Why does the derivative of x come with a restriction?

x is defined only for -1 x 1, and at x = 1 the bottom of 11 - x2 is 0. So the derivative exists only for -1 < x < 1.

Unit 3 · Topic 3.5 · about 30 minutes · 7 parts · 10 scored checks, 3 of them multiple choice, plus 2 from earlier lessons

AP Calculus AB 3.5: Selecting Procedures for Calculating Derivatives

Q: Is it wrong to use the quotient rule when a rewrite would be shorter?

No. Any correct rule gives the same derivative; the rewrite just has fewer steps. Two correct answers can look different before they are simplified, so if you want to compare them, evaluate both at the same input.

Q: How do I spot a composition?

Ask whether some function's input is something other than x alone. x2 + 1 and ( x) are compositions; x2 x is a product of two functions of x, not a composition.

Q: When do I need implicit differentiation?

When x and y are related by an equation you cannot easily solve for y, such as x3 + y3 = 6xy. Some equations can be solved for y first, but differentiating implicitly avoids that work.

Q: Do I have to simplify a derivative before I use it?

Only as far as the question needs. To find a value at one input, substitute into the unsimplified derivative; every extra line of algebra is another place to lose a sign.

Built by Tanner Crow from College Board's course framework · Every answer key independently checked

Unit 3: 5 to 10% of the multiple-choice section

Key vocabulary

Term	What it means	The exam trap
Composite function	A function that uses one function's output as another's input, $f (g (x))$ , where $g$ is the inside function and $f$ the outside function.	A composition can hide in shorthand notation; rewrite with full parentheses to see which function acts last.
Chain rule	$\frac{d}{d x} f (g (x)) = f^{'} (g (x)) g^{'} (x)$ : the derivative of the outside function, evaluated at the inside, times the derivative of the inside.	Keep the inside function unchanged inside $f^{'}$ ; the factor $g^{'} (x)$ is the only new piece.
Product rule	$\frac{d}{d x} (u (x) v (x)) = u^{'} (x) v (x) + u (x) v^{'} (x)$ for differentiable functions $u$ and $v$ .	Write the two-term structure before substituting any values, so neither term goes missing.
Implicit differentiation	Differentiating both sides of an equation in $x$ and $y$ with respect to $x$ , treating $y$ as a differentiable function of $x$ .	Solve for $\frac{d y}{d x}$ only after every term on both sides has been differentiated.

Graphs of y = sin(x^2) and y = (sin x)^2 for x from 0 to 2.2, built from the same two pieces in opposite orders, with dashed tangent lines at x = 1 whose slopes are 1.081 and 0.909.

Part 1 Think first How fast is the medicine level changing?

Make a guess before the lesson starts. Even a wrong guess helps the right answer stick.

Predict first

After a dose, the concentration of a medicine in a patient's blood is $C (t) = \frac{10 t}{t^{2} + 4}$ milligrams per liter, $t$ hours after the dose. One hour after the dose, how fast is the concentration changing?

In short. Selecting a procedure for calculating a derivative means reading how a function is built and choosing the rule, or the rewrite, that matches the last operation in it.

The last operation names the first rule: term by term for a sum, the product or quotient rule, or the chain rule for a composition.
Rewriting with powers, identities or a split fraction can replace a long rule with a short one, and every correct route gives the same derivative.
When no earlier rule fits directly, as with an implicit equation, an inverse function or an inverse trigonometric function, the chain rule still does the work.

Part 2 Learn it Read the function, then choose the rule

5a. The last operation names the first rule

Each derivative rule answers one question: how was this function built? Imagine evaluating the function at a number. The operation you would do last tells you which rule to start with.

A sum, difference or constant multiple: differentiate term by term.
A product $u (x) v (x)$ : the product rule, $u^{'} (x) v (x) + u (x) v^{'} (x)$ .
A quotient $\frac{u (x)}{v (x)}$ : the quotient rule, $\frac{v (x) u^{'} (x) - u (x) v^{'} (x)}{(v (x))^{2}}$ .
A function of a function, $f (g (x))$ : the chain rule, $f^{'} (g (x)) g^{'} (x)$ .

