Lesson 11: Unit 2 Foundations Review

(a) Opening Question

Before you read anything, try this mixed warm-up. It quietly touches four different Unit 2 rules. No calculator. Differentiate each, naming the rule you used:

(i)    f(x) = 3x⁴ − 2x² + 7
(ii)   g(x) = x² sin x
(iii)  h(x) = (2x + 1)/(x − 3)
(iv)   p(x) = 5 sec x

Then one conceptual question: a function w has a sharp corner at x = 2 but no break there. Is w continuous at x = 2? Is w differentiable at x = 2?

Take three minutes. Jot answers before peeking. (i) is power + constant + sum rules. (ii) is the Product Rule with a trig factor. (iii) is a genuine Quotient Rule. (iv) is a trig derivative with a constant multiple. The corner question is the heart of Lesson 10. Full solutions live in the Answer Key as Warm-Up — but notice already that this single warm-up spans almost the entire unit. That is exactly what the exam does.

(b) Core Concepts — The Big Picture of Unit 2

You now own a toolbox of differentiation rules. The exam will not tell you which tool to grab — recognizing the structure of the function is the skill. This section is a synthesis, not a re-teach. Keep it as your one-page map for differentiation through trig.

The "which rule do I use?" decision guide

Look at the outermost structure of the function and match it to a rule. (The chain rule for composites arrives in Lesson 12 — every function here is built from plain x, so you will not need it yet.)

Is it a sum or difference of terms? Differentiate term by term. The derivative of a sum is the sum of the derivatives, and a constant multiple comes straight out: d/dx[c·f] = c·f'. Example: 3x⁴ − 2x² + 7 → handle 3x⁴, −2x², 7 separately.

Is each term a power of x? Use the Power Rule: d/dx[xⁿ] = n·xⁿ⁻¹, valid for every real n. Rewrite roots and reciprocals as powers first: √x = x^(1/2), 1/x³ = x⁻³. A constant alone has derivative 0.

Is it two genuine factors multiplied? Use the Product Rule. ("Genuine" means you can't easily expand or simplify it away first.)

Is it one expression divided by another? Glance at the denominator. A single monomial (like x or x²) → split the fraction and use the Power Rule. A genuine binomial/expression in the denominator → Quotient Rule.

Is it a trig function? Use the memorized trig derivative. If it's a trig function multiplied by, or divided by, something else, combine with the Product or Quotient Rule.

None of the above — or you're asked "from the definition"? Use the limit definition of the derivative.

> Always ask "simplify first?" before reaching for Product or Quotient. Expanding (x²)(x³) to x⁵, or splitting (x³ + 2x)/x into x² + 2, turns a rule into a one-line Power Rule.

### The derivative-rules summary table (everything through trig)

| Structure | Rule | Formula |

|---|---|---|

| Constant | Constant Rule | d/dx[c] = 0 |

| Power of x | Power Rule | d/dx[xⁿ] = n·xⁿ⁻¹ |

| Constant multiple | Constant Multiple | d/dx[c·f] = c·f' |

| Sum / difference | Sum Rule | d/dx[f ± g] = f' ± g' |

| Product | Product Rule | d/dx[f·g] = f'·g + f·g' |

| Quotient | Quotient Rule | d/dx[f/g] = (f'·g − f·g')/g² |

| sin x | Trig | d/dx[sin x] = cos x |

| cos x | Trig | d/dx[cos x] = −sin x |

| tan x | Trig | d/dx[tan x] = sec²x |

| sec x | Trig | d/dx[sec x] = sec x · tan x |

| csc x | Trig | d/dx[csc x] = −csc x · cot x |

| cot x | Trig | d/dx[cot x] = −csc²x |

Two anchors for the trig block: the two derivatives that gain a minus sign are exactly the three "co-" functions — cos, csc, cot — and sec and csc each differentiate into a product of two trig functions. Memorize sin → cos → −sin first; the rest hang off it. (All trig derivatives assume x is in radians; they come from lim_{x→0} (sin x)/x = 1, the limit you met in Unit 1.)

### The limit definition (still fair game)

The derivative is, at root, the limit of a difference quotient — the instantaneous slope:

                f(x + h) − f(x)                       f(x) − f(a)
f'(x) = lim_{h→0} ─────────────────     f'(a) = lim_{x→a} ───────────────
                       h                                     x − a

The exam still asks you to recognize a difference quotient. If you see lim_{h→0} [(2 + h)² − 4]/h, that is f'(2) for f(x) = x² in disguise — its value is 4. Connecting a limit to "this is a derivative" is a Practice-2 skill the MCQ loves.

