Lesson 7: Basic Derivative Rules

(a) Opening Question

In Lesson 6 you found derivatives the hard way — through the limit definition:

f'(x) = lim_{h→0} [f(x+h) − f(x)] / h

Use it once more, by hand, for f(x) = x³.

Expand (x+h)³ = x³ + 3x²h + 3xh² + h³, so the difference quotient is

[(x³ + 3x²h + 3xh² + h³) − x³] / h = 3x² + 3xh + h²

Let h → 0 and the last two terms vanish, leaving f'(x) = 3x².

Now look at the pattern across what you have already proven this way:

What would you bet d/dx[x⁴] is, without touching a single limit? Write down your guess and a one-line reason. By the end of this lesson you will be differentiating these in seconds — and trusting it.

(b) Core Concepts

The limit definition is the foundation, but nobody differentiates a polynomial from scratch every time. The pattern you spotted in the opening is a theorem.

The Power Rule

Power Rule. For any real number n,

`

d/dx[xⁿ] = n·xⁿ⁻¹

`

Multiply by the old exponent, then subtract 1 from the exponent.

This holds for every real exponent — positive integers, negative integers, and fractions alike. That is the rule's superpower, and it is exactly where students leave points on the table.

Positive integers are the easy case:

d/dx[x⁷] = 7x⁶

Negative exponents work identically — keep the sign on the exponent:

d/dx[x⁻³] = −3x⁻⁴

The new exponent is −3 − 1 = −4, not −2. Subtracting 1 always moves the exponent down, even when it is already negative.

Fractional exponents work too:

d/dx[x^(2/3)] = (2/3)·x^(2/3 − 1) = (2/3)·x^(−1/3)

Rewrite First — the habit that makes everything work

The power rule needs the function in the form xⁿ. Radicals and reciprocals are disguised powers. Your first move, every time, is to undisguise them:

You cannot apply the power rule to √x written as a radical — there is no exponent to grab. Rewrite, then differentiate. For example:

d/dx[√x] = d/dx[x^(1/2)] = (1/2)x^(−1/2) = 1/(2√x)

d/dx[1/x²] = d/dx[x⁻²] = −2x⁻³ = −2/x³

Notice the answers are usually rewritten back into radical/fraction form at the end. Powers are the working form; radicals are the presentation form.

The Constant Rule

Constant Rule. For any constant c,

`

d/dx[c] = 0

`

A constant function is a horizontal line. Its slope is zero everywhere, so its derivative is 0 — not c. The derivative of 7 is 0, not 7. (You can see this from the power rule too: 7 = 7x⁰, and the power rule would multiply by the exponent 0.)

The Constant Multiple Rule

Constant Multiple Rule. For any constant c,

`

d/dx[c·f(x)] = c·f'(x)

`

A constant coefficient comes along for the ride. Differentiate the function, keep the coefficient:

d/dx[5x⁴] = 5·(4x³) = 20x³

The classic error is to drop the coefficient or to differentiate it. The 5 is not a separate term — it multiplies, so it stays.

The Sum and Difference Rule

Sum/Difference Rule.

`

d/dx[f(x) ± g(x)] = f'(x) ± g'(x)

`

You may differentiate term by term. This is what makes polynomials painless — hit each term with the power rule and the constant-multiple rule, keep the signs:

d/dx[3x⁴ − 5x² + 2x − 9]
   = 12x³ − 10x + 2 − 0
   = 12x³ − 10x + 2

The constant −9 differentiates to 0 and simply disappears.

Worked Example: A Tangent Line (full habit on display)

Find the equation of the tangent line to f(x) = x² at x = 3.

A tangent line needs two things: a point and a slope.

• Point: f(3) = 3² = 9, so the point is (3, 9).
• Slope: f'(x) = 2x, so the slope at x = 3 is f'(3) = 2(3) = 6.
• Line (point-slope): y − 9 = 6(x − 3), i.e. y = 6x − 9.

The derivative is the slope of the tangent line — that is the whole point of Unit 2. A normal line is the line perpendicular to the tangent at the same point; its slope is the negative reciprocal. Here the normal slope would be −1/6, giving y − 9 = −(1/6)(x − 3).

Where is the tangent horizontal?

A horizontal tangent has slope 0. So the tangent is horizontal exactly where

f'(x) = 0

Solve that equation. These x-values are where the graph levels off — they will become the candidates for maxima and minima in Unit 5. For now, just find them.

