Difficulty: Intermediate | Prerequisites: Part 1 of these notes (definition of derivative, tangent lines, basic rules).
Once you have the basic differentiation rules down, the next step is handling functions built from combinations: products, quotients, and piecewise definitions. This half of Unit 2 also asks you to read derivative information from graphs and tables rather than formulas, which is a different skill from computing derivatives algebraically. The AP exam tests both fluently, so comfort with tables, graphs, and rules working together is essential before moving into the chain rule in Unit 3.
The product rule handles derivatives of functions multiplied together. You also need to combine multiple rules when given function values in a table. Differentiability requires continuity plus matching slopes from both sides. On graph-reading questions, horizontal tangent lines correspond to where f'(x) = 0, and sharp corners mean the derivative does not exist.
Product rule
If h(x) = f(x)·g(x), then h'(x) = f'(x)·g(x) + f(x)·g'(x). Think of it as "derivative of the first times the second, plus the first times the derivative of the second."
Differentiable
A function is differentiable at a point if its derivative exists there. This requires the function to be continuous at that point and for the left-hand and right-hand derivative limits to agree. In simple terms, the graph must be smooth (no sharp corners, no jumps, no vertical tangents) at that point.
Piecewise function
A function defined by different formulas on different intervals. At the boundary between pieces, you need to check both continuity and matching derivatives to determine differentiability.
Sharp turn (corner or cusp)
A point on a graph where the curve changes direction abruptly. The derivative does not exist at a sharp turn because the slope from the left does not equal the slope from the right.
Horizontal tangent line
A tangent line with slope zero. This occurs where f'(x) = 0, which on a graph of f' corresponds to the x-intercepts.
When two functions are multiplied, you cannot simply multiply their derivatives. The product rule states:
If h(x) = f(x)·g(x), then h'(x) = f'(x)·g(x) + f(x)·g'(x).
Example from graphs: Given graphs of f(x) and g(x), with h(x) = f(x)·g(x), find h'(2).
Read from the graphs: f'(2) = −1, g(2) = −3, f(2) = 2, g'(2) = 2.
h'(2) = f'(2)·g(2) + f(2)·g'(2) = (−1)(−3) + (2)(2) = 3 + 4 = 7.
The AP exam often provides graphs of f and g with piecewise-linear segments, so the slopes (derivatives) are found by counting rise over run on the straight portions.
When h(x) = g(x)/f(x), apply the quotient rule and substitute known values.
Example: h(x) = g(x)/f(x) with g(2) = 3, g'(2) = −1, f(2) = 5, f'(2) = −2.
h'(x) = [g'(x)·f(x) − g(x)·f'(x)] / [f(x)]²
h'(2) = [(−1)(5) − (3)(−2)] / 5² = (−5 + 6) / 25 = 1/25.
A common AP format gives a table of f(x), g(x), f'(x), and g'(x) at specific x-values. You then evaluate derivatives of combinations.
Sum/difference rule example: For 3f(x) − g(x) at x = −1:
Derivative is 3f'(−1) − g'(−1) = 3(2) − (1) = 5.
Product rule example: For 3f(x)·g(x) at x = −1:
Derivative is 3[f'(−1)·g(−1) + f(−1)·g'(−1)] = 3[(2)(−1) + (0)(1)] = 3(−2) = −6.
Quotient rule example: For f(x)/[g(x) + 2] at x = 0:
Derivative is [f'(0)·(g(0) + 2) − f(0)·g'(0)] / [g(0) + 2]²
= [(−2)(−3 + 2) − (−1)(4)] / (−3 + 2)²
= [(−2)(−1) − (−4)] / 1 = (2 + 4) / 1 = 6.
The key skill here is reading the table accurately and plugging values in the right places. Rushing is where the errors happen.
For a piecewise function to be differentiable at a boundary point, two conditions must hold:
Continuity: The left-hand limit and right-hand limit of f(x) must equal f at that point.
Matching derivatives: The derivative from the left and the derivative from the right must be equal at that point.
Example: For f(x) = { ax, x ≤ 1; bx² + x + 1, x > 1 }, find a and b so that f is differentiable at x = 1.
Step 1, continuity: Left limit = a(1) = a. Right limit = b(1) + 1 + 1 = b + 2. So a = b + 2.
Step 2, matching derivatives: f'(x) = a for x ≤ 1, and f'(x) = 2bx + 1 for x > 1. At x = 1: left derivative = a, right derivative = 2b + 1. So a = 2b + 1.
Solving the system: b + 2 = 2b + 1, which gives b = 1 and a = 3.
Always do both checks. Continuity alone is not sufficient for differentiability.
If a graph has a sharp corner (a "V" shape or an abrupt change in direction), the derivative does not exist there. The slopes approaching from the left and right are different, so no single tangent line can be drawn.
Example: If the graph of g(x) has a sharp turn at x = 3, then g'(3) does not exist. On the AP exam, this is a free mark if you recognise the visual cue.
When you are shown the graph of f'(x) and asked where f(x) has horizontal tangent lines:
Horizontal tangent means slope = 0, which means f'(x) = 0.
On the graph of f', this corresponds to where the curve crosses or touches the x-axis.
Example: If the graph of f'(x) crosses the x-axis at x = 0, x = 2, and x = 4, then f has horizontal tangent lines at x = 0, 2, and 4.
Be careful to distinguish between the f' graph crossing the axis (f'(x) = 0) and the f' graph simply being positive or negative (f increasing or decreasing). The question asks about f, and the graph shows f'.
When a table gives accumulated values over time, the average rate of change is the total change divided by the total time. The instantaneous rate at a point between two table entries is estimated by the slope between the nearest data points.
