How much work do I need to show on a calculus FRQ?

Show enough that a reader can follow your method. For derivatives, write the rule setup (for example u'v+uv' with the pieces identified) before simplifying, and substitute values before reducing when you only need f'(a). Answers alone can earn answer points but forfeit method points if the work is missing or wrong.

When should I use the product rule versus the quotient rule?

Look at the outermost operation. If two x-dependent expressions are multiplied, use the product rule. If one is divided by another, use the quotient rule. If the denominator is just a constant, factor it out and use linearity instead — it is faster and less error-prone.

Do I have to simplify my derivative on the exam?

Usually no. A correct unsimplified derivative earns full credit unless the problem specifically asks you to evaluate or simplify. That said, simplifying often makes it easier to substitute a value accurately in a later part, so weigh the time trade-off.

How do I justify that a function is differentiable at a point?

State two things explicitly: that the function is continuous there (values match), and that the left- and right-hand derivatives are equal. Then write the conclusion in words. Both conditions and the concluding sentence are needed to earn the justification points.

AP-CALCBC-2-FRQ

U2 FRQ Practice

Master AP Calculus BC Unit 2 free-response questions: combine the power, product, quotient, and trig/exponential derivative rules across multi-part FRQs.

What you'll do in this lesson

A voice-first session with the Crimsora tutor on U2 FRQ Practice, then targeted practice and FRQs — with the tutor adapting to where you get stuck.

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What this lesson covers

Unit 2 FRQs rarely test a single rule in isolation. Instead, they hand you a function built from powers, products, quotients, trig, exponentials, and logs, then ask you to differentiate, evaluate, write a tangent line, and justify a claim about continuity or differentiability. The challenge is not knowing one rule — it's choosing the right rule quickly and organizing your work so a reader can follow every step.

This guide shows you how to attack Unit 2 free-response problems: how to read the parts, which rule each cue signals, how to present derivatives cleanly, and how to earn justification points. You already know the individual rules from U2.1 through U2.9; here you learn to deploy them together under exam conditions.

How Unit 2 FRQs Are Structured

A typical Unit 2 free-response question gives you one or two functions and then breaks into three or four parts that escalate in difficulty. Part (a) usually asks for a derivative or its value at a point. Part (b) often applies that derivative — an equation of a tangent line, an instantaneous rate of change, or a slope comparison. Part (c) may ask you to justify continuity or differentiability, or to interpret a derivative in context with units.

Because the parts build on each other, an early arithmetic slip can cost points downstream. Graders award method points and answer points separately, so showing the rule you applied protects you even when a number is wrong. A useful habit is to write

f'(x)=

before you simplify, so the reader sees your setup.

Part cue	What it wants	Rule to reach for
"Find $f'(x)$ "	General derivative	Power, product, quotient
"Find the slope at $x=a$ "	Evaluate $f'(a)$	Same, then substitute
"Write the tangent line"	Point-slope form	$y-f(a)=f'(a)(x-a)$
"Is $f$ differentiable at..."	Justification	Continuity + matching one-sided derivatives

Read all parts before starting; the later parts often tell you which form of the derivative will be most useful.

Choosing the Right Rule Fast

The biggest time-saver on Unit 2 FRQs is recognizing structure instantly. Look at the outermost operation of the expression, because that tells you the primary rule.

If the function is a sum of terms, apply linearity and differentiate term by term with the power rule. If it is one expression multiplied by another where both depend on

x

, use the product rule

\frac{d}{dx}[uv]=u'v+uv'

. If it is a ratio, use the quotient rule

\frac{d}{dx}\left[\frac{u}{v}\right]=\frac{u'v-uv'}{v^2}

. Memorize the standard derivatives:

\frac{d}{dx}\sin x=\cos x

\frac{d}{dx}\cos x=-\sin x

\frac{d}{dx}e^x=e^x

\frac{d}{dx}\ln x=\frac{1}{x}

\frac{d}{dx}\tan x=\sec^2 x

, and

\frac{d}{dx}\sec x=\sec x\tan x

.

