Lesson 16: Derivative Mastery Review

(a) Opening Question

No new rule today — just a warm-up that quietly uses three of them at once. Differentiate:

f(x) = x² · sin(2x)

Don't reach for a single rule. Read the structure first. The outermost operation is a product: x² times sin(2x). So the spine of the answer is the Product Rule, f'·g + f·g'. But the second factor, sin(2x), is itself a composite — a sine of 2x — so its derivative needs the chain rule: cos(2x) · 2.

Putting it together:

f'(x) = (2x)·sin(2x) + x²·[cos(2x)·2]
      = 2x·sin(2x) + 2x²·cos(2x)

If you wrote x²·cos(2x) and forgot the ·2 from the inside, you just met the single most common derivative error on the AP exam — the missing inside derivative. This lesson is about making sure that never happens again, no matter how many rules are stacked.

(b) Core Concepts

This is a review lesson. You already learned each rule in Lessons 7–15; the job now is to hold them all in one frame, choose between them fast, and stop making cross-rule mistakes. We won't re-derive anything from scratch — we'll organize.

The master derivative-rules reference table

Memorize this table cold. On a non-calculator section, recall is the whole game.

Structural rules (how to combine functions):

Library of specific functions (the "outside derivatives" the chain rule plugs into):

Two structural ideas tie the table together:

Chain rule, stated precisely. If g is differentiable at x and f is differentiable at g(x), then (f ∘ g)'(x) = f'(g(x))·g'(x). Differentiate the outside, leave the inside alone, then multiply by the derivative of the inside. Every entry in the function library above is really a chain-rule entry waiting for an inside — d/dx[sin(u)] = cos(u)·u'.

Implicit differentiation. When y is defined implicitly by an equation in x and y, differentiate both sides with respect to x, treating y as a function of x. Every time you differentiate a y, the chain rule forces a dy/dx factor. Then solve algebraically for dy/dx.

Inverse-function derivative. If f is one-to-one and differentiable with f(a) = b, then (f⁻¹)'(b) = 1/f'(a) = 1/f'(f⁻¹(b)), provided f'(a) ≠ 0. You never need a formula for f⁻¹ itself — just the matching point and f' there.

"Which rule(s) do I need?" — a decision strategy

The mistake students make under time pressure isn't not knowing a rule; it's applying it to the wrong layer. Use this top-down read every time:

1. Look at the outermost operation. Ignore the details. Is the whole thing a sum, a product, a quotient, a power, or a single composed function? That picks your spine rule.
2. Descend one layer. Whatever pieces the spine rule hands you (the f and g, or the inside g(x)), ask the same question of each. A piece that is itself composite needs the chain rule; a piece that's a product needs the product rule; and so on.
3. Stop at library functions. When you hit a sin, eˣ, ln, arctan, or a power of x, you're at the bottom — read its derivative straight off the table.
4. Reassemble bottom-up, multiplying inside-derivatives as you climb back out.

Worked diagnosis, no computation: for y = [ln(x²+1)] / [sec(3x)], the outermost operation is a quotient → spine is the quotient rule. Its top ln(x²+1) is a chain (ln of something). Its bottom sec(3x) is also a chain (sec of 3x). So the full job is quotient on the outside, chain inside each piece — three rules, but only one is the spine.

The most error-prone combinations

Knowing the rules isn't enough; the points are lost at the seams between rules.

• Chain inside product (or product inside chain). In x²·sin(2x), the product rule is the spine, but the sin(2x) factor still needs its own ·2. Stacking rules does not let you skip the inside derivative.
• General power rule masquerading as the plain power rule. d/dx[(x²+1)⁵] is not 5(x²+1)⁴. It's 5(x²+1)⁴·(2x). The base isn't x, so the chain rule applies.
• Quotient where simplifying first is faster. (x³+x)/x = x²+1 differentiates to 2x instantly — no quotient rule, no error surface. Always glance for an algebraic simplification before committing.
• Implicit terms that are products of x and y. A term like 6xy needs the product rule (6y + 6x·y'), and the y part drags a dy/dx. Forgetting either piece is the classic implicit slip.
• Inverse-trig vs. reciprocal-trig confusion. arcsin x (an inverse function, derivative 1/√(1−x²)) is not csc x = 1/sin x (a reciprocal, derivative −csc x cot x). The notation sin⁻¹x means the inverse; (sin x)⁻¹ means the reciprocal. Read carefully.

