Unit 3: Differentiation: Composite, Implicit, and Inverse Functions
Unit 3 extends differentiation to functions built from other functions. It covers the chain rule for composites, implicit differentiation for equations that mix x and y, the derivative of an inverse function, the derivatives of the inverse trigonometric functions, and higher-order derivatives.
How to use this guide
Read it in order the first time. The chain rule comes first because implicit differentiation is the chain rule applied to y-terms, and the inverse-function rule is best understood as the chain rule applied to f(f−1(x)) = x. Work each example on paper before reading the solution. The exam tests whether you apply the inner derivative every time, not whether you can state the rule.
What this unit is worth. This unit is about 4 to 7 percent of the AP Calculus BC exam. Like Unit 2, it is a tools unit, and the chain rule in particular shows up inside nearly every later topic, including related rates, differential equations, and series.
3.1 The Chain Rule
A composite function is one function applied to the output of another, written f(g(x)) or (f ∘ g)(x). For example, (3x2 + 1)4 is the fourth-power function applied to the polynomial 3x2 + 1, and sin(2x) is the sine function applied to 2x.
The chain rule says that for f(g(x)), the derivative is f′(g(x)) · g′(x). In words, take the derivative of the outside, then multiply by the derivative of the inside. The inner derivative is the part students skip, and the exam counts on that.
Example: differentiate (3x2 + 1)4. The outside is u4, whose derivative is 4u3. The inside is u = 3x2 + 1, whose derivative is 6x. Multiply: 4(3x2 + 1)3 · 6x = 24x(3x2 + 1)3.
Example: differentiate sin(2x). The outside is sin u, whose derivative is cos u. The inside is u = 2x, whose derivative is 2. So the result is cos(2x) · 2 = 2cos(2x).
Trap. The answer 4(3x2 + 1)3 is wrong for the first example above. It is the derivative of the outside only. The chain rule is not finished until you multiply by the derivative of the inside. Every distractor on chain-rule questions is built from answers that skip that step.
3.2 Implicit Differentiation
Implicit differentiation handles equations where y is not solved for, such as x2 + y2 = 25. Treat y as a function of x. When you differentiate a y-term, the chain rule produces a dy/dx factor. Then solve the resulting equation for dy/dx.
Example: x2 + y2 = 25. Differentiating term by term gives 2x + 2y · dy/dx = 0, because the derivative of y2 with respect to x is 2y times dy/dx by the chain rule. Solving: dy/dx = −2x / 2y = −x/y.
Example: xy + y2 = 1. The product xy needs the product rule: its derivative is y + x · dy/dx. The y2 term gives 2y · dy/dx. So y + x · dy/dx + 2y · dy/dx = 0. Collect the dy/dx terms: dy/dx (x + 2y) = −y, and dy/dx = −y / (x + 2y).
Trap. Differentiating y2 as 2y is the single most common error here. The derivative is 2y · dy/dx. Any y-term always picks up a dy/dx factor because y is a function of x. If your answer has no dy/dx in it, you missed one.
3.3 Differentiating Inverse Functions
The derivative of an inverse function follows from the definition. If f(a) = b and f is invertible, then (f−1)′(b) = 1 / f′(a), provided f′(a) ≠ 0. The slope of the inverse at the point b is the reciprocal of the slope of the original function at the point a.
The way to use it: find the x-value a that maps to b, evaluate f′ there, and take the reciprocal. The evaluation happens at a, the input to f, not at b.
Example: f(x) = x3 + 2. Find (f−1)′(3). First find a with f(a) = 3. Since 13 + 2 = 3, a = 1. Then f′(x) = 3x2, so f′(1) = 3. Therefore (f−1)′(3) = 1/3.
Trap. The formula needs f′(a), evaluated at the x-value of the original function. Computing 1 / f′(3) instead of 1 / f′(1) in the example above gives 1/27, which is wrong. The first step of every inverse-derivative problem is finding a, the value with f(a) = b.
3.4 Differentiating Inverse Trigonometric Functions
The three inverse trigonometric derivatives come from implicit differentiation of the inverse relationship, and they are worth memorizing because the exam treats them as known rules.
| Function | Derivative |
|---|---|
| arcsin x | 1 / √(1 − x2) |
| arccos x | −1 / √(1 − x2) |
| arctan x | 1 / (1 + x2) |
When the argument is itself a function, apply the chain rule. Example: differentiate arcsin(2x). The derivative of arcsin u is 1 / √(1 − u2), and the inner derivative is 2, so the result is 2 / √(1 − 4x2).
Trap. The arcsin and arccos derivatives differ only by a sign. The arcsin rule has no negative; the arccos rule does. Students also mix up the square roots, writing 1 / √(1 + x2) for arcsin. The plus belongs to arctan, which has no square root at all.
3.6 Calculating Higher-Order Derivatives
Note. Topic 3.5 (selecting procedures for calculating derivatives) has no separate section here. Choosing the right rule is practiced in every example and question above.
The second derivative is the derivative of the derivative: f″(x), y″, or d2y/dx2. It measures how the rate of change itself changes. For f(x) = x4, f′(x) = 4x3 and f″(x) = 12x2.