A common misconception is forcing the quotient rule when the denominator is a constant. If

f(x)=\frac{x^2+3}{5}

, rewrite as

\frac{1}{5}(x^2+3)

and use linearity — much cleaner. Another trap: the product rule sign is always plus, while the quotient rule numerator is always minus, and order matters. Write

u

v

u'

v'

in a small table before assembling if the expression is dense.

Presenting Work That Earns Every Point

Free-response scoring rewards clear communication. When you differentiate, label the rule if the algebra is nontrivial, then substitute before simplifying so a grader can trace your logic. For a tangent line, state both the point and the slope explicitly, then write point-slope form; you do not have to simplify to slope-intercept unless asked.

Units and interpretation matter when the function models something physical. If

C(t)

gives concentration in milligrams per liter and

t

is in hours, then

C'(3)

carries units of milligrams per liter per hour, and a sentence of interpretation should name that rate and its sign. Missing units or a vague interpretation commonly forfeits an easy point.

For justification parts, connect ideas explicitly. To argue differentiability at a point where a piecewise function changes rule, first show the two pieces agree in value (continuity, from U2.4), then show the one-sided derivatives are equal. State the theorem-level reasoning in words: "Since

f

is continuous at

x=2

and the left and right derivatives both equal 4,

f

is differentiable there." Do not leave the reader to infer your conclusion — write it.

Common Errors and Time Management

Under time pressure students lose points to predictable mistakes. The quotient rule numerator order is reversed most often; remember it mirrors the product rule but with a subtraction, so the derivative of the top comes first. The derivative of

\cos x

is negative

\sin x

— dropping that minus sign is a frequent, costly slip. When differentiating

\ln x

, the answer is

\frac{1}{x}

, not

\frac{1}{x}\ln x

or anything more complex.

Mistake	Fix
Quotient rule terms swapped	Top-derivative first: $u'v-uv'$
Forgetting minus on $\frac{d}{dx}\cos x$	$-\sin x$ , always
Using quotient rule for constant denominator	Factor out the constant
Simplifying before substituting a value	Substitute first if only $f'(a)$ is needed
No units or interpretation sentence	Name the rate, its units, and its sign

Budget roughly a couple of minutes per part. If part (a) stalls, still attempt part (b) using a placeholder derivative — partial method credit is real. Always re-read what each part actually asks; answering "find the slope" with a full tangent line is fine, but the reverse leaves points on the table.

Key terms

Product Rule.: For a product of two functions, $\frac{d}{dx}[uv]=u'v+uv'$ . Used whenever both factors depend on $x$ .
Quotient Rule.: For a ratio, $\frac{d}{dx}\left[\frac{u}{v}\right]=\frac{u'v-uv'}{v^2}$ ; the derivative of the numerator comes first.
Linearity of the Derivative.: The derivative of a sum is the sum of derivatives, and constants factor out: $\frac{d}{dx}[af+bg]=af'+bg'$ .
Tangent line.: The line through $(a,f(a))$ with slope $f'(a)$ : $y-f(a)=f'(a)(x-a)$ .
Justification.: A written argument linking evidence (values, one-sided limits) to a conclusion, required to earn reasoning points on FRQs.
Differentiability.: A function is differentiable at a point if it is continuous there and its left- and right-hand derivatives are equal.
Rate of change with units.: The value $f'(a)$ interpreted in context, carrying units of the output quantity per unit of the input quantity.

Worked example

Let

f(x)=\dfrac{e^x}{x^2+1}

. (a) Find

f'(x)

. (b) Find the slope of the tangent line to

f

x=0

. (c) Write an equation of the tangent line at

x=0

Part (a): The outermost operation is a quotient, so use the quotient rule with

u=e^x

and

v=x^2+1

. Then

u'=e^x

and

v'=2x

f'(x)=\frac{u'v-uv'}{v^2}=\frac{e^x(x^2+1)-e^x(2x)}{(x^2+1)^2}.

