AP Calculus BC 3.2 Implicit Differentiation Study Notes - New Syllabus
AP Calculus BC 3.2 Implicit Differentiation Study Notes- New syllabus
AP Calculus BC 3.2 Implicit Differentiation Study Notes – AP Calculus BC- per latest AP Calculus BC Syllabus.
LEARNING OBJECTIVE
- Calculate derivatives of implicitly defined functions.
Key Concepts:
- Implicit Differentiation
Implicit Differentiation: Calculating Derivatives of Implicitly Defined Functions
Implicit Function
An implicit function is a function that is not expressed in the form \( y = f(x) \), but rather defined by an equation involving both \( x \) and \( y \), such as:
\( F(x, y) = 0 \quad \text{or} \quad G(x, y) = c \).
Here, \( y \) is not isolated as a function of \( x \). However, it can still be treated as a function locally in calculus (via implicit differentiation).
Implicit Differentiation
Implicit differentiation is used when a function is not given explicitly as \( y = f(x) \) but rather in a relationship involving \( x \) and \( y \), such as \( F(x, y) = 0 \). To find \( \dfrac{dy}{dx} \), we differentiate both sides of the equation with respect to \( x \), treating \( y \) as a function of \( x \), and then solve for \( \dfrac{dy}{dx} \).
Rule: Whenever you differentiate a term containing \( y \), multiply by \( \dfrac{dy}{dx} \) (because of the chain rule).
Example:
Find \( \dfrac{dy}{dx} \) if \( x^2 + y^2 = 25 \).
▶️ Answer/Explanation
Differentiate both sides with respect to \( x \):
\( \dfrac{d}{dx}(x^2) + \dfrac{d}{dx}(y^2) = \dfrac{d}{dx}(25) \).
\( 2x + 2y \dfrac{dy}{dx} = 0 \).
Solve for \( \dfrac{dy}{dx} \):
\( 2y \dfrac{dy}{dx} = -2x \) ⇒ \( \dfrac{dy}{dx} = -\dfrac{x}{y} \).
Answer: \( \dfrac{dy}{dx} = -\dfrac{x}{y} \).
Example:
Find \( \dfrac{dy}{dx} \) if \( xy + y^3 = 7 \).
▶️ Answer/Explanation
Differentiate each term with respect to \( x \):
\( \dfrac{d}{dx}(xy) + \dfrac{d}{dx}(y^3) = \dfrac{d}{dx}(7) \).
Product rule for \( xy \): \( (1)(y) + (x)(\dfrac{dy}{dx}) \).
Derivative of \( y^3 \): \( 3y^2 \dfrac{dy}{dx} \).
Combine terms:
\( y + x\dfrac{dy}{dx} + 3y^2\dfrac{dy}{dx} = 0 \).
Factor out \( \dfrac{dy}{dx} \):
\( (x + 3y^2)\dfrac{dy}{dx} = -y \).
Answer: \( \dfrac{dy}{dx} = -\dfrac{y}{x + 3y^2} \).
Example:
Find \( \dfrac{dy}{dx} \) if \( \sin(xy) = x \).
▶️ Answer/Explanation
Differentiate both sides:
\( \dfrac{d}{dx}[\sin(xy)] = \dfrac{d}{dx}[x] \).
Chain rule on \( \sin(xy) \): \( \cos(xy) \cdot \dfrac{d}{dx}(xy) \).
\( \dfrac{d}{dx}(xy) = y + x\dfrac{dy}{dx} \).
So \( \cos(xy)(y + x\dfrac{dy}{dx}) = 1 \).
Expand and solve:
\( \cos(xy)y + \cos(xy)x\dfrac{dy}{dx} = 1 \).
\( \cos(xy)x\dfrac{dy}{dx} = 1 – \cos(xy)y \).
\( \dfrac{dy}{dx} = \dfrac{1 – \cos(xy)y}{\cos(xy)x} \).
Answer: \( \dfrac{dy}{dx} = \dfrac{1 – y\cos(xy)}{x\cos(xy)} \).
Example:
Find \( \dfrac{dy}{dx} \) if \( x^2 + xy + \ln y = 5 \).
▶️ Answer/Explanation
Differentiate term by term:
\( \dfrac{d}{dx}(x^2) + \dfrac{d}{dx}(xy) + \dfrac{d}{dx}(\ln y) = 0 \).
\( 2x + (y + x\dfrac{dy}{dx}) + \dfrac{1}{y}\dfrac{dy}{dx} = 0 \).
Combine terms:
\( 2x + y + x\dfrac{dy}{dx} + \dfrac{1}{y}\dfrac{dy}{dx} = 0 \).
Factor \( \dfrac{dy}{dx} \):
\( \dfrac{dy}{dx}(x + \dfrac{1}{y}) = -(2x + y) \).
Answer: \( \dfrac{dy}{dx} = -\dfrac{2x + y}{x + \dfrac{1}{y}} \).