AP Calculus AB Unit 3: Composite, Implicit, & Inverse
Differentiation II
Composite, Implicit, and Inverse Functions
1. The Chain Rule
The Chain Rule is used for differentiating composite functions (functions inside other functions), typ-
ically written as f (g(x)).
Chain Rule Definition
If y = f (u) and u = g(x), then:
dy dy du
= ·
dx du dx
d
[f (g(x))] = f ′ (g(x)) · g ′ (x)
dx
"Derivative of the outer function (evaluated at the inner) × derivative of the inner function."
Example: Let y = (3x2 + 1)5 .
• Outer: (. . . )5 → 5(. . . )4
• Inner: 3x2 + 1 → 6x
• Result: 5(3x2 + 1)4 · (6x) = 30x(3x2 + 1)4
2. Implicit Differentiation
Used when y cannot be easily isolated (e.g., x2 + y 2 = 25).
Key Formula
Steps for Implicit Differentiation:
1. Differentiate both sides with respect to x.
dy
2. Whenever you differentiate a y term, multiply by dx (Chain Rule).
dy
3. Collect all terms with dx on one side.
dy
4. Factor out dx and solve for it.
Strategy Tip – Watch out for Product Rule mixed with Implicit Differentiation! Example:
d dy
dx (xy) = (1)(y) + (x)( dx ).
1
Differentiation II
Composite, Implicit, and Inverse Functions
1. The Chain Rule
The Chain Rule is used for differentiating composite functions (functions inside other functions), typ-
ically written as f (g(x)).
Chain Rule Definition
If y = f (u) and u = g(x), then:
dy dy du
= ·
dx du dx
d
[f (g(x))] = f ′ (g(x)) · g ′ (x)
dx
"Derivative of the outer function (evaluated at the inner) × derivative of the inner function."
Example: Let y = (3x2 + 1)5 .
• Outer: (. . . )5 → 5(. . . )4
• Inner: 3x2 + 1 → 6x
• Result: 5(3x2 + 1)4 · (6x) = 30x(3x2 + 1)4
2. Implicit Differentiation
Used when y cannot be easily isolated (e.g., x2 + y 2 = 25).
Key Formula
Steps for Implicit Differentiation:
1. Differentiate both sides with respect to x.
dy
2. Whenever you differentiate a y term, multiply by dx (Chain Rule).
dy
3. Collect all terms with dx on one side.
dy
4. Factor out dx and solve for it.
Strategy Tip – Watch out for Product Rule mixed with Implicit Differentiation! Example:
d dy
dx (xy) = (1)(y) + (x)( dx ).
1
