical points and intervals of increase and decrease

A critical point is a value of x in the domain where f'(x) = 0 or where f'(x) does not exist. College Board: critical points appear alongside the Extreme Value Theorem and local versus global extrema in Unit 5. The routine is: find f', find the critical points, then test the sign of f' on each interval between them.

Example: analyse g(x) = x^4 - 4x^3.

g'(x) = 4x^3 - 12x^2 = 4x^2(x - 3), which is zero at x = 0 and x = 3.
For x < 0, try x = -1: g'(-1) = 4(1)(-4) = -16, negative, so g is decreasing.
For 0 < x < 3, try x = 1: g'(1) = 4(1)(-2) = -8, negative, so g is decreasing.
For x > 3, try x = 4: g'(4) = 4(16)(1) = 64, positive, so g is increasing.

Conclusions: g decreases on (-∞, 3) and increases on (3, ∞). The point x = 0 is a critical point, but the sign of g' does not change there, so it is not an extremum; the graph just flattens briefly on the way down. At x = 3 the sign changes from negative to positive, so g has a local minimum with g(3) = 81 - 108 = -27. Because g falls all the way to 3 and rises afterwards, that is also the lowest point of the whole graph.

Optimisation

Optimisation problems ask for the largest or smallest value of something. College Board: Unit 5 includes solving optimisation problems. A reliable sequence:

Name the variable and the quantity to optimise, and draw a sketch if it helps.
Write that quantity as a function of one variable, and state the domain.
Differentiate and find the critical points.
Justify that you have a maximum or minimum, using a sign change in the derivative, the second derivative, or the endpoint values.
Answer the question that was asked, with units.

Example: you have 120 m of fencing to build a rectangular pen against a long straight wall, so only three sides need fencing. What dimensions give the largest area?

Let x be the width of the two sides that meet the wall. The side parallel to the wall is then 120 - 2x.
Both lengths must be positive, so the domain is 0 < x < 60.
Area: A(x) = x(120 - 2x) = 120x - 2x^2.
A'(x) = 120 - 4x, which is zero at x = 30.
Justification: A'(x) > 0 for x < 30 and A'(x) < 0 for x > 30, so A rises and then falls, and x = 30 gives the maximum. The second derivative A''(x) = -4 is negative, which agrees. The area tends to 0 as x approaches either end of the domain.
Answer: the width is 30 m, the length is 120 - 2(30) = 60 m, and the maximum area is 30 × 60 = 1800 square metres.

Check: 2(30) + 60 = 120, so the fencing is fully used.

Common derivative mistakes

Many errors with derivatives are slips rather than gaps in understanding. Each mistake below comes with a quick way to catch it.

One check works for almost all of them: the numerical slope test. Pick a value a, then compute (f(a + 0.01) - f(a - 0.01)) / 0.02 on a calculator. That number is very close to f'(a), so it should agree with your formula evaluated at a. College Board: the course overview says technology should be used to confirm written work, and the exam has parts where a graphing calculator is and is not permitted, so use this where your task allows it.

1. Applying the power rule incorrectly

Two versions are common:

Forgetting to lower the exponent: writing d/dx [x^4] = 4x^4 instead of 4x^3.
Using the power rule when the variable is in the exponent: writing d/dx [e^x] = x·e^(x-1) instead of e^x.

How to catch it: for x^4 at x = 2, the numerical slope test gives about 32.0008. The correct formula 4x^3 gives 32, while the faulty one gives 64. For e^x at x = 1, the test gives about 2.7183, which matches e^1 and not the faulty answer of 1. Also ask: is the variable in the base (power rule) or in the exponent (not the power rule)?

2. Forgetting the chain rule

The classic slip is d/dx [sin(3x)] = cos(3x). The correct answer is 3cos(3x), because the inside function 3x has its own derivative, 3.

How to catch it: whenever the thing inside a function is anything other than a plain x, ask "Did I multiply by the derivative of the inside?" The numerical slope test at x = 0 gives about 2.9996 for sin(3x), which supports 3 and rules out the faulty value of 1.