(c) Worked Examples

Example 1 — Polynomial derivative (NO CALC)

Problem. Differentiate f(x) = 3x⁵ − 7x³ + 2x − 9.

Strategy. Sum/difference rule term by term; power rule + constant multiple on each term; constant goes to 0.

Solution.

f'(x) = 3(5x⁴) − 7(3x²) + 2(1) − 0
      = 15x⁴ − 21x² + 2

Self-check (differentiate-check). Each exponent dropped by 1 (5→4, 3→2, 1→0); each coefficient is old-coefficient × old-exponent (3·5=15, 7·3=21, 2·1=2); the constant vanished. ✓

Example 2 — Rewrite, then differentiate (NO CALC)

Problem. Differentiate g(x) = √x + 4/x².

Strategy. Neither term is in xⁿ form yet. Rewrite both as powers first, then apply the power rule.

Solution. Rewrite:

g(x) = x^(1/2) + 4x⁻²

Differentiate term by term:

g'(x) = (1/2)x^(−1/2) + 4(−2)x⁻³
      = (1/2)x^(−1/2) − 8x⁻³

Rewrite back into radical/fraction form for the final answer:

g'(x) = 1/(2√x) − 8/x³

Self-check. For 4x⁻²: new exponent −2 − 1 = −3 ✓, coefficient 4·(−2) = −8 ✓. The most common slip here is writing the new exponent as −1 instead of −3 — subtracting 1 from −2 goes down.

Example 3 — Tangent line (NO CALC)

Problem. Find the equation of the line tangent to f(x) = x³ − 4x at x = 2.

Strategy. Get the point from f, the slope from f', then point-slope.

Solution.

• Point: f(2) = 2³ − 4(2) = 8 − 8 = 0 → (2, 0).
• Derivative: f'(x) = 3x² − 4.
• Slope: f'(2) = 3(4) − 4 = 12 − 4 = 8.
• Tangent line: y − 0 = 8(x − 2), i.e.

y = 8x − 16

Justification language (AP-style). "Because the derivative gives the slope of the tangent line, the slope at x = 2 is f'(2) = 8. Using the point (2, 0), the tangent line is y = 8(x − 2)."

Example 4 — Where is the tangent horizontal? (NO CALC)

Problem. For f(x) = x³ − 12x + 5, find all x-values where the graph has a horizontal tangent line.

Strategy. Horizontal tangent ⟺ f'(x) = 0. Differentiate, set equal to zero, solve.

Solution.

f'(x) = 3x² − 12
3x² − 12 = 0
x² = 4
x = ±2

The graph has horizontal tangents at x = −2 and x = 2.

Self-check. f'(−2) = 3(4) − 12 = 0 ✓ and f'(2) = 3(4) − 12 = 0 ✓.

[GRAPH: f(x) = x³ − 12x + 5 on [−4, 4] × [−15, 25]
- Horizontal tangent (slope 0) at x = −2, point (−2, 21)
- Horizontal tangent (slope 0) at x = 2, point (2, −11)
- Local max at (−2, 21), local min at (2, −11)
- Increasing on (−∞, −2) ∪ (2, ∞), decreasing on (−2, 2)]

(d) Common Mistakes

Applying the power rule before rewriting.

What students do: leave √x or 1/x³ as-is and guess a derivative.

Why it's wrong: there is no visible exponent to "bring down," so the rule literally doesn't apply.

Fix: rewrite first — √x = x^(1/2), 1/x³ = x⁻³ — then differentiate.

Mishandling negative and fractional exponents.

What students do: write d/dx[x⁻²] = −2x⁻¹ or d/dx[x^(1/2)] = (1/2)x^(1/2).

Why it's wrong: they forgot to subtract 1 from the exponent. −2 − 1 = −3; 1/2 − 1 = −1/2.

Fix: always compute the new exponent as n − 1 explicitly, even (especially) when n is negative or fractional.

Dropping or differentiating the constant multiple.

What students do: d/dx[5x⁴] = 20 or = x³.

Why it's wrong: a coefficient is a multiplier. It stays attached and is not itself differentiated.

Fix: d/dx[5x⁴] = 5·4x³ = 20x³. Keep the coefficient, differentiate only the power.

Thinking d/dx[c] = c.

What students do: d/dx[9] = 9.

Why it's wrong: a constant function is flat; its slope is 0 everywhere.