Factor

e^x

in the numerator to simplify:

f'(x)=\dfrac{e^x(x^2-2x+1)}{(x^2+1)^2}=\dfrac{e^x(x-1)^2}{(x^2+1)^2}

.

Part (b): Substitute

x=0

. The numerator is

e^0(0-1)^2=1\cdot 1=1

and the denominator is

(0+1)^2=1

, so

f'(0)=1

. The slope is

1

.

Part (c): Find the point.

f(0)=\dfrac{e^0}{0+1}=1

, so the point is

(0,1)

. Using point-slope form with slope

1

y-1=1(x-0)

, which simplifies to

y=x+1

. Because the slope came from a correctly applied quotient rule and the point was evaluated separately, every step is traceable for full credit.

Practice questions

g(x)=x^3\sin x

, what is

g'(x)

$3x^2\cos x$
$3x^2\sin x+x^3\cos x$
$3x^2\sin x-x^3\cos x$
$x^3\cos x-3x^2\sin x$

Answer: $3x^2\sin x+x^3\cos x$

This is a product of

u=x^3

and

v=\sin x

, so use the product rule

u'v+uv'

. Here

u'=3x^2

and

v'=\cos x

, giving

3x^2\sin x+x^3\cos x

. The plus sign is required; the product rule never subtracts, which rules out the options with minus signs, and the first option forgets to differentiate the second factor.

Let

h(x)=\dfrac{2x+1}{\cos x}

. Find

h'(x)

and evaluate

h'(0)

. Show your setup.

Answer: $h'(x)=\dfrac{2\cos x+(2x+1)\sin x}{\cos^2 x}$ , and $h'(0)=2$ .

Apply the quotient rule with

u=2x+1

v=\cos x

u'=2

v'=-\sin x

. Then

h'(x)=\dfrac{2\cos x-(2x+1)(-\sin x)}{\cos^2 x}=\dfrac{2\cos x+(2x+1)\sin x}{\cos^2 x}

. Watch the double negative:

-uv'

with

v'=-\sin x

becomes

+(2x+1)\sin x

. At

x=0

\cos 0=1

and

\sin 0=0

, so the numerator is

2(1)+1(0)=2

and the denominator is

1

, giving

h'(0)=2

A tank's water temperature is modeled by

T(t)=40+15e^{-0.5t}

degrees Celsius, where

t

is in minutes. Find

T'(2)

and interpret it with units.

Answer: $T'(2)=-7.5e^{-1}\approx -2.76$ degrees Celsius per minute; the water is cooling at about 2.76 degrees Celsius per minute at $t=2$ .

Differentiate: the constant 40 has derivative 0, and

\frac{d}{dt}[15e^{-0.5t}]=15\cdot(-0.5)e^{-0.5t}=-7.5e^{-0.5t}

. At

t=2

T'(2)=-7.5e^{-1}\approx-2.76

. The negative sign means temperature is decreasing, and the units — degrees Celsius per minute — must be stated to earn the interpretation point. Naming the rate, its sign, and its units is exactly what FRQ graders look for.

FAQ

How much work do I need to show on a calculus FRQ?: Show enough that a reader can follow your method. For derivatives, write the rule setup (for example $u'v+uv'$ with the pieces identified) before simplifying, and substitute values before reducing when you only need $f'(a)$ . Answers alone can earn answer points but forfeit method points if the work is missing or wrong.
When should I use the product rule versus the quotient rule?: Look at the outermost operation. If two $x$ -dependent expressions are multiplied, use the product rule. If one is divided by another, use the quotient rule. If the denominator is just a constant, factor it out and use linearity instead — it is faster and less error-prone.
Do I have to simplify my derivative on the exam?: Usually no. A correct unsimplified derivative earns full credit unless the problem specifically asks you to evaluate or simplify. That said, simplifying often makes it easier to substitute a value accurately in a later part, so weigh the time trade-off.
How do I justify that a function is differentiable at a point?: State two things explicitly: that the function is continuous there (values match), and that the left- and right-hand derivatives are equal. Then write the conclusion in words. Both conditions and the concluding sentence are needed to earn the justification points.

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