Example: Duck-crossing data over 7 hours, with 0 ducks at t = 0 and 28 at t = 7.
Average rate over [0, 7]: (28 − 0)/(7 − 0) = 4 ducks per hour.
Estimated instantaneous rate at t = 4.5: use the values at t = 4 (12 ducks) and t = 5 (21 ducks). f'(4.5) ≈ (21 − 12)/(5 − 4) = 9 ducks per hour.
The interpretation sentence matters on free response: "At t = 4.5 hours, ducks are crossing the road at a rate of approximately 9 ducks per hour."
Rule | Formula |
|---|---|
Product rule | d/dx [f·g] = f'g + fg' |
Quotient rule | d/dx [f/g] = [f'g − fg'] / g² |
Differentiability conditions | Continuity at the point AND left derivative = right derivative |
The product rule appears whenever two changing quantities are multiplied. For example, revenue = price × quantity, and if both price and quantity change over time, the rate of change of revenue uses the product rule. Differentiability matters in physics and engineering because a non-differentiable point in a position function would imply an instantaneous, physically impossible jump in velocity.
Students assume continuity guarantees differentiability. It does not. A function can be continuous at a sharp corner but still fail to have a derivative there.
When applying the product rule, a common error is writing f'(x)·g'(x) instead of f'(x)·g(x) + f(x)·g'(x). You do not simply multiply the derivatives.
On graph questions, students sometimes confuse the graph of f with the graph of f'. If you are shown f' and asked about f, you are reading zeros of f' (not peaks of f') to find horizontal tangents of f.
In table problems, students misread which row or column gives f' vs. f. Take an extra moment to check the column headers.
⚠️ Product and quotient rule problems with tables of values appear on nearly every AP exam. Practise reading a table and plugging values cleanly.
⚠️ Differentiability of piecewise functions is a classic free-response topic. You must show both the continuity check and the derivative check for full marks.
⚠️ "The derivative does not exist" is a valid and expected answer when a graph has a corner, cusp, vertical tangent, or discontinuity. Do not try to force a numerical answer.
⚠️ Applied rate-of-change problems (like the duck table) require units and an interpretation sentence on free response. Forgetting the sentence costs you a point.
⚠️ When the AP gives you the graph of f' and asks about f, read carefully. The x-intercepts of f' correspond to horizontal tangents of f, not vice versa.
True or false: If f and g are both differentiable, then (fg)' = f'·g'.
Fill in the blank: A function is differentiable at x = a only if it is _______ at x = a and the left and right derivatives _______.
True or false: A sharp corner on a graph means the derivative exists but equals zero.
Fill in the blank: On the graph of f'(x), horizontal tangent lines of f(x) correspond to points where f'(x) = _______.
True or false: To estimate f'(4.5) from a table, you can use the slope between the nearest data points on either side of 4.5.
Answers: 1. False (the product rule gives f'g + fg'). 2. Continuous; are equal. 3. False (the derivative does not exist at a sharp corner). 4. 0. 5. True.
Q: Suppose h(x) = g(x)/f(x) with g(2) = 3, g'(2) = −1, f(2) = 5, f'(2) = −2. Find h'(2).
A: h'(2) = [g'(2)·f(2) − g(2)·f'(2)] / [f(2)]² = [(−1)(5) − (3)(−2)] / 25 = (−5 + 6)/25 = 1/25.
Q: Let h(x) = f(x)·g(x). From the graphs, f'(2) = −1, g(2) = −3, f(2) = 2, g'(2) = 2. Find h'(2).
A: h'(2) = f'(2)·g(2) + f(2)·g'(2) = (−1)(−3) + (2)(2) = 3 + 4 = 7.
Q: What is g'(3) if the graph of g(x) has a sharp turn at x = 3?
A: g'(3) does not exist. The sharp turn means the left-hand and right-hand slopes disagree.
Q: For f(x) = { ax, x ≤ 1; bx² + x + 1, x > 1 }, find a and b for differentiability at x = 1.
A: Continuity gives a = b + 2. Matching derivatives gives a = 2b + 1. Solving: b = 1, a = 3.
Q: The graph of f'(x) crosses the x-axis at x = 0, 2, and 4. Where does f have horizontal tangent lines?
A: At x = 0, x = 2, and x = 4. These are the points where f'(x) = 0.
Q: Using the table { t: 0,1,2,3,4,5,6,7; ducks: 0,3,8,11,12,21,24,28 }, what is the average rate over [0,7] and the estimated instantaneous rate at t = 4.5?
A: Average rate = (28 − 0)/7 = 4 ducks per hour. f'(4.5) ≈ (21 − 12)/(5 − 4) = 9 ducks per hour.
Q: Given f(−1) = 0, g(−1) = −1, f'(−1) = 2, g'(−1) = 1, and f(0) = −1, g(0) = −3, f'(0) = −2, g'(0) = 4, find the derivative of f(x)/[g(x)+2] at x = 0.
A: Using the quotient rule: [f'(0)(g(0)+2) − f(0)·g'(0)] / [g(0)+2]² = [(−2)(−1) − (−1)(4)] / (−1)² = (2+4)/1 = 6.
The product and quotient rules extend naturally once you learn the chain rule in Unit 3; many problems require all three together. Differentiability conditions reappear in the Mean Value Theorem (Unit 5), which requires the function to be differentiable on an open interval. Reading graphs of f' to analyse f is foundational for curve sketching and the first/second derivative tests, also in Unit 5.
product rule, quotient rule, differentiability, piecewise function, continuous, sharp turn, corner, cusp, horizontal tangent, f prime graph, table of values, derivative from graph, AP Calculus Unit 2, sections 2.7 2.8 2.9 2.10, combining derivative rules, derivative does not exist, DNE, rate of change interpretation