Compositions are the ones that hide. $\sin^{2} x$ means $(\sin x)^{2}$ , so squaring comes last and $\sin x$ is inside: the derivative is $2 \sin x \cos x$ . In $e^{\sqrt{x}}$ the exponent is a function of $x$ , so the derivative is $e^{\sqrt{x}} \cdot \frac{1}{2 \sqrt{x}}$ , not $e^{\sqrt{x}}$ . In $\cos (4 x + 1)$ the input is $4 x + 1$ , which brings a factor of 4.

Most functions need more than one rule, in layers. For $x^{2} \sin (5 x)$ the last operation is multiplication, so the product rule comes first, and the chain rule shows up inside it when you differentiate $\sin (5 x)$ :

\frac{d}{d x} (x^{2} \sin (5 x)) = 2 x \sin (5 x) + x^{2} \cdot 5 \cos (5 x) .

Quick check · Sort it

No calculator. Sort each function: is it a composition, one function inside another, so that the chain rule is needed? Tap a card, then tap its bin.

A composition: chain rule needed

Not a composition

5b. Rewrite when it saves work

The way a function is written is not always the easiest way to differentiate it. A little algebra first can turn a long rule into a short one.

Roots and reciprocals as powers. $\frac{6}{x^{3}}$ is $6 x^{- 3}$ , so its derivative is $- 18 x^{- 4}$ . A constant on top never needs the quotient rule.

A fraction over a single term splits. For $x \neq 0$ , $\frac{x^{3} - 5 x}{x^{2}} = x - 5 x^{- 1}$ , and its derivative is $1 + 5 x^{- 2}$ . The function and its derivative are still not defined at $x = 0$ ; rewriting does not change the domain.

Trigonometric functions through identities. $\tan x = \frac{\sin x}{\cos x}$ and $\sec x = (\cos x)^{- 1}$ hand the work to the quotient rule, or to the power and chain rules, using derivatives you already know.

Every correct route gives the same derivative. Rewriting only shortens the algebra, and a shorter route has fewer places to drop a sign or a factor. What no rewrite allows is differentiating the top and the bottom of a fraction on their own: the derivative of $\frac{u (x)}{v (x)}$ is not $\frac{u^{'} (x)}{v^{'} (x)}$ .

Quick check · Calculate

No calculator. Let $f (x) = \frac{x^{2} + 4}{\sqrt{x}}$ for $x > 0$ . Find $f^{'} (4)$ . Enter your answer as a decimal.

Your answer

5c. When no basic rule fits directly

Some derivatives start from an equation or an inverse rather than a formula built from the basic rules. The chain rule still does the work; what changes is the setup.

An equation in $x$ and $y$ . Differentiate both sides with respect to $x$ , treating $y$ as a function of $x$ . For $\sin y + y = x^{3}$ , the derivative of $\sin y$ is $\cos y \frac{d y}{d x}$ , so $(\cos y + 1) \frac{d y}{d x} = 3 x^{2}$ and $\frac{d y}{d x} = \frac{3 x^{2}}{\cos y + 1}$ wherever $\cos y \neq - 1$ . Every term with $y$ in it picks up $\frac{d y}{d x}$ , and a term like $x y$ is a product.

An inverse function. If $g$ is the inverse of $f$ and $f (a) = b$ , then $g^{'} (b) = \frac{1}{f^{'} (a)}$ , provided $f^{'} (a) \neq 0$ . Find the input $a$ first; $f^{'} (b)$ is the wrong slope.

