Worksheets · AP Calculus AB · Unit 3

AP Calculus AB Unit 3 FRQ Practice: Chain Rule, Implicit, Inverse

Four exam-style free-response questions on the chain rule, implicit differentiation and inverse functions, all without a calculator. Each is scored out of 9, with a scoring guide and a printable PDF with room to write, free to use in class. Good practice before the Unit 3 Progress Check in AP Classroom. Multiple choice for this unit: Unit 3 worksheet.

The exam rarely builds a whole free-response question on Unit 3, so this set is practice with its skills in the free-response format. The mixed set is the closest to a real exam.

Sigma Prep · sigmaprep.io/worksheets/ap-calculus-ab

Name

AP Calculus AB Unit 3 Free Response: Chain Rule, Implicit, Inverse

Date Version A

Directions: Show your work in the space under each part: the setup, the values you use and your reasoning. A justification needs a mathematical reason, with the conditions of any theorem or test you use checked. An answer with no supporting work may earn no credit. Give decimal answers correct to three decimal places unless a part says otherwise. Each question is worth 9 points. Unless a question says otherwise, the domain of a function f is every real number x for which f(x) is a real number.

This set is practice with Unit 3 skills in the free-response format: the exam almost never builds a whole question on Unit 3 alone. The mixed free-response set is the closest to a real exam.

Part B: No calculator is allowed.

Questions 1–4

1.
The functions $f$ and $g$ are twice differentiable everywhere. Some of their values, and of $f^{'}$ and $g^{'}$ , are recorded below.
$\begin{array}{ccccc} x & 1 & 2 & 3 & 4 \\ f (x) & 2 & 0 & 3 & 4 \\ f^{'} (x) & - 2 & - 3 & 3 & 1 \\ g (x) & - 1 & 1 & 3 & 4 \\ g^{'} (x) & 3 & 2 & 1 & 2 \end{array}$
1. (a)
  Let $h (x) = f (g (x))$ . Must there be a value $c$ , for $2 < c < 4$ , at which $h (c) = 3$ ? Justify your answer.
2. (b)
  Let $k (x) = g (f (x))$ . Write an equation for the line tangent to the graph of $k$ at $x = 3$ . Show the work that leads to your answer.
1. (c)
  Let $r (x) = \frac{g (x)}{f (x)}$ . Find $r^{'} (3)$ . Show the work that leads to your answer.
2. (d)
  Find $\lim_{x \to 4} \frac{f (x) - 4}{g (x) - 4}$ . Show the work that leads to your answer.
2.
Throughout this question, the curve is the set of points $(x, y)$ with $x^{2} - x y - y^{2} = 5$ .
1. (a)
  Differentiate implicitly to show that $\frac{d y}{d x} = \frac{2 x - y}{x + 2 y}$ .
2. (b)
  The curve passes through $(- 3, - 1)$ . Use the tangent line there to estimate the $y$ -value of the nearby point of the curve with $x = - 3.2$ .
1. (c)
  Determine every point $(x, y)$ of the curve where the tangent line is vertical. Justify your answer.
2. (d)
  Compute $\frac{d^{2} y}{d x^{2}}$ where the curve passes through $(- 3, - 1)$ . Show the work that leads to your answer.
3.
The graph of the continuous function $f$ , which consists of line segments, is shown for $0 \leq x \leq 6$ . Let $g$ be the function given by $g (x) = x^{2} + 2 x - 5$ .
1. (a)
  Let $h$ be the function defined by $h (x) = f (g (x))$ . Find $h^{'} (2)$ . Show the work that leads to your answer.
2. (b)
  For $0 \leq x \leq 6$ , let $k (x) = g (f (x))$ . Find the average rate of change of $k$ over the interval $0 \leq x \leq 4$ . Show the work that leads to your answer.
1. (c)
  With $k$ as in part (b), is $k$ differentiable at $x = 4$ ? Give a reason for your answer.
2. (d)
  With $k$ as in part (b), at which $x$ -values in $0 < x < 6$ does the graph of $k$ have a horizontal tangent line? Justify your answer.
4.
Let $f$ be the function defined by $f (x) = {(x^{2} + 1)}^{3}$ , where the domain of $f$ is restricted to $x \geq 0$ .
1. (a)
  Find $f^{'} (x)$ . Use $f^{'} (x)$ to explain why $f$ has an inverse function.
2. (b)
  Let $g$ be the inverse function of $f$ . Find $g^{'} (8)$ . Show the work that leads to your answer.
1. (c)
  For constants $k$ and $m$ , let $p (x) = {\begin{cases} k x^{2} + m & for x < 1 \\ f (x) & for x \geq 1 \end{cases}$ . Find the values of $k$ and $m$ for which $p$ is differentiable at $x = 1$ . Show the work that leads to your answer.
2. (d)
  Let $h$ be the function defined by $h (x) = x g (x)$ , with $g$ as in part (b). Find $h^{'} (8)$ . Show the work that leads to your answer.
  End of Version A