3. Misusing the product or quotient rule

The product mistake is differentiating each factor and multiplying: for (x^2 + 1)(x^3 - 2x) that gives 2x(3x^2 - 2) = 6x^3 - 4x. The correct derivative is 5x^4 - 3x^2 - 2.

How to catch it: the original function has degree 5, so its derivative should have degree 4, but the faulty answer has degree 3. Expanding first, when the product is short, also gives an independent answer.

The quotient mistake is swapping the numerator terms, which flips the sign. Test your version of the rule on x^2 / x, which simplifies to x and so has derivative 1. The correct order gives (2x·x - x^2·1) / x^2 = 1. The swapped order gives -1, so you know it is wrong.

4. Confusing a function value with a derivative value

For f(x) = x^2 + 3x, we have f(2) = 10 but f'(2) = 7. The first is a height on the graph and the second is a steepness. A tangent line at x = 2 is y - 10 = 7(x - 2). Using 10 as the slope, or 7 as the y-coordinate, is a very common error.

How to catch it: label them as you write, “point: (2, f(2))” and “slope: f’(2)”. With graphs and tables, first ask whether the question is about the height of f or the height of f'. Units help here too: height has the output's units, while slope has the output's units per input unit.

5. Ignoring domain restrictions

The formula d/dx [ln x] = 1/x only makes sense for x > 0. Similarly, √x is defined at 0, but its derivative 1/(2√x) is not, so f'(0) does not exist.

Example: find the critical points of f(x) = x^2 - 4 ln x. The domain is x > 0. Then f'(x) = 2x - 4/x = (2x^2 - 4)/x, which is zero when x = √2 or x = -√2. Only x = √2 is in the domain, so x = -√2 must be rejected.

How to catch it: write the domain at the top of the problem, and after solving f'(x) = 0, check that every solution belongs to it. Also check where f' is undefined but f is defined.

6. Making algebra mistakes after differentiating

Typical slips are losing a minus sign when distributing, and cancelling a term that is not a factor. In the quotient-rule example earlier, the numerator 2x(x + 1) - x^2 becomes 2x^2 + 2x - x^2. Writing + x^2 instead of - x^2 is easy to do when you rush.

How to catch it: substitute a simple number into both your unsimplified and your simplified answers. At x = 1, the unsimplified form (2x(x + 1) - x^2) / (x + 1)^2 gives (4 - 1)/4 = 3/4, and the simplified form x(x + 2)/(x + 1)^2 also gives 3/4. A wrong "simplification" such as (x + 2)/(x + 1) gives 3/2, so the mismatch exposes it. Factor wherever you can, because factored forms make zeros and signs easy to read.

A practical derivative practice method

A repeatable process stops you from staring at a problem and guessing. Here it is, with one function run through every step: y = √(x^2 + 1).

Identify the type of function. Is it a polynomial, a trigonometric, exponential or logarithmic function, a product, a quotient, or a composite? Here the square root is applied to the expression x^2 + 1, so it is a composite function.
Select the appropriate rule. Read from the outside in. The outer function is a square root, which is a power of 1/2, and the inner function is x^2 + 1. That calls for the chain rule with the power rule.
Differentiate carefully. Write each piece down: y = (x^2 + 1)^(1/2), so y' = (1/2)(x^2 + 1)^(-1/2)·2x.
Simplify and check the result. Tidy up to y' = x / √(x^2 + 1). Then run a check: at x = 1 the formula gives 1/√2 ≈ 0.7071, and the numerical slope test gives about 0.7071 too.
Interpret the answer in context if required. If y were a cost in dollars and x the number of units made, y' ≈ 0.71 at x = 1 would mean the cost rises by about 71 cents for each extra unit near that point. State the units and the direction, as in the rates-of-change section.
Review incorrect attempts and practise similar problems. Keep an error log. For every miss, write down what you did, what was correct and which type of slip it was (rule choice, chain rule, algebra, domain or units). A few days later, do two or three similar problems without looking at the log.