An inverse trigonometric function. $\frac{d}{d x} \arcsin x = \frac{1}{\sqrt{1 - x^{2}}}$ for $- 1 < x < 1$ , and $\frac{d}{d x} \arctan x = \frac{1}{1 + x^{2}}$ . With an inside function, the chain rule multiplies by its derivative: $\frac{d}{d x} \arctan (5 x) = \frac{1}{1 + (5 x)^{2}} \cdot 5 = \frac{5}{1 + 25 x^{2}}$ .

Quick check · Multiple choice

No calculator. The differentiable function $y$ of $x$ satisfies $x^{2} + x y + y^{2} = 7$ . Which expression is $\frac{d y}{d x}$ , at points where $x + 2 y \neq 0$ ?

Common mistakes

The mistake	Why it is wrong	Do this instead
Differentiating a product or quotient piece by piece, as in $(u v)^{'} = u^{'} v^{'}$ or ${(\frac{u}{v})}^{'} = \frac{u^{'}}{v^{'}}$ .	Neither is a rule. The product rule has two terms, and the quotient rule is a difference over the square of the bottom.	Name the last operation, write that rule's structure, and only then fill in each piece.
Dropping the derivative of the inside function, as in writing $e^{x^{2}}$ as the derivative of $e^{x^{2}}$ .	The chain rule multiplies the outside derivative, still evaluated at the inside, by the derivative of the inside.	For every composition, say out loud what is inside, and multiply by its derivative before you simplify.
Treating $y$ as a constant when differentiating an equation in $x$ and $y$ .	$y$ depends on $x$ , so $\frac{d}{d x} (y^{3}) = 3 y^{2} \frac{d y}{d x}$ , not $3 y^{2}$ .	Give every term that contains $y$ its $\frac{d y}{d x}$ factor, and use the product rule on terms such as $x^{2} y$ .

Part 3 Watch it done Choosing the rules, one layer at a time

Open the steps in order. The last one is yours to finish.

Let $h (x) = x^{3} e^{1 - x^{2}}$ . Find $h^{'} (1)$ .

Predict first

Before the steps open: which rule should you write down first for $h (x) = x^{3} e^{1 - x^{2}}$ ?

Step 1 · Name the last operation
To evaluate $h (x)$ you would find $x^{3}$ , find $e^{1 - x^{2}}$ , and then multiply them. Multiplication is last, so the product rule comes first, with $u (x) = x^{3}$ and $v (x) = e^{1 - x^{2}}$ . The factor $v$ is itself a composition: $e$ raised to the inside function $1 - x^{2}$ .
Step 2 · Write the rule's structure
$h^{'} (x) = u^{'} (x) v (x) + u (x) v^{'} (x)$ , with $u^{'} (x) = 3 x^{2}$ .
Step 3 · Differentiate the composition
The outside function is the exponential, which is its own derivative, and the inside function $1 - x^{2}$ has derivative $- 2 x$ . So $v^{'} (x) = e^{1 - x^{2}} \cdot (- 2 x)$ , and
$h^{'} (x) = 3 x^{2} e^{1 - x^{2}} + x^{3} e^{1 - x^{2}} (- 2 x) .$
Why this step?
Why does $v^{'} (x)$ carry the factor $- 2 x$ , when the derivative of $e^{x}$ is just $e^{x}$ ?
Because the exponent is $1 - x^{2}$ , not $x$ . The chain rule multiplies the outside derivative, $e^{1 - x^{2}}$ , by the derivative of the inside, $- 2 x$ . Without that factor you would have the rate of change with respect to $1 - x^{2}$ , not with respect to $x$ .
Your step · Calculate

No calculator. Finish it. For $h (x) = x^{3} e^{1 - x^{2}}$ , what is $h^{'} (1)$ ?

Your answer

Part 4 Try it Two derivatives with no formula to start from

Try it · Decide and justify

No calculator. Let $f (x) = x^{3} + 2 x + 1$ . Since $f^{'} (x) = 3 x^{2} + 2 > 0$ , $f$ is increasing and has an inverse function $g$ . A classmate says $g^{'} (4)$ cannot be found without first finding a formula for $g (x)$ . Is the classmate right?