Free AP Calculus AB worksheets at sigmaprep.io/worksheets/ap-calculus-abOriginal questions written by Sigma Prep. AP is a registered trademark of the College Board, which is not affiliated with this sheet.

Scoring guide · Version A

Question 1Composite functions and derivative rules from a table · no calculator · 9 points

(a) 2 points IVT justification · Topic 1.16

1 point: $h (2) = f (g (2)) = f (1) = 2$ and $h (4) = f (g (4)) = f (4) = 4$ , so $3$ is between them
1 point: yes, by the Intermediate Value Theorem, with $h$ continuous

Solution

A composite of continuous functions is continuous, so the Intermediate Value Theorem applies to $h$ on $2 \leq x \leq 4$ .
$h (2) = f (g (2)) = f (1) = 2$ and $h (4) = f (g (4)) = f (4) = 4$ , and $3$ is between them.

(b) 3 points Tangent line or tangent-line approximation · Topic 3.1

1 point: chain rule:
$k^{'} (3) = g^{'} (f (3)) f^{'} (3)$
1 point: $k^{'} (3) = 3$
1 point: tangent line $y = 3 + 3 (x - 3)$ , through $(3, 3)$

Solution

$k (3) = g (f (3)) = g (3) = 3$ .
By the chain rule, $k^{'} (x) = g^{'} (f (x)) f^{'} (x)$ , so $k^{'} (3) = g^{'} (3) f^{'} (3) = (1) (3) = 3$ .
Point-slope form through $(3, 3)$ : $y = 3 + 3 (x - 3)$ .

(c) 2 points Quotient rule · Topic 2.8

1 point: quotient rule:
$r^{'} (x) = \frac{f (x) g^{'} (x) - g (x) f^{'} (x)}{(f (x))^{2}}$
1 point: answer $- \frac{2}{3}$

Solution

By the quotient rule, $r^{'} (x) = \frac{f (x) g^{'} (x) - g (x) f^{'} (x)}{(f (x))^{2}}$ .
From the table at $x = 3$ :
$r^{'} (3) =$ $\frac{(3) (1) - (3) (3)}{3^{2}} =$ $- \frac{2}{3}$ .

(d) 2 points Limit read as a derivative · Topic 2.1

1 point: dividing the numerator and the denominator by $x - 4$ :
$\lim_{x \to 4} \frac{\frac{f (x) - 4}{x - 4}}{\frac{g (x) - 4}{x - 4}}$ , a quotient of difference quotients with limits $f^{'} (4) = 1$ and $g^{'} (4) = 2$
Also acceptable: $\lim_{x \to 4} (f (x) - 4) = 0$ and $\lim_{x \to 4} (g (x) - 4) = 0$ (since $f (4) = 4$ , $g (4) = 4$ ), then L'Hospital's Rule, $\lim_{x \to 4} \frac{f^{'} (x)}{g^{'} (x)}$ .
1 point: answer $\frac{f^{'} (4)}{g^{'} (4)} = \frac{1}{2}$

Solution

From the table, $f (4) = 4$ and $g (4) = 4$ , so the numerator $f (x) - f (4)$ and the denominator $g (x) - g (4)$ both have limit $0$ : the form $\frac{0}{0}$ .
By L'Hospital's Rule (or by dividing both by $x - 4$ , which makes each a difference quotient), the limit is $\frac{f^{'} (4)}{g^{'} (4)}$ , with $f^{'} (4) = 1$ and $g^{'} (4) = 2$ .
The limit is $\frac{1}{2}$ .