(c) Worked Examples

Four examples, each a different corner of the toolkit. Structure: Problem → Strategy → Solution → Justification/Check.

Example 1 — Chain + product + trig [NO CALC]

Problem. Differentiate y = (3x² + 1)⁴ · cos x.

Strategy. Outermost operation: a product of (3x²+1)⁴ and cos x. Spine = product rule. The first factor (3x²+1)⁴ is a power of a function → general power rule (chain). The second factor cos x is a library function.

Solution.

Let f = (3x² + 1)⁴,   g = cos x.
f' = 4(3x² + 1)³·(6x) = 24x(3x² + 1)³     ← general power rule, inside derivative 6x
g' = −sin x

y' = f'·g + f·g'
   = 24x(3x² + 1)³·cos x + (3x² + 1)⁴·(−sin x)
   = 24x(3x² + 1)³·cos x − (3x² + 1)⁴·sin x

You can factor (3x²+1)³ out front, but on the AP exam the un-factored line above earns full credit.

Check. Both terms keep one factor undifferentiated, as the product rule demands, and the inside derivative 6x is present in the first term — the seam between product and chain is intact. ✓

Example 2 — Implicit differentiation [NO CALC]

Problem. The curve x³ + y³ = 6xy (the folium of Descartes) passes through (3, 3). Find dy/dx, and evaluate it at (3, 3).

Strategy. y is not isolated, so differentiate both sides with respect to x, treating y as a function of x. The 6xy term is a product of x and y → product rule. Every y differentiated emits a dy/dx.

Solution.

d/dx[x³ + y³] = d/dx[6xy]
3x² + 3y²·(dy/dx) = 6y + 6x·(dy/dx)      ← right side: product rule on 6xy

Collect the dy/dx terms on one side:

3y²·(dy/dx) − 6x·(dy/dx) = 6y − 3x²
(3y² − 6x)·(dy/dx) = 6y − 3x²
dy/dx = (6y − 3x²)/(3y² − 6x) = (2y − x²)/(y² − 2x)

At (3, 3):

dy/dx = (2·3 − 3²)/(3² − 2·3) = (6 − 9)/(9 − 6) = −3/3 = −1

Justification/Check. The tangent at (3, 3) has slope −1 — consistent with the curve's symmetry about the line y = x at that point (reflecting a curve across y = x negates reciprocal slopes; a slope of −1 is its own image). ✓

Example 3 — Exponential/log chain [NO CALC]

Problem. Differentiate y = ln(x² + 1) · e^(3x), and find y'(0).

Strategy. Outermost: a product. First factor ln(x²+1) is a chain (ln of something). Second factor e^(3x) is a chain (eˣ with inside 3x).

Solution.

Let f = ln(x² + 1),   g = e^(3x).
f' = 1/(x² + 1) · (2x) = 2x/(x² + 1)       ← chain: derivative of ln(u) is u'/u
g' = e^(3x)·3 = 3e^(3x)                      ← chain: inside derivative 3

y' = f'·g + f·g'
   = [2x/(x² + 1)]·e^(3x) + ln(x² + 1)·3e^(3x)

At x = 0: x² + 1 = 1, so f'(0) = 0/1 = 0 and ln(0²+1) = ln 1 = 0. Both terms vanish:

y'(0) = (0)·e⁰ + (0)·3e⁰ = 0

Check. ln(u) differentiates to u'/u, not 1/u — the 2x on top is the inside derivative. At x = 0 both the log value and the log's derivative-numerator are zero, so y'(0) = 0. ✓

Example 4 — Inverse-function derivative [NO CALC]

Problem. Let f(x) = x³ + 2x + 1. The function f is one-to-one (its derivative is always positive). Find (f⁻¹)'(4).

