AP Calculus BC - Unit 3 Study Guide
AP Calculus BC • Unit 3

Composite, Implicit & Inverse Functions

Exam Weight: 4-7% • Topics 3.1 – 3.6
3.0

Unit 3 “Get a 5” Plan

Unit 3 is where students start losing points from structure mistakes (forgetting chain rule, missing y', wrong inverse input). Fix those, and Unit 3 becomes free points.

✅ Must-Memorize
  • Chain Rule: d/dx [f(g(x))] = f'(g(x)) · g'(x)
  • Implicit: “every y term pays y'”
  • Inverse slope: (f−1)'(b) = 1f'(a) where f(a)=b
  • Inverse trig derivatives table
🎯 FRQ Scoring Habits
  • Write the correct setup first (formula/structure), then simplify.
  • At a point? Plug numbers after differentiating.
  • Include units/meaning if problem is contextual (Unit 4 overlaps).
  • If asked for a tangent line, you need: slope + point.
Fast self-check (10 seconds): If your derivative has a nested expression (like (3x²+1)⁵, ex², sin(5x)), you should see a multiplier from the inside derivative.
3.1

The Chain Rule

The Chain Rule is the #1 rule in Unit 3. Any time you see “function inside a function,” think: outside derivative × inside derivative.

The Formula
d/dx [ f(g(x)) ] = f'(g(x)) · g'(x)
Outside stays, then multiply by the inside slope.
Example 1 (Power + Polynomial)

Problem: Differentiate y=(3x²+1)⁵.

  1. Outside: (□)⁵ → 5(□)⁴
  2. Inside: 3x²+1 → 6x
  3. Answer:
    y' = 5(3x²+1)⁴ · 6x = 30x(3x²+1)⁴
Example 2 (Trig + Linear)

Problem: Differentiate y=sin(5x).

y' = cos(5x) · 5 = 5cos(5x)

Common miss: writing just cos(5x) (forgot the inside derivative).

Example 3 (Exponential)

Problem: Differentiate y=ex²−4x.

y' = ex²−4x · (2x−4)
AP trap: Don’t simplify too early. Get the structure right first (outside × inside). You can always clean it later.
3.2

Implicit Differentiation

Use implicit differentiation when y is mixed with x (like x²+y²=25). Treat y as y(x), so every y term triggers the chain rule.

The “Chain Rule Tax”

Whenever you differentiate a y-expression, multiply by y'.

d/dx (y³) = 3y² · y'
1

Differentiate
Both sides in x.
Add y' when touching y.

2

Gather
Move all y' terms to one side.

3

Solve
Factor out y' and divide.

Full Example (Classic Circle)

Problem: Find dy/dx if x²+y²=25.

  1. Differentiate: 2x + 2y·y' = 0
  2. Solve: 2y·y' = −2x → y' = −x/y

AP trap: forgetting the y' on y².

Point evaluation rule: If asked for the slope at a point, find y' first, then substitute (x,y).
3.3

Differentiating Inverse Functions

Inverse functions reflect across y=x. That reflection flips slopes into reciprocals, but you must match the correct input/output pair.

Reciprocal Slope Relationship
(f−1)'(b) = 1f'(a)
where f(a)=b

Inverse derivative at b uses original derivative at the matching a.

AP Table Method (Most Common)

Problem: If f(2)=5 and f'(2)=3, find (f−1)'(5).

  1. Match: f(2)=5 → point (2,5) on f, so (5,2) on f−1.
  2. Reciprocal: (f−1)'(5)=1/f'(2)=1/3
Top mistake: Using f'(5) instead of f'(2).
3.4

Inverse Trigonometric Functions

You don’t need to derive these on the exam — but you must memorize them and apply the chain rule correctly.

Function Derivative Chain Rule Version
arcsin(x) 1 / √(1−x²) u' / √(1−u²)
arccos(x) −1 / √(1−x²) −u' / √(1−u²)
arctan(x) 1 / (1+x²) u' / (1+u²)
Memory Aids
  • Sin has a Square root: √(1−x²)
  • Tan has a +: 1+x² (no root)
  • “Co” function (arccos) is the negative version of arcsin
Example (Chain Rule Included)

Problem: Differentiate y=arctan(3x).

y' = 31+9x²
3.5

Tangent Line Skills (Implicit + Inverse)

College Board loves “find the equation of the tangent line” because it combines skills: differentiate correctly, evaluate at a point, then use point-slope form.

Tangent Line Template
y − y1 = m(x − x1)
You need a slope and a point.
FRQ-Style Example (Implicit Tangent Line)

Problem: For x²+y²=25, find the tangent line at (3,4).

  1. From implicit diff: y' = −x/y
  2. Slope at (3,4): m = −3/4
  3. Tangent line: y − 4 = (−3/4)(x − 3)
FRQ-Style Example (Inverse Derivative Tangent)

Problem: If f(2)=5 and f'(2)=3, write the tangent line to y=f−1(x) at x=5.

  1. Point on inverse: (5,2)
  2. Slope: (f−1)'(5)=1/3
  3. Tangent line: y − 2 = (1/3)(x − 5)
3.6

Higher-Order Derivatives

Higher-order derivatives measure how the rate of change itself changes. This is a bridge into Unit 4 (motion + concavity).

First Derivative

f'(x)

Slope / Velocity

Second Derivative

f''(x)

Concavity / Acceleration

Notation

d²y/dx²

(Not “dy squared”)

Quick Example

Problem: If f(x)=x⁴−2x², find f'(x) and f''(x).

f'(x)=4x³−4x    and    f''(x)=12x²−4
Interpretation: If f''(x)>0, concave up. If f''(x)<0, concave down.
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