Question 2An implicitly defined curve · no calculator · 9 points

(a) 2 points Implicit differentiation · Topic 3.2

1 point: implicit differentiation
1 point: verifies $\frac{d y}{d x} = \frac{2 x - y}{x + 2 y}$

Solution

Differentiate both sides with respect to $x$ (the chain rule on each $y$ term, the product rule on each product of $x$ and $y$ ):
$2 x - y - x \frac{d y}{d x} - 2 y \frac{d y}{d x} = 0$ .
Collect the $\frac{d y}{d x}$ terms:
$(- x - 2 y) \frac{d y}{d x} = - 2 x + y$ , so $\frac{d y}{d x} = \frac{2 x - y}{x + 2 y}$ .

(b) 2 points Tangent line or tangent-line approximation · Topic 4.6

1 point: tangent line $y = - 1 + (x + 3)$ (slope $1$ )
1 point: approximation $- 1.2$

Solution

Substitute $x = - 3$ and $y = - 1$ into $\frac{d y}{d x} = \frac{2 x - y}{x + 2 y}$ : the slope is $1$ , and the tangent line is $y = - 1 + (x + 3)$ .
Evaluate the tangent line at $x = - 3.2$ :
$y \approx - 1 + (- 3.2 + 3) = - 1.2$ .

(c) 2 points Horizontal, vertical or given-slope tangent · Topic 5.12

1 point: sets $x + 2 y = 0$ and substitutes into the equation of the curve: $5 y^{2} = 5,$ that is, $y^{2} = 1$
1 point: answer $(- 2, 1)$ and $(2, - 1)$ , with the justification that the numerator $2 x - y$ is not $0$ there: $- 5 \neq 0$ at $(- 2, 1)$ and $5 \neq 0$ at $(2, - 1)$

Solution

The tangent line is vertical where the denominator of $\frac{d y}{d x}$ is $0$ and the numerator is not: $x + 2 y = 0$ , that is, $x = - 2 y$ .
Substitute into the equation of the curve: $5 y^{2} = 5,$ that is, $y^{2} = 1$ , so $y = 1$ or $y = - 1$ .
The points are $(- 2, 1)$ and $(2, - 1)$ ; there $2 x - y \neq 0$ , so the tangent line is vertical.

(d) 3 points Second derivative at a point · Topic 3.6

1 point: implicit differentiation of $\frac{d y}{d x} = \frac{2 x - y}{x + 2 y}$ (quotient rule, a factor $\frac{d y}{d x}$ on each $y$ term)
1 point: substitutes $x = - 3$ , $y = - 1$ and $\frac{d y}{d x} = 1$
1 point: answer $\frac{2}{5}$

Solution

By the quotient rule, $\frac{d^{2} y}{d x^{2}} = \frac{D \frac{d N}{d x} - N \frac{d D}{d x}}{D^{2}}$ with $N = 2 x - y$ and $D = x + 2 y$ ; each $y$ term contributes a factor $\frac{d y}{d x}$ when differentiated.
At $(- 3, - 1)$ with $\frac{d y}{d x} = 1$ : $N = - 5$ , $D = - 5$ , $\frac{d N}{d x} = 1$ , $\frac{d D}{d x} = 3$ .
$\frac{d^{2} y}{d x^{2}} =$ $\frac{(- 5) (1) - (- 5) (3)}{{(- 5)}^{2}} =$ $\frac{2}{5}$ .

Question 3Composites with a function given by its graph · no calculator · 9 points

(a) 2 points Chain rule · Topic 3.1

1 point: $h^{'} (2) = f^{'} (g (2)) g^{'} (2)$ , with $g (2) = 3$
1 point: answer $12$

Solution

By the chain rule, $h^{'} (x) = f^{'} (g (x)) g^{'} (x)$ , where $g^{'} (x) = 2 x + 2$ .
$g (2) = 3$ and $g^{'} (2) = 6$ ; $f^{'} (3) = 2$ , the slope of the segment through $x = 3$ .
$h^{'} (2) = (2) (6) = 12$ .

(b) 2 points Average rate of change · Topic 2.1

1 point: $k (0) = g (f (0)) = g (0) = - 5$ and $k (4) = g (f (4)) = g (2) = 3$
1 point: $\frac{k (4) - k (0)}{4} = \frac{3 - (- 5)}{4} = 2$

Solution

From the graph, $f (0) = 0$ and $f (4) = 2$ :
$k (0) = g (f (0)) = g (0) = - 5$ and $k (4) = g (f (4)) = g (2) = 3$ .
$\frac{k (4) - k (0)}{4} = \frac{3 - (- 5)}{4} = 2$ .