A derivative is a slope: tangent lines

f'(a) is the slope of the tangent line to y = f(x) at x = a. The tangent line itself is point-slope through (a, f(a)):

y = f(a) + f'(a)(x − a)

You need two numbers: the point f(a) and the slope f'(a). Near x = a, this line approximates the curve, which is why f(a + Δx) ≈ f(a) + f'(a)·Δx.

The differentiability checklist

This is the conceptual spine of Lesson 10. State it precisely:

Differentiability ⇒ continuity, but NOT the reverse. If f is differentiable at x = c, then f is continuous at x = c. The converse fails: a function can be continuous yet not differentiable.

A function fails to be differentiable at x = c in exactly these situations:

1. Discontinuity — any break (jump, hole, asymptote). No continuity ⇒ no derivative.
2. Corner — the left- and right-hand slopes both exist but disagree (e.g. |x| at 0). Continuous, not differentiable.
3. Cusp — the slopes approach +∞ on one side and −∞ on the other.
4. Vertical tangent — the slope approaches ±∞ (e.g. x^(1/3) at 0). The tangent line is vertical, so f' is undefined there.

To check differentiability at a seam of a piecewise function, verify two things in order: (1) the function is continuous there (the pieces meet), and (2) the one-sided derivatives match (the slopes agree). Both are required.

Must-know facts to have memorized

• d/dx[xⁿ] = n·xⁿ⁻¹ for all real n — including roots and reciprocals once rewritten as powers.
• Product Rule adds; Quotient Rule subtracts and the order is f'g − fg' ("Lo·D-Hi first"), all over g².
• The six trig derivatives, with the "co-" functions carrying the minus sign.
• Differentiable ⇒ continuous; continuous does not ⇒ differentiable.
• The tangent line needs both the point and the slope.

(c) Worked Examples

Example 1 — A derivative from the definition · [NO CALC]

Problem. Use the limit definition to find f'(x) for f(x) = x² − 3x.

Strategy. The phrase "from the definition" forbids the Power Rule shortcut. Build the difference quotient [f(x + h) − f(x)]/h, expand, cancel the h, then take h → 0.

Solution.

f(x + h) = (x + h)² − 3(x + h) = x² + 2xh + h² − 3x − 3h

f(x + h) − f(x) = (x² + 2xh + h² − 3x − 3h) − (x² − 3x)
                = 2xh + h² − 3h

Divide by h (valid since h ≠ 0 inside the limit), then take the limit:

f(x + h) − f(x)      2xh + h² − 3h
───────────────── = ─────────────── = 2x + h − 3
       h                   h

f'(x) = lim_{h→0} (2x + h − 3) = 2x − 3

Answer. f'(x) = 2x − 3.

Check. The Power Rule on x² − 3x gives 2x − 3 directly. ✓ The definition must agree with the shortcut — and it does.

Example 2 — Quotient Rule with a trig function · [NO CALC]

Problem. Differentiate y = (cos x)/(1 + x).

Strategy. A genuine quotient (the denominator 1 + x can't be split off a single cos x), so use the Quotient Rule with a trig derivative inside. Top f = cos x, bottom g = 1 + x.

Solution.

f = cos x       f' = −sin x
g = 1 + x       g' = 1

     f'·g − f·g'      (−sin x)(1 + x) − (cos x)(1)
y' = ───────────  =  ──────────────────────────────
        g²                      (1 + x)²

     −(1 + x) sin x − cos x
   = ────────────────────────
            (1 + x)²

Answer. y' = [−(1 + x) sin x − cos x] / (1 + x)².

Justification. The minus sign in the numerator comes from the Quotient Rule subtraction and the f' = −sin x derivative — keep them straight by writing f' out before substituting. (sympy confirms y' = [(−x − 1) sin x − cos x]/(x + 1)², the same expression.) ✓ The denominator stays factored as (1 + x)²; never multiply it out.

Example 3 — Differentiability of a piecewise function (justification) · [NO CALC]

Problem. Let

        ⎧  x² + 1,     x ≤ 2
w(x) =  ⎨
        ⎩  3x − 1,     x > 2

Is w continuous at x = 2? Is w differentiable at x = 2? Justify each answer.

Strategy. Run the two-step differentiability check: first continuity (do the pieces meet?), then matching one-sided derivatives (do the slopes agree?).

Solution & justification.

Continuity. From the left, lim_{x→2⁻} w(x) = 2² + 1 = 5, and w(2) = 5. From the right, lim_{x→2⁺} w(x) = 3(2) − 1 = 5. Since the left limit, the right limit, and w(2) all equal 5, w is continuous at x = 2.