Strategy. Don't invert f — you can't solve x³ + 2x + 1 = y for x cleanly anyway. Use the inverse-function formula (f⁻¹)'(b) = 1/f'(a) where f(a) = b. Here b = 4, so first find the a with f(a) = 4.

Solution.

Find a:  f(a) = 4  ⇒  a³ + 2a + 1 = 4  ⇒  a³ + 2a − 3 = 0  ⇒  a = 1  (since 1 + 2 − 3 = 0).
So f(1) = 4, meaning f⁻¹(4) = 1.

f'(x) = 3x² + 2,   so   f'(1) = 3 + 2 = 5.

(f⁻¹)'(4) = 1/f'(1) = 1/5.

Justification/Check. Why f' > 0 matters: f'(x) = 3x² + 2 ≥ 2 > 0 for all x, so f is strictly increasing and therefore one-to-one — the inverse exists and is differentiable, and f'(1) ≠ 0 guarantees the formula is valid. The graphs of f and f⁻¹ are reflections across y = x, and reflection swaps (1, 4) with (4, 1) while taking the slope to its reciprocal — exactly 1/5. ✓

(d) Common Mistakes

The missing inside derivative (the #1 AP derivative error). Writing d/dx[sin(2x)] = cos(2x) and stopping. ✗ The chain rule demands the inside derivative: cos(2x)·2 = 2cos(2x). This shows up most often when a chain is buried inside another rule (a product or quotient factor), so you're already busy and forget the ·u'. Fix: after every composite, ask out loud "times the derivative of the inside?"

Product/quotient order and sign. Quotient rule written as (f·g' − f'·g)/g² (top differentiated first) flips the sign of the whole answer. ✗ The order is f'·g − f·g² on top — "Lo·D-Hi first." And the product rule is a sum (f'g + fg'), the quotient a difference — don't cross them. Fix: anchor the quotient on f'·g leading and always square the denominator.

Forgetting dy/dx in implicit differentiation. Differentiating y³ as 3y² instead of 3y²·(dy/dx). ✗ Because y is a function of x, the chain rule applies to every y. Dropping the dy/dx makes the equation unsolvable for the slope. Fix: every time your pencil leaves a y, immediately attach ·(dy/dx).

General power rule treated as the plain power rule. d/dx[(x²+1)⁵] = 5(x²+1)⁴. ✗ The base is x²+1, not x, so the chain rule applies: 5(x²+1)⁴·(2x). Fix: the plain power rule is only for xⁿ — a literal x in the base. Anything else triggers the general power rule.

Inverse-trig vs. reciprocal-trig. Treating arcsin x (i.e., sin⁻¹x) as (sin x)⁻¹ = csc x. ✗ arcsin x is the inverse function (derivative 1/√(1−x²)); csc x is the reciprocal (derivative −csc x cot x). Fix: read sin⁻¹x as "the angle whose sine is x," never as 1/sin x.

(e) Practice Problems

Attempt before checking section (g). [NO CALC] unless marked [CALC].

(short answer) Differentiate y = cos³x (that is, (cos x)³). Show the rule(s) you use.

(justify) A student computes d/dx[(x² + 4)³] = 3(x² + 4)² and stops. Identify the error by name, give the correct derivative, and in one sentence justify why the extra factor is required.

(justify) For the curve x³ + y³ = 6xy, a classmate differentiates the right-hand side as d/dx[6xy] = 6·(dy/dx). State whether this is correct and justify your answer using the appropriate rule.

(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. Both f and g are differentiable for all real numbers.