A template you can copy for the error log:

Problem:
My answer:
Correct answer:
Where it went wrong (rule / chain rule / algebra / domain / units):
What I will do differently:
Similar problem to try again in a few days:

College Board: what to practise with. Its guidance on practising for AP Exams starts in your own AP class: ask your teacher about the practice resources in AP Classroom, which include progress checks with multiple-choice and free-response questions, the AP Question Bank and AP Daily videos. It also points students to the free-response questions, scoring information and sample student responses from past exams on AP Central, and to Bluebook test previews for digital exams. Check with your teacher about which of these apply to your exam.

College Board: why writing your reasoning matters. The course page lists “justify reasoning and solutions” as a mathematical practice, weighted at 35%–60% of the free-response section, and “use correct notation, language, and mathematical conventions” at 10%–25%. So showing your reasoning and using correct notation are not optional extras: in a free-response answer, both are part of what is assessed.

Worked practice questions

These five problems are original to this article. They mix procedure and reasoning, and each uses a different combination of rules. Work each one on paper before you look at the solutions, which start after the divider below.

The problems

Problem 1 (product rule and chain rule). Find dy/dx for y = x^2·e^(3x).

Problem 2 (quotient rule and trigonometry). Let f(x) = sin x / x^2 for x ≠ 0.

(a) Find f'(x).
(b) Find f'(π).

Problem 3 (tangent lines). Let f(x) = 2x^3 - 5x^2 + 4.

(a) Write an equation of the tangent line to the graph of f at x = 1.
(b) Find every value of x at which the tangent line is horizontal.

Problem 4 (rates of change in context). A drone’s height above the ground is A(t) = 40t - 5t^2 metres, t seconds after launch, for 0 ≤ t ≤ 8.

(a) Find A'(3) and interpret it with units.
(b) Find A'(6) and interpret it with units.

Problem 5 (reasoning from a derivative). The derivative of a function f is f'(x) = (x - 1)(x + 3)^2.

(a) Find the critical points of f.
(b) On which intervals is f increasing, and on which is it decreasing?

Stop here and attempt the problems first. Solutions are below.

The solutions

Solution 1. The function is a product, so use the product rule, with the chain rule for the exponential factor.

Let u = x^2 and v = e^(3x). Then u' = 2x, and by the chain rule v' = e^(3x)·3 = 3e^(3x).
dy/dx = u'v + uv' = 2x·e^(3x) + x^2·3e^(3x)
Factor: dy/dx = e^(3x)(2x + 3x^2) = x·e^(3x)(3x + 2).

Check: at x = 1 the formula gives 5e^3 ≈ 100.43, and the numerical slope test on x^2·e^(3x) gives about 100.43 as well.

Solution 2. (a) Use the quotient rule with f = sin x on top and g = x^2 on the bottom, so the top's derivative is cos x and the bottom's derivative is 2x.

f'(x) = (x^2·cos x - sin x·2x) / (x^2)^2 = (x^2 cos x - 2x sin x) / x^4
Every term in the numerator has a factor of x, so for x ≠ 0 you can divide the top and bottom by x: f'(x) = (x cos x - 2 sin x) / x^3.

(b) Substitute x = π, using cos π = -1 and sin π = 0:

f'(π) = (π(-1) - 2(0)) / π^3 = -π/π^3 = -1/π^2, which is about -0.101.

The slope is negative, so the graph of f is falling at x = π.

Solution 3. (a) You need a point and a slope.

Point: f(1) = 2 - 5 + 4 = 1, so the point is (1, 1).
Slope: f'(x) = 6x^2 - 10x, so f'(1) = 6 - 10 = -4.
Tangent line: y - 1 = -4(x - 1), which simplifies to y = -4x + 5.

(b) A horizontal tangent has slope 0, so solve f'(x) = 0:

6x^2 - 10x = 2x(3x - 5) = 0, so x = 0 or x = 5/3.
The corresponding points on the graph are (0, 4) and (5/3, -17/27).

Solution 4. First differentiate: A'(t) = 40 - 10t.