Which reason supports your answer?

Try it · Multiple choice

No calculator. For $- 1 < x < 1$ , what is $\frac{d}{d x} \arcsin (x^{2})$ ?

Part 5 Explain it Write it out, then check it

On the exam, free response is written by hand, so work this one on paper. Nothing you type here leaves the page. Then open the model answer and mark yourself against each point.

Free response is 50% of the exam score. This one is shaped like Question 6, from free-response Part B (calculator not permitted).

Free-response practice · Question 6 style

No calculator. A drone's height above the ground is modeled by $H (t) = 8 t e^{- t / 4}$ meters, where $t$ is measured in seconds and $0 \leq t \leq 12$ .

(a) Find $H^{'} (t)$ . Show the work that leads to your answer, and name each derivative rule you use.

(b) Find $H^{'} (2)$ . Indicate units of measure.

(c) A classmate writes $H (t) = \frac{8 t}{e^{t / 4}}$ and uses the quotient rule. Verify that the classmate's method gives the same expression for $H^{'} (t)$ as your answer to part (a).

Your answer stays on this page. It is never sent or saved.

Model answer

(a) $H (t)$ is a product of $u (t) = 8 t$ and $v (t) = e^{- t / 4}$ , so the product rule comes first. $v$ is a composition, so by the chain rule $v^{'} (t) = e^{- t / 4} \cdot (- \frac{1}{4})$ . Then

H^{'} (t) = u^{'} (t) v (t) + u (t) v^{'} (t) = 8 e^{- t / 4} + 8 t (- \frac{1}{4} e^{- t / 4}) = (8 - 2 t) e^{- t / 4} .

(b) $H^{'} (2) = (8 - 4) e^{- 2 / 4} = 4 e^{- 1 / 2}$ meters per second.

(c) The top is $8 t$ , with derivative 8, and the bottom is $e^{t / 4}$ , with derivative $\frac{1}{4} e^{t / 4}$ by the chain rule. The quotient rule gives

H^{'} (t) = \frac{e^{t / 4} \cdot 8 - 8 t \cdot \frac{1}{4} e^{t / 4}}{{(e^{t / 4})}^{2}} = \frac{(8 - 2 t) e^{t / 4}}{e^{t / 2}} = (8 - 2 t) e^{- t / 4},

which is the expression from part (a).

Score your own answer. This is a self-check, not a grade. Tick each point your answer makes.

Names the product rule and writes its structure, $u^{'} (t) v (t) + u (t) v^{'} (t)$ , with $u (t) = 8 t$ and $v (t) = e^{- t / 4}$ .
Uses the chain rule to get $v^{'} (t) = - \frac{1}{4} e^{- t / 4}$ , and arrives at $H^{'} (t) = (8 - 2 t) e^{- t / 4}$ or an equivalent expression.
Gets $H^{'} (2) = 4 e^{- 1 / 2}$ , with units of meters per second.
Applies the quotient rule with the bottom squared, ${(e^{t / 4})}^{2} = e^{t / 2}$ , and simplifies to $(8 - 2 t) e^{- t / 4}$ .

Part 6 Check yourself Two functions known only by a table

Both case questions use this table. Write the rule's structure first, then look up each value it asks for.

The functions $f$ and $g$ are differentiable for all real numbers. Selected values of $f$ , $f^{'}$ , $g$ and $g^{'}$ are given in the table.

Selected values of f, g and their derivatives
$x$	$f (x)$	$f^{'} (x)$	$g (x)$	$g^{'} (x)$
1	3	-2	2	4
2	1	5	3	-2
3	4	3	1	2
4	2	3	4	5

Case question 1 · Calculate

No calculator. Let $h (x) = f (g (x))$ . Find $h^{'} (2)$ .