Higher-order derivatives continue the pattern: f‴(x), f(4)(x), and in general f(n)(x) for the nth derivative. Just differentiate again, applying the same rules each time. There is no shortcut that skips a step.
Example: f(x) = x4 − 2x2. Then f′(x) = 4x3 − 4x and f″(x) = 12x2 − 4. Each pass uses the power rule on the result of the previous pass.
Trap. The second derivative is not the derivative of f′(x) with a fresh power rule shortcut. It is f′ differentiated normally, which for f(x) = x4 − 2x2 gives 12x2 − 4, not 4x3 − 4x written again. Always do both passes on paper.
Confusions That Cost Points
| Pair | How to keep them straight |
|---|---|
| Chain rule with and without the inner derivative | d/dx[(3x2 + 1)4] = 24x(3x2 + 1)3. Without the 6x from the inside, the answer is incomplete. |
| 2y · dy/dx vs 2y in implicit differentiation | Any y-term picks up dy/dx because y is a function of x. An answer with no dy/dx missed a chain-rule step. |
| 1 / f′(a) vs 1 / f′(b) for inverses | The inverse rule evaluates f′ at a, the input with f(a) = b. Find a first, then differentiate, then reciprocate. |
| arcsin vs arccos vs arctan derivatives | arcsin: 1/√(1 − x2). arccos: same with a negative. arctan: 1/(1 + x2), no root. |
| f″(x) vs f′(x) | The second derivative requires a full second pass of differentiation. Write both passes out. |
Practice Questions
Original questions written for this guide in the style of the AP exam. Answers and explanations are on the next page, so complete the questions before checking them.
1. d/dx[(2x3 − 1)4] =
- 4(2x3 − 1)3
- 24x2(2x3 − 1)3
- 24x2(2x3 − 1)2
- 4(6x2)3
2. For x2 + 3xy = 1, dy/dx =
- −(2x + 3y) / (3x)
- −(2x + 3y) / 3
- 2x + 3y
- (−2x − 3y) / (3x + 3y)
3. Let f(x) = x3 + 2x. Then (f−1)′(0) =
- 2
- 1/2
- 0
- −1/2
4. If f(x) = x4 − 2x2, then f″(x) =
- 4x3 − 4x
- 12x2 − 4
- 12x3 − 4
- 24x
Answer Key
1. B. The outside derivative is 4(2x3 − 1)3 and the inside derivative is 6x2. Multiply: 4(2x3 − 1)3 · 6x2 = 24x2(2x3 − 1)3. A is the outside derivative alone, the classic skipped inner derivative. C keeps the wrong power, which happens when the outside is differentiated carelessly. D differentiates the inside first and then cubes it, which is not how the chain rule works.
2. A. Differentiating: 2x + 3y + 3x · dy/dx = 0, since d/dx[3xy] = 3y + 3x · dy/dx by the product rule. Collect: 3x · dy/dx = −2x − 3y, so dy/dx = −(2x + 3y) / (3x). B drops the x from the denominator, which comes from collecting the dy/dx terms incorrectly. C never solves for dy/dx at all. D adds a 3y to the denominator, which is not a term that multiplies dy/dx.
3. B. Find a with f(a) = 0. Since f(0) = 0, a = 0. Then f′(x) = 3x2 + 2, so f′(0) = 2, and (f−1)′(0) = 1/2. A reports f′(0) without taking the reciprocal. C confuses the input b = 0 with the answer. D adds a negative sign that appears nowhere in the reciprocal rule.
4. B. First pass: f′(x) = 4x3 − 4x. Second pass: f″(x) = 12x2 − 4. A stops after one derivative. C differentiates the power wrong on the second pass (3 · 4x3 would need an x4 term, which is not there). D is the third derivative, one pass too many.
One-Page Recall Check
- Write the chain rule for f(g(x)) in words and in symbols.
- Differentiate (5x2 − 3)6 from scratch, showing the inner derivative step.
- Explain why differentiating y2 with respect to x gives 2y · dy/dx and not 2y.
- Find dy/dx for x2 + y2 = 16, showing the solve step.
- State the inverse-function derivative rule and explain why f′ is evaluated at a, not b.
- For f(x) = x3 + x with f(0) = 0, compute (f−1)′(0) by hand.
- Write the derivatives of arcsin x, arccos x, and arctan x from memory.
- Differentiate arctan(3x), showing the chain-rule factor.
- Explain how the second derivative differs from the first, and compute f″(x) for f(x) = x5.
- Name two errors from the trap boxes that you have made before, and say how you will catch them next time.
Where to go next. Turn every missed item above into flashcards and drill them spaced out over several days rather than in one sitting. In Rycal, open the AP Calculus BC deck. The deck covers the terms in this guide, and its practice questions target the same traps named here. If you have a test date, add it in the Test Planner. You can also start your next review with a Brain Dump, then check what you missed against this guide.
Key terms for this unit
Chain Rule, Composite function, Implicit differentiation, Derivative of an inverse function, Derivative of arcsin x, Derivative of arctan x, Derivative of arccos x, Second derivative, Higher-order derivatives.
About this guide. Written for Rycal and aligned to the College Board AP Calculus BC course framework, Unit 3. All questions and explanations are original Rycal writing. Rycal is independent and is not affiliated with or endorsed by the College Board.