(a) A'(3) = 40 - 30 = 10. At t = 3 seconds, the drone's height is increasing at 10 metres per second.
(b) A'(6) = 40 - 60 = -20. At t = 6 seconds, the drone's height is decreasing at 20 metres per second, so it is descending.
© Set A'(t) = 0, which gives t = 4. Justification: A'(t) > 0 for t < 4 and A'(t) < 0 for t > 4, so A increases and then decreases, which means t = 4 is a maximum. The endpoint values A(0) = 0 and A(8) = 0 are lower, and A(4) = 160 - 80 = 80. The drone is highest at t = 4 seconds, at a height of 80 metres.

Solution 5. The derivative is a polynomial, so it exists for every x.

(a) Set (x - 1)(x + 3)^2 = 0. The critical points are x = 1 and x = -3.
(b) The factor (x + 3)^2 is never negative, and it is zero only at x = -3. So the sign of f' follows the sign of x - 1. For x < 1, f'(x) < 0, so f is decreasing on (-∞, 1). For x > 1, f'(x) > 0, so f is increasing on (1, ∞).
© At x = 1, f' changes from negative to positive, so f has a local minimum there. At x = -3, f' is negative on both sides (for example, f'(-4) = -5 and f'(-2) = -3), so the sign does not change and f has neither a local maximum nor a local minimum there. The graph of f just flattens out for an instant as it keeps falling.

A derivative revision checklist

Copy this into your notes or print it, and tick items off as you can do them without looking anything up.

DERIVATIVE REVISION CHECKLIST

Rule recognition
☐ I can state the power, constant multiple, sum/difference, product, quotient and chain rules from memory.
☐ I know the derivatives of sin x, cos x, tan x, cot x, sec x, csc x, e^x and ln x.
☐ I read each expression from the outside in to decide which rule comes first.
☐ I rewrite roots and reciprocals as powers before differentiating.Accurate differentiation
☐ I write down u, v, u' and v' before using the product or quotient rule.
☐ I multiply by the derivative of the inside whenever the inside is not plain x.
☐ I check my answer with a simple value of x or the numerical slope test.
☐ I factor my answer when I need to find where it equals zero.Graph and table interpretation
☐ I can say what f' > 0, f' < 0 and f' = 0 mean for the graph of f.
☐ I use a sign change in f' to decide between a local maximum, a local minimum, or neither.
☐ I estimate a derivative from a table using the closest values on either side, and I call it an estimate.
☐ I can spot corners, vertical tangents and breaks, where a derivative may not exist.Application problems
☐ I find the point and the slope separately, then write the tangent line.
☐ I state the domain and reject any critical point that lies outside it.
☐ I interpret a rate of change with the quantity, the moment, the direction and the units.
☐ I justify every maximum and minimum, and I check endpoints when the domain is closed.Error review
☐ I keep an error log that records the type of slip for each mistake.
☐ I redo missed problems a few days later without looking at the solution.
☐ I check that I am not confusing f(a) with f'(a).

Conclusion: a more systematic way to study AP Calculus AB derivatives

Preparing for AP Calculus AB derivatives becomes more systematic when you build on three habits. First, know what a derivative means, as a rate of change and as the slope of a tangent line, so you can judge whether an answer makes sense. Second, choose the rule by reading each expression from the outside in, and write out every piece before you combine them. Third, review your mistakes by type, so that each slip becomes a pattern you can recognise and fix instead of a one-off.

None of this guarantees any particular result on any exam. What it does is turn practice into something you can track: for each topic you can say which rules you recognise, which checks you run and which slips you still make.

If you would like guided practice built around habits like these, you can explore EduShaale’s AP Calculus AB course support alongside the official College Board resources listed below. A good next step is to take the checklist above, pick the one section you ticked least, and spend your next three practice sessions on it.

Official sources

The College Board pages below were read on 28 October 2026. Statements labelled “College Board” in this article come from them.

For the complete, current list of topics and skills, see the AP Calculus AB and BC Course and Exam Description, which College Board links from the course page above.

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