(c) 2 points Differentiability at a point · Topic 2.4

1 point: the one-sided derivatives of $k$ at $x = 4$ :
$g^{'} (f (4)) \cdot (2) = 12$ from the left and $g^{'} (f (4)) \cdot (- \frac{1}{2}) = - 3$ from the right
1 point: no, because the one-sided derivatives are not equal

Solution

$k^{'} (x) = g^{'} (f (x)) f^{'} (x)$ wherever $f$ is differentiable, with $g^{'} (x) = 2 x + 2$ .
At the corner $x = 4$ , $f (4) = 2$ and $g^{'} (2) = 6$ ; the one-sided slopes of $f$ are $2$ and $- \frac{1}{2}$ , so the one-sided derivatives of $k$ are $12$ and $- 3$ .
$12 \neq - 3$ , so $k^{'} (4)$ does not exist.

(d) 3 points Horizontal, vertical or given-slope tangent · Topic 3.1

1 point: $k^{'} (x) =$ $g^{'} (f (x)) f^{'} (x) =$ $(2 f (x) +$ $2) f^{'} (x)$
1 point: $f^{'} (x) \neq 0$ wherever it exists, so $k^{'} (x) = 0$ only where $f (x) = - 1$
1 point: $x = 1$ and $x = \frac{5}{2}$

Solution

By the chain rule, $k^{'} (x) = g^{'} (f (x)) f^{'} (x)$ with $g^{'} (x) = 2 x + 2$ :
$k^{'} (x) = (2 f (x) + 2) f^{'} (x)$ .
$f^{'} (x)$ is the slope of a segment, never $0$ , so $k^{'} (x) = 0$ exactly where $f (x) = - 1$ (not at a corner, where $f (x) \neq - 1$ ).
On each segment that crosses height $- 1$ , solve for $x$ : on $0 \leq x \leq 2$ , $f (x) = - x$ , and $- x = - 1$ gives $x = 1$ ; on $2 \leq x \leq 4$ , $f (x) = - 2 + 2 (x - 2)$ , and $- 2 + 2 (x - 2) = - 1$ gives $x = \frac{5}{2}$ . On the other segment, $f (x)$ stays above $- 1$ .

Question 4The chain rule on a given function · no calculator · 9 points

(a) 2 points Chain rule · Topic 3.1

1 point: $f^{'} (x) = 6 x {(x^{2} + 1)}^{2}$
1 point: $f^{'} (x) > 0$ for every $x$ inside the domain, so $f$ is increasing and therefore one-to-one
Also acceptable: $f^{'} (x) \geq 0$ on the domain with $f^{'} (x) = 0$ only at an endpoint

Solution

Chain rule: the derivative of the outside function, evaluated at the inside function, times the derivative of the inside function, $2 x$ : $f^{'} (x) = 6 x {(x^{2} + 1)}^{2}$ .
For $x > 0$ , $6 x > 0$ and ${(x^{2} + 1)}^{2} > 0$ , so $f^{'} (x) > 0$ there: $f$ is increasing, so no two inputs give the same output, and $f$ has an inverse.

(b) 2 points Derivative of an inverse · Topic 3.3

1 point: $f (1) = 8$ , so $g (8) = 1$ and $g^{'} (8) = \frac{1}{f^{'} (1)}$
1 point: answer:
$\frac{1}{24}$

Solution

Find the input of $f$ that gives $8$ : $f (1) = 8$ , with $x = 1$ in the domain $x \geq 0$ , so $g (8) = 1$ .
Differentiate $f (g (x)) = x$ by the chain rule: $f^{'} (g (x)) g^{'} (x) = 1$ , so $g^{'} (x) = \frac{1}{f^{'} (g (x))}$ .
$g^{'} (8) = \frac{1}{f^{'} (1)} = \frac{1}{24} = \frac{1}{24}$ .

(c) 3 points Constants for differentiability · Topic 2.4

1 point: continuity at $x = 1$ : $k + m = 8$
1 point: equal one-sided derivatives: $2 k = 24$
1 point: answer $k = 12$ , $m = - 4$

Solution

Continuity at $x = 1$ : the left piece at $x = 1$ equals $f (1) = 8$ , so $k + m = 8$ .
Equal one-sided derivatives: the derivative of the left piece is $2 k x$ , and $f^{'} (1) = 24$ , so $2 k = 24$ .
Solve: $k = 12$ and $m = - 4$ .