Differentiability. The left piece x² + 1 has derivative 2x, giving a left-hand slope of 2(2) = 4. The right piece 3x − 1 has derivative 3, giving a right-hand slope of 3. Since the one-sided derivatives 4 and 3 are not equal, w has a corner at x = 2, so w is not differentiable at x = 2.

Takeaway. Continuity is necessary but not sufficient. Here the graph is unbroken yet bends sharply — the classic "continuous but not differentiable" corner. Always check the slopes, not just the values.

Example 4 — Tangent line application · [NO CALC]

Problem. Find the equation of the line tangent to y = x sin x at x = π/2.

Strategy. Need the point f(π/2) and the slope f'(π/2). The function is a product x · sin x, so differentiate with the Product Rule, then evaluate.

Solution.

Point. f(π/2) = (π/2)·sin(π/2) = (π/2)(1) = π/2. So the point is (π/2, π/2).

Slope. With f = x (f' = 1) and g = sin x (g' = cos x):

y' = f'·g + f·g' = (1)(sin x) + (x)(cos x) = sin x + x cos x

At x = π/2: y'(π/2) = sin(π/2) + (π/2)cos(π/2) = 1 + (π/2)(0) = 1.

Tangent line through (π/2, π/2) with slope 1:

y = π/2 + 1·(x − π/2)     i.e.     y = x

Answer. The tangent line is y = x.

Check. cos(π/2) = 0 kills the x cos x term, leaving slope 1 — a clean result, and the line y = x does pass through (π/2, π/2). ✓

(d) Common Mistakes — Unit 2's Top Traps

Multiplying derivatives in a product. Writing d/dx[f·g] = f'·g'. ✗ The Product Rule keeps one factor undifferentiated in each term: f'·g + f·g'. Gut-check on x²·x³: the shortcut gives 6x³, but the truth is 5x⁴. Fix: two terms, two factors, each takes its turn.

Quotient Rule sign/order — and forgetting to square the bottom. The numerator is f'·g − f·g' ("Lo·D-Hi first"), and the denominator is always g². ✗ Reversing the numerator flips the sign; dividing by g instead of g² is just wrong. Fix: write f' and g' explicitly, anchor on "Lo·D-Hi first, all over Lo squared."

Sign errors in trig derivatives. Writing d/dx[cos x] = sin x (missing the minus) or d/dx[sec x] = sec²x (that's tan's derivative). ✗ Fix: the "co-" functions — cos, csc, cot — carry the minus sign; tan → sec²x, sec → sec x tan x, cot → −csc²x.

Assuming continuous means differentiable. Seeing an unbroken graph and concluding f' exists everywhere. ✗ Corners, cusps, and vertical tangents are continuous but not differentiable. Fix: differentiability is the stronger condition — check that the one-sided slopes actually agree.

Tangent line with only the slope. Computing f'(a) and stopping, or writing y = f'(a)·x. ✗ A tangent line needs the point f(a) too, and must pass through it: y = f(a) + f'(a)(x − a). Fix: always compute both f(a) and f'(a) before writing the line.

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## (e) Practice Problems

Mix of [CALC] and [NO CALC], spanning all of Unit 2. Items 13–15 are short-answer / free-response with required justification. Full solutions in the Answer Key.

Short-answer / free-response (justify your work)

(justify) Let

        ⎧  ax + 1,      x ≤ 1
f(x) =  ⎨
        ⎩  x² + b,      x > 1

Find values of a and b so that f is differentiable at x = 1. Show the two conditions you use and justify why both are required. (3 pts)

(justify) A student claims that because y = |x − 2| + 5 is continuous everywhere, it must be differentiable everywhere. Identify the one point where the claim fails and explain, in precise language, why y is not differentiable there. (2 pts)

Let g(x) = x² − 4x.

(a) Find g'(x). (1 pt)

(b) Find all x where the tangent line to g is horizontal, and justify. (1 pt)

(c) Write the equation of the tangent line to g at x = 1. (1 pt)

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## (f) AP Exam Focus — Free Response

> Free Response Question (No Calculator) — 9 points total

>

> The table below gives values of the differentiable functions f and g and their derivatives f' and g' at selected values of x.

>

> | x | f(x) | f'(x) | g(x) | g'(x) |

> |---|---|---|---|---|

> | 1 | 2 | 3 | 4 | −1 |

> | 2 | 5 | −2 | 1 | 3 |

> | 3 | 6 | 4 | 2 | 1 |

> | 4 | 3 | 1 | −2 | 2 |

>

> (a) (2 points) Let P(x) = f(x)·g(x). Find P'(2). Show the setup that leads to your answer.

