Unit 3Second Derivative

f''(x)

Concept 4.216Built for May 2027 AP Exam

Second Derivative $f''(x)$

Reviewed by ACEpath AP Math Editorial Team

Last Updated: September 2026

30-second summary

The second derivative is the derivative of the first derivative, so no new rule set is introduced and everything already learned gets used twice. Write $f''(x) = \frac{d}{dx}\left[f'(x)\right]$ , differentiate your own answer, and keep the rules straight, since the second pass is chosen by the shape of your first answer rather than by the shape of the original function. On a graph of $f'$ , the second derivative is the slope of that graph, which is why $f''$ fails wherever $f'$ has a corner. In context, $f''$ answers a different question than $f'$ : $f'$ says which way a quantity moves, and $f''$ says whether that movement is speeding up or slowing down. Get the first pass right, since every error in it survives the second.

On the AP Exam

What you’ll see: A polynomial or a trigonometric function differentiated twice, a relation with $\frac{dy}{dx}$ given and $\frac{d^2y}{dx^2}$ requested, a graph of $f'$ with the slope of one segment to be read, or a stated sign of $f''$ used to decide whether a rate is growing or shrinking.
What you’ll do: Finish the first derivative completely, then treat that answer as the new function and differentiate it with the rule its own structure demands.
Memory hook: Differentiate the derivative. The second pass starts from your own handwriting, not from the original function.
Use it on the exam: For MCQs: Count the passes. An option that still contains an undifferentiated variable came from a first pass, and an option equal to the square of $f'$ came from no pass at all. For FRQs: The scoring splits the two passes. On 2025 Q5(A) the guideline awards the first derivative, then a product-rule point and a chain-rule point for $\frac{d^2y}{dx^2}$ , then the numerical value, so a correct $f''$ with no visible first pass earns nothing.

Concept Review

The Second Derivative Is the Slope of the Slope

Figure 8 — Three stacked panels showing a function, its derivative and its second derivative aligned on the same input, with one vertical marker and sign notes.

The top panel is the height, the middle panel is its slope, and the bottom panel is the rate at which that slope changes. The orange marker sits at the same input in all three, so the three panels are three readings of one point. Where the middle panel is falling, the bottom panel is negative, which is the whole content of the idea.

Three stacked panels showing a function, its derivative and its second derivative aligned on the same input, with one vertical marker and sign notes

Four Names for One Object

Notation	Reads as	Where it comes from
$f''(x)$	the derivative of $f'$	one more pass on $f'$
$y''$	the same thing in prime form	shorthand when $y$ is already named
$\frac{d^2y}{dx^2}$	the derivative of $\frac{dy}{dx}$ with respect to $x$	Leibniz notation
$\frac{d^2}{dx^2}\left[f\right]$	the operator applied twice	operator notation

The exponent in $\frac{d^2y}{dx^2}$ is not a power. It counts how many times the operation was applied, which is why $\frac{d^2y}{dx^2}$ cannot be simplified to $\frac{d^2y}{d^2x}$ .

Two Passes on a Formula

For $f(x) = x^3 - 3x^2 + 1$ the first pass gives $f'(x) = 3x^2 - 6x$ and the second pass gives $f''(x) = 6x - 6$ . The degree dropped by one each time, and nothing else happened.

Trigonometric and exponential functions behave the same way, except that the constant factor reappears doubled:

$\frac{d^2}{dx^2}\left[\sin 2x\right] = \frac{d}{dx}\left[2\cos 2x\right] = -4\sin 2x$

The $2$ from the chain rule was joined by a second $2$ , which is why the amplitude of the second derivative is four times the original and not two.

The Second Pass Often Needs a Different Rule

For $h(x) = x^2e^{x}$ the first pass uses the product rule:

$h'(x) = 2xe^{x} + x^2e^{x} = e^{x}\left(x^2 + 2x\right)$

Factoring before the second pass is what keeps the work short, because $\frac{d}{dx}\left[x^2 + 2x\right] = 2x + 2$ needs no rule at all:

$h''(x) = e^{x}\left(x^2 + 2x\right) + e^{x}\left(2x + 2\right) = e^{x}\left(x^2 + 4x + 2\right)$

Skipping the factorisation produces the same answer through three cross terms instead of two, and three cross terms is where a lost factor hides.

Reading $f''$ Off a Graph of $f'$

When the graph of $f'$ is given, $f''$ is the slope of that graph, not its height. A linear piece of $f'$ therefore has a constant second derivative equal to that line's slope, and a corner in $f'$ is a place where $f''$ does not exist. On the 2017 exam the graph of $f'$ had a straight segment from $(-6, 2)$ to $(-2, 0)$ , so $f''(-5) = \frac{0-2}{-2-(-6)} = -\frac{1}{2}$ , and at the corner $x = 3$ the left and right slopes disagreed, so $f''(3)$ did not exist.

Implicit Second Derivatives

When $\frac{dy}{dx}$ contains both $x$ and $y$ , the second pass produces a $\frac{dy}{dx}$ term that must then be replaced. Starting from $x^2 + y^2 = 16$ :

$\frac{dy}{dx} = -\frac{x}{y} \quad\Longrightarrow\quad \frac{d^2y}{dx^2} = -\frac{y - x\frac{dy}{dx}}{y^2} = -\frac{y + \frac{x^2}{y}}{y^2} = -\frac{16}{y^3}$

The middle expression still contains $\frac{dy}{dx}$ , and the final one does not. An answer that stalls in the middle has finished the differentiation but not the problem.