(d) 2 points Product rule · Topic 2.8

1 point: $h^{'} (x) = g (x) + x g^{'} (x)$
1 point: answer:
$h^{'} (8) = \frac{4}{3}$ , using $g (8) = 1$ and $g^{'} (8) = \frac{1}{24}$ from part (b)

Solution

Product rule:
$h^{'} (x) = 1 \cdot g (x) + x g^{'} (x)$ .
At $x = 8$ : $g (8) = 1$ and $g^{'} (8) = \frac{1}{24}$ (part (b)), so $h^{'} (8) = 1 + (8) (\frac{1}{24}) = \frac{4}{3}$ .

What does the AP Calculus AB free response ask on Unit 3?

Not much on its own: Unit 3 is about 4% of the free-response points since 2012.

Composite functions from a table: about 33% of Unit 3 free-response points; 1 on this set.
An implicitly defined curve: about 27% of Unit 3 free-response points; 1 on this set.
Composites with a function given by its graph: about 18% of Unit 3 free-response points; 1 on this set.
The chain rule on a given function: about 12% of Unit 3 free-response points; 1 on this set.

This set has 4 questions in that mix, all without a calculator (about 91% of Unit 3 free-response points are in the no-calculator section), each scored out of 9 with a scoring guide.

Counted from the official AP Calculus AB free-response questions, 2012–2026. No official question is reproduced here.

AP Calculus AB Unit 3 free-response settings by share of exam points

Free-response settings on this set and their share of official AP Calculus AB free-response points, 2012–2026
Setting	Questions on this set	Share of Unit 3 free-response points	Share of all free-response points	Last on the exam
An implicitly defined curveNo calculator	1 of 4	27%	–	2025
Composite functions and derivative rules from a tableNo calculator	1 of 4	33%	–	2026
The chain rule on a given functionNo calculator	1 of 4	12%	–	2017
Composites with a function given by its graphNo calculator	1 of 4	18%	–	2014

A part counts toward a setting when its primary skill is in Unit 3, in the same section, and the setting asks that skill (in the part’s own setting where there is one), on the main exams 2012–2026, by its points; a part that fits two settings is shared between them.

Questions about this free-response set

Unit 3, Differentiation: Composite, Implicit, and Inverse Functions, has 6 topics in the College Board's course description: the chain rule; implicit differentiation; the derivative of an inverse function; the derivatives of the inverse trig functions; choosing a procedure for a derivative; and higher-order derivatives. The course description weights the unit at 5–10% of the multiple-choice section.

Each question is worth 9 points, spread over its three or four parts. Points are earned for specific pieces of work: a setup, an antiderivative, a value with its units, a conclusion with its reason. The scoring guide lists every point and what earns it, then a worked solution for each part. On the exam the six free-response questions are half of the score.

Only where the exam allows one. On this set, all four are in Part B, without a calculator, as almost every Unit 3 free-response part on the exam has been.

We tagged every free-response part from 2012 to 2026 by the skill it tests and the setting of its question. The Unit 3 settings the exam leans on most get the most questions here, each in its own context with its own parts.

No. Every question here is original, written by Sigma Prep in the style of the exam: the settings, functions, numbers and wording are our own, and no official question or scoring guideline is copied or reworded. Every value and statement in the scoring guide was checked by an independent computer-algebra check before the set was published.

Yes. Each version has its own: open it under the questions, or download the PDF, which prints the questions with room to write and then the scoring guide on its own pages. There is also a PDF without the scoring guide to hand out.

AP is a registered trademark of the College Board, which was not involved in making these questions and does not endorse them. All questions and scoring guides are original to Sigma Prep. Frozen 2026-10-24.

AP Calculus AB Unit 3 FRQ Practice: Chain Rule, Implicit, Inverse

Part B: No calculator is allowed.

Scoring guide · Version A

Question 1Composite functions and derivative rules from a table · no calculator · 9 points

Question 2An implicitly defined curve · no calculator · 9 points

Question 3Composites with a function given by its graph · no calculator · 9 points

Question 4The chain rule on a given function · no calculator · 9 points

What does the AP Calculus AB free response ask on Unit 3?

AP Calculus AB Unit 3 free-response settings by share of exam points

More AP Calculus AB free response

Questions about this free-response set

What is on AP Calculus AB Unit 3?+

How is an AP Calculus AB free-response question scored?+

Can students use a calculator?+

How were the questions chosen?+

Are these real AP exam questions?+

Is there a scoring guide?+