>

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

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

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

> | 2 | 3 | −1 | 1 | 6 |

> | 3 | 1 | 5 | 2 | −3 |

>

> (a) (2 points) Let P(x) = f(x)·g(x). Find P'(2).

>

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

>

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

>

> (d) (2 points) Let H(x) = f(g(x)). The function g satisfies g(2) = 1. Determine whether H is increasing or decreasing at x = 2. Justify your answer.

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)
      = (−1)(1) + (3)(6)
      = −1 + 18 = 17

P'(2) = 17.

(b) Q'(1) — 2 points

By the Chain Rule, Q'(x) = f'(g(x))·g'(x). At x = 1, g(1) = 3, so:

Q'(1) = f'(g(1))·g'(1) = f'(3)·g'(1) = (5)(2) = 10

Q'(1) = 10.

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

First the point. R(3) = f(3)/g(3) = 1/2.

Then the slope, by the Quotient Rule R'(x) = [f'(x)·g(x) − f(x)·g'(x)] / [g(x)]²:

        f'(3)·g(3) − f(3)·g'(3)      (5)(2) − (1)(−3)      10 + 3      13
R'(3) = ─────────────────────────  = ───────────────────  = ─────────  = ───
              [g(3)]²                       (2)²               4         4

Tangent line through (3, ½) with slope 13/4:

y = ½ + (13/4)(x − 3)

Approximate R(3.1):

R(3.1) ≈ ½ + (13/4)(3.1 − 3) = ½ + (13/4)(0.1) = ½ + 13/40 = 20/40 + 13/40 = 33/40 = 0.825

Tangent line: y = ½ + (13/4)(x − 3); R(3.1) ≈ 33/40 = 0.825.

(d) Is H increasing or decreasing at x = 2? — 2 points

By the Chain Rule, H'(x) = f'(g(x))·g'(x). At x = 2, g(2) = 1, so:

H'(2) = f'(g(2))·g'(2) = f'(1)·g'(2) = (4)(6) = 24

Since H'(2) = 24 > 0, the function H is increasing at x = 2. Justification: H is increasing at x = 2 because H'(2) > 0; by the chain rule the rate H'(2) = f'(g(2))·g'(2) is the product of two positive quantities, f'(1) = 4 > 0 and g'(2) = 6 > 0, so the composite rises there.

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 17. Common loss: the "product of derivatives" error f'(2)·g'(2) = (−1)(6) = −6 earns 0/2.

Part (b) — 2 pts: 1 pt for the chain-rule setup f'(g(1))·g'(1) with the correct inner evaluation g(1) = 3; 1 pt for 10. Common loss: using f'(1)·g'(1) = (4)(2) = 8 — failing to feed g(1) = 3 into f' — loses both points; the chain rule requires f' evaluated at the inside output, not at x.

Part (c) — 3 pts: 1 pt for R(3) = ½ (the point); 1 pt for R'(3) = 13/4 (the slope, requires correct Quotient-Rule order and squared denominator); 1 pt for a correct tangent equation and the approximation 0.825. Common loss: a sign slip on f(3)·g'(3) = (1)(−3) = −3 turning 10 − (−3) into 10 − 3 = 7/4; or forgetting to square g(3) (dividing by 2 instead of 4). A bare 0.825 with no line shown earns only the points the written work supports — write the line.

Part (d) — 2 pts: 1 pt for H'(2) = 24 (correct chain rule, correct inner g(2) = 1); 1 pt for the justification that H is increasing because H'(2) > 0. Common loss: getting 24 but writing only "it's positive" without the word increasing or without tying the sign of H' to the behavior of H loses the reasoning point. Evaluating f' at x = 2 instead of at g(2) = 1 gives f'(2)·g'(2) = (−1)(6) = −6 and the wrong conclusion.

Notation reminder: On table FRQs, AP readers reward the written rule with values substituted — f'(2)·g(2) + f(2)·g'(2) — over a lone final number. Always show the substitution line.