Common Misconception Check

Myth 1: "The second derivative needs its own rules." Truth: It needs the same rules applied to a different input, namely the first derivative.
Myth 2: " $f''(a) = 0$ means the graph switches concavity at $a$ ." Truth: Zero is a candidate, not a verdict. For $f(x) = x^4$ the second derivative vanishes at $0$ and the concavity never changes.
Myth 3: " $f'' > 0$ means the function is positive there." Truth: The sign of $f''$ describes how the slope behaves. The sign of $f$ itself is a separate question with a separate test.

Topic-Specific Study Method

Differentiate Your Own Answer

Borrowed idea: Go one pass deeper. The second derivative is the derivative of what you just wrote, so the first pass must be finished and correct before the second begins. Differentiate your answer once more with the same rules. Watch the notation change: $f'(x)$ becomes $f''(x)$ and $\frac{dy}{dx}$ becomes $\frac{d^2y}{dx^2}$ . A sign error in the first derivative survives into the second one. Check the first pass before you start the second. When the answer is a product or a quotient, the second pass usually needs a different rule than the first.

First pass	Second pass	Result
$x^4$	power rule again	$12x^2$
$\cos 3x$	chain rule twice	$-9\cos 3x$
$xe^{2x}$	product rule, then chain	$e^{2x}\left(4x+4\right)$
$-\frac{x}{y}$	quotient rule, then substitute $\frac{dy}{dx}$	$-\frac{16}{y^3}$
slope of a straight piece of $f'$	no pass at all, just a slope	a number

① Differentiate $x^3e^{2x}$ twice without factoring, then again after factoring the first derivative. The two answers must match, and the second route shows why factoring early is not cosmetic.

② Read $f''(-5)$ and $f''(3)$ from the graph of $f'$ in Section 5 above. One is a slope and the other is a refusal, and telling them apart is the skill.

③ Take one implicit relation, find $\frac{dy}{dx}$ , and then circle every $\frac{dy}{dx}$ in your second pass before substituting. The circles are the substitutions you owe.

Training & Error Diagnosis

Core Archetype (Simplest Form): For $f(x) = x^3 - 3x^2 + 1$ , find $f''(x)$ and describe where $f''$ changes sign.

Standard Execution:

First pass. $f'(x) = 3x^2 - 6x$ .

Second pass. $f''(x) = 6x - 6$ .

Factor and solve. $6x - 6 = 6(x-1)$ , so $f''(x) = 0$ exactly at $x = 1$ .

Sign test. $f''(x) < 0$ for $x < 1$ and $f''(x) > 0$ for $x > 1$ , so the sign changes at $x = 1$ .

Exam Variation (Context / Parameter Shift): A culture's population is modeled by $P(t) = 200e^{-0.5t} + 500$ for $t \ge 0$ . Find $P''(t)$ and explain what its sign says about the population.

Solution: The first pass gives $P'(t) = -100e^{-0.5t}$ , and the second pass gives $P''(t) = 50e^{-0.5t}$ . Since the exponential is positive, $P''(t) > 0$ for every $t \ge 0$ , while $P'(t) < 0$ . The population is therefore falling, and the rate at which it falls is itself increasing toward zero, so the decline is slowing down rather than accelerating.

Common Traps & Failure Signals:

The one-pass answer (Signal: an undifferentiated variable still present): the second derivative was requested and the first derivative was delivered.
The squared derivative (Signal: $f''$ written as $\left(f'\right)^2$ ): the derivative was multiplied by itself instead of being differentiated again.
The height mistake (Signal: $f''$ set equal to the value of $f'$ at a point): a graph of $f'$ was read for its value when the question asked for its slope.

6D Cognitive Error Diagnosis (Where Your Thinking Broke):

Pass Skipped: You stopped after one derivative.
Notation Drift: You wrote $f''$ while the expression still contained an undifferentiated variable, which means the second pass was begun and abandoned, and the leftover term is exactly the one a reader checks first.
Squared Instead: You multiplied the derivative by itself rather than differentiating it again, so a linear first derivative produced a parabola where a constant was expected.

Past 10-Year AP Exam Appearance

Testing Frequency: Most Years; Five released exams score a second derivative, and two of those hand the expression over and score only your reading of its sign
10-Year Total Count: 5 of 9 Exams · 7 FRQ Parts; Counted where a part names $f''$, $\frac{d^2y}{dx^2}$, $h''$ or a second derivative of a model, and the value or sign of that object is the scored answer
FRQ Years Appeared: 2017 (Q3d, Q4b), 2023 (Q3c), 2024 (Q1d, Q4c), 2025 (Q5A), 2026 (Q3C)

Practice Examples

MCQ 1

Two Passes on a Product

The function $f$ is defined by $f(x) = x^2e^{3x}$ for all real numbers $x$ . What is $f''(x)$ ?

MCQ 2

Reading a Slope Off the Graph of $f'$

The graph of $f'$ on the interval $-6 \le x \le -2$ is the line segment joining the points $(-6, 2)$ and $(-2, 0)$ . What is the value of $f''(-5)$ ?

MCQ 3

What the Sign of $f''$ Actually Says

A cup of coffee cools on a table, and its temperature $C(t)$ , in degrees Celsius, satisfies $C''(t) < 0$ for $12 < t < 20$ . Which statement is supported by this information alone?

FRQ 1 • A Second Derivative From a Differential Equation

Let $y = f(x)$ be the particular solution to the differential equation $\frac{dy}{dx} = xy^3 - 2$ with initial condition $f(1) = -1$ .

(a) Find the value of $\frac{dy}{dx}$ at the point $(1, -1)$ .

(b) Find $\frac{d^2y}{dx^2}$ in terms of $x$ and $y$ .

(c) Use the results of parts (a) and (b) to find $f''(1)$ .

(d) Explain why the value of $f''$ must be recomputed at each new point even though the formula from part (b) does not change.

Cheatsheet