Fix: d/dx[9] = 0. Any standalone constant term disappears.

Confusing f(a) with f'(a) in tangent-line problems.

What students do: use f'(a) as the point's y-coordinate, or f(a) as the slope.

Why it's wrong: the point comes from f, the slope comes from f'. They are different functions.

Fix: write both, labeled: point (a, f(a)), slope f'(a).

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## (e) Practice Problems

(Short answer, NO CALC) Differentiate h(x) = ∛x + 2/x². Show the rewrite step.

(Short answer, NO CALC) A particle moves along a line with position s(x) = x³ − 6x² + 9x (in meters, x in seconds). Find s'(x), then find all times when the velocity is zero. (Velocity is the derivative of position.)

(Justification, NO CALC) A student claims that because f(x) = √x + 3 and g(x) = √x − 100 differ by a constant, they have the same derivative. Justify whether the student is correct, referencing the appropriate rule.

(Interpretation, NO CALC) For f(x) = x³ − 12x + 5 (Example 4), you found horizontal tangents at x = ±2. Explain the meaning of f'(2) = 0 in terms of the graph of f at the point x = 2.

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## (f) AP Exam Focus — Free Response

> FRQ (9 points total). No calculator.

>

> A particle moves along a horizontal line. For 0 ≤ t ≤ 5, its position (in meters) at time t seconds is given by

> `

> s(t) = t³ − 6t² + 9t + 2

> `

>

> (a) (2 points) Find the velocity v(t) of the particle. Find the velocity at t = 4.

>

> (b) (3 points) Find all times t in [0, 5] at which the particle is at rest (velocity zero), and find the particle's position at each of those times.

>

> (c) (2 points) Find an equation of the line tangent to the graph of s(t) at t = 4.

>

> (d) (2 points) The acceleration is a(t) = v'(t). Find a(2), and interpret the meaning of a(2) in the context of the particle's motion, including units.

### Model Solution

(a) Velocity is the derivative of position:

v(t) = s'(t) = 3t² − 12t + 9

At t = 4:

v(4) = 3(16) − 12(4) + 9 = 48 − 48 + 9 = 9

So v(4) = 9 meters per second.

(b) The particle is at rest when v(t) = 0:

3t² − 12t + 9 = 0
3(t² − 4t + 3) = 0
3(t − 1)(t − 3) = 0
t = 1  or  t = 3

Both are in [0, 5]. Positions:

s(1) = 1 − 6 + 9 + 2 = 6
s(3) = 27 − 54 + 27 + 2 = 2

The particle is at rest at t = 1 (position 6 m) and at t = 3 (position 2 m).

(c) Tangent line to s(t) at t = 4 needs a point and a slope.

• Point: s(4) = 64 − 96 + 36 + 2 = 6 → (4, 6).
• Slope: s'(4) = v(4) = 9 (from part (a)).
• Tangent line:

s − 6 = 9(t − 4)   ⟹   s = 9t − 30

(d) Acceleration:

a(t) = v'(t) = 6t − 12
a(2) = 6(2) − 12 = 0

Interpretation: At t = 2 seconds, the acceleration is 0 meters per second per second, meaning the particle's velocity is momentarily neither increasing nor decreasing — the velocity is, instantaneously, not changing at t = 2.

Scoring Commentary — where students lose points

• (a) 1 point for the correct derivative 3t² − 12t + 9; 1 point for v(4) = 9. Lost most often by arithmetic at t = 4 or by writing s(4) instead of s'(4).
• (b) 1 point for setting v(t) = 0; 1 point for both solutions t = 1, 3 (a single root earns no point — both required); 1 point for the two positions 6 and 2. Lost most often by solving the quadratic incorrectly or stopping at the times and never finding positions.
• (c) 1 point for the slope 9 (must come from s', not s); 1 point for a correct tangent equation through (4, 6). Lost most often by using s(4) as the slope or forgetting the point.
• (d) 1 point for a(2) = 0; 1 point for an interpretation that is in context and states units. Lost most often by giving a bare number with no sentence, or omitting units (m/s²). A vague "the speed isn't changing" without referencing velocity/instant can be marked down.

Note on justification: On the real exam, "at rest" answers must show v(t) = 0 was solved, not merely asserted. Always exhibit the equation you set to zero.

h'(x) = (1/3)x^(−2/3) + 2(−2)x⁻³ = (1/3)x^(−2/3) − 4x⁻³