Choosing a Differentiation Rule — AP® Calculus AB

Part 1 · Hook

Why this matters

A mechanic does not grab a random wrench: one look at the bolt tells which tool fits. Differentiation is the same. You already own every rule you need; this topic trains the one look that tells you which rule to use first, so that long expressions stop being scary.

Part 2 · Before you start

What this builds on

Part 3 · Prerequisite check

Quick check before you start

1. What is $\frac{d}{dx}\left(x e^x\right)$ ?

$e^x + xe^x$
$e^x$
$xe^x$
$e^x \cdot 1$

Show the answer

Product rule: $1 \cdot e^x + x \cdot e^x$ .

Correct: $e^x + xe^x$ :
$e^x$ :
$xe^x$ :
$e^x \cdot 1$ :

2. What is $\frac{d}{dx}\frac{\sin x}{x}$ ?

$\frac{x\cos x - \sin x}{x^2}$
$\cos x$
$\frac{\sin x - x\cos x}{x^2}$
$\frac{\cos x}{1}$

Show the answer

Quotient rule: $\frac{(\sin x)' x - \sin x \cdot (x)'}{x^2}$ .

Correct: $\frac{x\cos x - \sin x}{x^2}$ :
$\cos x$ :
$\frac{\sin x - x\cos x}{x^2}$ :
$\frac{\cos x}{1}$ :

3. What is $\frac{d}{dx}e^{5x}$ ?

$5e^{5x}$
$e^{5x}$
$5xe^{5x - 1}$
$e^5$

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Chain rule: $e^{5x}$ times the inside derivative 5.

Correct: $5e^{5x}$ :
$e^{5x}$ :
$5xe^{5x - 1}$ :
$e^5$ :

Part 4 · See it

See it first

Decision diagram: what is the last operation? A sum or difference: differentiate term by term. A product: product rule u′v + uv′. A quotient: quotient rule or rewrite with a power. A function of a function: chain rule f′(g(x))·g′(x). Then repeat inside each piece. — Read the last operation to choose the first rule, then repeat inside each piece. LevlPrep original diagram.

Part 5 · Step by step

How it works, step by step

Ask what you would compute last to evaluate the expressionthat last operation names the first rule: sum, product, quotient or composition
Apply that ruleit leaves smaller pieces, each needing its own derivative
Repeat the question on each pieceone rule per layer, until only basic derivatives remain
Rewrite first when it removes a rule $\frac{5}{x^3} = 5x^{-3}$ needs only the power rule

Part 6 · Key ideas

Key ideas

The last operation decides the first rule: $x^2\sin(3x)$ is a product; $\sin^2(x^3)$ is a composition (a square).
Inside a product or quotient, any factor may still need the chain rule.
Equations mixing $x$ and $y$ call for implicit differentiation; an inverse calls for $\frac{1}{f'(g(x))}$ .
Rewrite before differentiating when it saves a rule, but do not simplify more than the question asks.

Part 7 · Misconception

A common mistake

The wrong idea: The derivative of a product or a quotient is the product or quotient of the derivatives.

What actually happens: $(fg)' = f'g + fg'$ and $\left(\frac{f}{g}\right)' = \frac{f'g - fg'}{g^2}$ . Multiplying or dividing the derivatives gives a different, wrong function.

Part 8 · Check yourself

Check yourself

Exam-style questions. Anything you miss goes into your review queue.

Table

Values of $f$ , $g$ and their derivatives

Selected values of the differentiable functions f and g and their derivatives
$x$	$f(x)$	$f'(x)$	$g(x)$	$g'(x)$
1	2	3	3	$-1$
2	$-1$	5	1	4
3	4	$-2$	2	6

1. Let $P(x) = f(x)g(x)$ . Find $P'(2)$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

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Product rule: $P'(2)$ $= f'(2)g(2)$ $+ f(2)g'(2)$ $= 5(1)$ $+ (-1)(4)$ $= 1$ .

Answer: 1

2. Using the table, find the derivative of the quotient $Q(x) = \frac{f(x)}{g(x)}$ at $x = 3$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Quotient rule: $Q'(3)$ $= \frac{f'(3)g(3) - f(3)g'(3)}{[g(3)]^2}$ $= \frac{(-2)(2) - (4)(6)}{4}$ $= \frac{-28}{4}$ $= -7$ .

Answer: -7

3. Let $C(x) = f(g(x))$ . What is $C'(3)$ ?

$30$
$-12$
$5$
$20$

Show the answer

The last operation is applying $f$ to $g(x)$ : chain rule. $C'(3)$ $= f'(g(3)) \cdot g'(3)$ $= f'(2) \cdot 6$ $= 5 \cdot 6$ $= 30$ .

Correct: $30$ : $f'(g(3))g'(3)$ $= f'(2) \cdot 6$ $= 30$ .
$-12$ : This is $f'(3)g'(3)$ ; read $f'$ at the inside value $g(3) = 2$ .
$5$ : This is $f'(g(3))$ without the inside derivative $g'(3) = 6$ .
$20$ : This is $f'(2)g'(2)$ : the inside derivative must be read at $x = 3$ .

4. Let $D(x) = x \cdot f(x^2)$ . Find $D'(1)$ .

No calculator. Type the exact value (a fraction, π, e or a root is fine).

Show the answer

Product rule first, chain rule inside: $D'(x)$ $= f(x^2) + x \cdot f'(x^2) \cdot 2x$ . So $D'(1)$ $= f(1)$ $+ 1 \cdot f'(1) \cdot 2$ $= 2$ $+ 6$ $= 8$ .

Answer: 8

5. Find the derivative of $y = x^2\sin(3x)$ .

Type an expression for $\frac{dy}{dx} =$ .

Show the answer

Product first: $(x^2)'\sin(3x) + x^2(\sin(3x))'$ . The second factor needs the chain rule: $(\sin(3x))' = 3\cos(3x)$ . So $2x\sin(3x) + 3x^2\cos(3x)$ .

Answer: $\frac{dy}{dx} =$ $2x\,\sin(3x) + 3x^{2}\,\cos(3x)$

6. What is $\frac{d}{dx}\sin^2\left(x^3\right)$ ?

$2\sin(x^3)\cos(x^3)$
$6x^2\sin(x^3)\cos(x^3)$
$6x^2\sin(x^3)$
$-6x^2\sin(x^3)\cos(x^3)$

Show the answer

Three layers: the square, the sine, and $x^3$ . Multiply one derivative for each: $2\sin(x^3) \cdot \cos(x^3) \cdot 3x^2$ .

$2\sin(x^3)\cos(x^3)$ : The innermost layer $x^3$ has derivative $3x^2$ , which is missing.
Correct: $6x^2\sin(x^3)\cos(x^3)$ : One factor per layer: $2\sin(x^3)$ , $\cos(x^3)$ , $3x^2$ .
$6x^2\sin(x^3)$ : The middle layer, the sine, contributes $\cos(x^3)$ , which is missing.
$-6x^2\sin(x^3)\cos(x^3)$ : The derivative of sine is cosine, with no minus sign.

Part 9 · Summary

Summary

Choose the first rule from the last operation: a sum term by term, a product by the product rule, a quotient by the quotient rule (or a rewrite), a function of a function by the chain rule. Then ask the same question of each piece, one rule per layer.

Part 10 · Up next

What comes next

Part 11 · Connections

Why this matters

What this builds on

Quick check before you start

See it first

How it works, step by step

Key ideas

A common mistake

Check yourself

Values of ff, gg and their derivatives

Summary

What comes next

Connections

Values of $f$ , $g$ and their derivatives