Your answer

Case question 2 · Multiple choice

No calculator. Let $k (x) = \frac{f (x)}{g (x)}$ . What is $k^{'} (4)$ ?

The rest stand on their own and do not use the case.

Check yourself · Spot the error

No calculator. A student finds $\frac{d}{d x} \sec x$ by rewriting it as a power. Tap the first line with a mistake.

Check yourself · Match

No calculator. Each function below is built from $e$ , but they call for different rules. Tap a function, then tap its derivative.

Part 7 Keep it Earlier ideas this lesson stands on

Every rule here works only where the derivative exists, and that rests on limits. Both come back now.

From 2.4 Connecting Differentiability and Continuity: Determining When Derivatives Do and Do Not Exist · Decide and justify

No calculator. Let $f (x) = | x - 1 | + 2 x$ . A classmate says $f^{'} (1)$ exists, because $f$ is continuous at $x = 1$ . Is the classmate right?

Which reason supports your answer?

From 1.10 Exploring Types of Discontinuities · Calculate

No calculator. Let $h (x) = \frac{x^{2} - 3 x - 10}{x - 5}$ for $x \neq 5$ , and $h (5) = 4$ . The function $h$ has a removable discontinuity at $x = 5$ . What is $\lim_{x \to 5} h (x)$ ?

Your answer

In three lines

Name the last operation first: it says whether the product, quotient or chain rule leads.
Rewrite roots, reciprocals and trigonometric functions when it shortens the work; every correct route gives the same derivative.
Implicit equations, inverse functions and inverse trigonometric functions all run on the chain rule.

Exam strategy for this topic

Section I Part A, Multiple Choice (calculator not permitted): 29 questions in 62 minutes.
Section I Part B, Multiple Choice (graphing calculator required): 13 questions in 38 minutes.
Section II Part A, Free Response (graphing calculator required): 2 questions in 30 minutes.
Section II Part B, Free Response (calculator not permitted): 4 questions in 60 minutes.

On a no-calculator derivative, write the rule's structure before you substitute: $f^{'} (g (2)) g^{'} (2)$ for a composition, or $u^{'} (2) v (2) + u (2) v^{'} (2)$ for a product. The structure shows which table value goes where, and a missing factor is easy to see.

Questions students ask

Is it wrong to use the quotient rule when a rewrite would be shorter?

No. Any correct rule gives the same derivative; the rewrite just has fewer steps. Two correct answers can look different before they are simplified, so if you want to compare them, evaluate both at the same input.

How do I spot a composition?

Ask whether some function's input is something other than $x$ alone. $\sqrt{x^{2} + 1}$ and $\ln (\cos x)$ are compositions; $x^{2} \cos x$ is a product of two functions of $x$ , not a composition.

When do I need implicit differentiation?

When $x$ and $y$ are related by an equation you cannot easily solve for $y$ , such as $x^{3} + y^{3} = 6 x y$ . Some equations can be solved for $y$ first, but differentiating implicitly avoids that work.

Do I have to simplify a derivative before I use it?

Only as far as the question needs. To find a value at one input, substitute into the unsimplified derivative; every extra line of algebra is another place to lose a sign.

Why does the derivative of

\arcsin x

come with a restriction?

$\arcsin x$ is defined only for $- 1 \leq x \leq 1$ , and at $x = \pm 1$ the bottom of $\frac{1}{\sqrt{1 - x^{2}}}$ is 0. So the derivative exists only for $- 1 < x < 1$ .

Course alignment, for teachers

AP Calculus AB topic 3.5, Unit 3: Differentiation: Composite, Implicit, and Inverse Functions.

Skill 1.C: Identify an appropriate mathematical rule or procedure based on the classification of a given expression (e.g., Use the chain rule to find the derivative of a composite function).

This lesson maps the topic's 0 learning objectives, since the course framework gives it no essential knowledge statements, to the part that teaches each one: 0 are checked by a scored question. The map is in the lesson file.