>

> (b) (2 points) Let k(x) = f(x)/g(x). Find k'(1). Show the setup that leads to your answer.

>

> (c) (3 points) Write an equation for the line tangent to the graph of P(x) = f(x)·g(x) at x = 3, and use it to approximate P(3.1).

>

> (d) (2 points) A different function m is continuous on [0, 4] and is defined piecewise so that m(x) = f(2)·x for x ≤ 2 and m(x) = 5x − 5 for x > 2. Determine whether m is differentiable at x = 2. Justify your answer.

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### MODEL SOLUTION

(a) P'(2) — 2 points

By the Product Rule, P'(x) = f'(x)·g(x) + f(x)·g'(x). At x = 2:

P'(2) = f'(2)·g(2) + f(2)·g'(2)
      = (−2)(1) + (5)(3)
      = −2 + 15 = 13

P'(2) = 13.

(b) k'(1) — 2 points

By the Quotient Rule, k'(x) = [f'(x)·g(x) − f(x)·g'(x)] / [g(x)]². At x = 1:

        f'(1)·g(1) − f(1)·g'(1)      (3)(4) − (2)(−1)      12 + 2      14      7
k'(1) = ─────────────────────────  = ────────────────────  = ─────────  = ────  = ───
              [g(1)]²                       (4)²               16       16      8

k'(1) = 7/8.

(c) Tangent line to P at x = 3; approximate P(3.1) — 3 points

First the point: P(3) = f(3)·g(3) = (6)(2) = 12.

Then the slope, by the Product Rule at x = 3:

P'(3) = f'(3)·g(3) + f(3)·g'(3) = (4)(2) + (6)(1) = 8 + 6 = 14

Tangent line through (3, 12) with slope 14:

y = 12 + 14(x − 3)

Approximate P(3.1) using the tangent line:

P(3.1) ≈ 12 + 14(3.1 − 3) = 12 + 14(0.1) = 12 + 1.4 = 13.4

Tangent line: y = 12 + 14(x − 3); P(3.1) ≈ 13.4.

(d) Differentiability of m at x = 2 — 2 points

Here f(2) = 5, so m(x) = 5x for x ≤ 2 and m(x) = 5x − 5 for x > 2.

Continuity at x = 2: from the left, m(2) = 5(2) = 10; from the right, lim_{x→2⁺} m(x) = 5(2) − 5 = 5. Since 10 ≠ 5, m is not continuous at x = 2 — contradicting the stated setup, so first test the slopes against the continuity requirement.

Because m is not continuous at x = 2 (the left value 10 and right limit 5 disagree), m is not differentiable at x = 2. Differentiability requires continuity, and continuity fails here, so the derivative cannot exist at x = 2.

SCORING COMMENTARY — where students lose points

Part (a) — 2 pts: 1 pt for the correct Product-Rule setup f'(2)g(2) + f(2)g'(2); 1 pt for the answer 13. Common loss: using f'(2)·g'(2) = (−2)(3) = −6 (the "product of derivatives" error) earns 0/2.

Part (b) — 2 pts: 1 pt for the Quotient-Rule expression with [g(1)]² in the denominator and the terms in the right order; 1 pt for 7/8. Common loss: reversing the numerator to f·g' − f'·g flips the sign; forgetting to square g(1) (dividing by 4 instead of 16) loses the answer point.

Part (c) — 3 pts: 1 pt for P(3) = 12 (the point); 1 pt for P'(3) = 14 (the slope, which requires a correct Product Rule); 1 pt for a correct tangent equation and the approximation 13.4. Common loss: computing the slope but never writing an actual line equation, or a sign slip in P'(3). A bare 13.4 with no tangent line shown earns only the work it is supported by — show the line.

Part (d) — 2 pts: 1 pt for testing continuity at x = 2 and finding the left value 10 and right limit 5 disagree; 1 pt for the justification that because continuity fails, differentiability fails (differentiable ⇒ continuous, so no continuity ⇒ no derivative). Common loss: only comparing the two slopes (5 and 5) and concluding "differentiable" — that ignores the continuity prerequisite and earns 0/2. The slopes matching is irrelevant when the pieces don't even meet.

Notation reminder: throughout, write f'(2)·g(2) + f(2)·g'(2) in full before substituting numbers. AP readers reward the labeled setup over a lone final number, and it protects you from sign slips.

a + 1 = 1 + b   ⟹   a = b

a = 2