2.8 – Implicit Differentiation
Learning Objectives
By the end of this section, you should be able to:
- Understand the difference between explicit and implicit functions
- Apply implicit differentiation to find derivatives of implicitly defined functions
- Use the Chain Rule correctly when differentiating terms involving y
- Solve for \( \frac{dy}{dx} \) after implicit differentiation
- Find slopes of tangent lines for curves defined implicitly
- Apply implicit differentiation to related rates problems
What is Implicit Differentiation?
Explicit functions are written in the form \( y = f(x) \), where y is explicitly defined in terms of x.
Implicit functions are relationships between x and y that aren't solved for y, such as \( x^2 + y^2 = 25 \).
Implicit differentiation is a technique for finding \( \frac{dy}{dx} \) when we can't (or don't want to) solve for y explicitly.
Explicit vs. Implicit Functions
| Type | Example | How to Differentiate |
|---|---|---|
| Explicit | \( y = x^2 + 3x - 2 \) | Direct differentiation: \( \frac{dy}{dx} = 2x + 3 \) |
| Implicit | \( x^2 + y^2 = 25 \) | Differentiate both sides with respect to x, treating y as a function of x |
The Key Idea: y is a Function of x
When we see y in an equation, we treat it as \( y(x) \) - a function of x. This means:
- \( \frac{d}{dx}[y] = \frac{dy}{dx} \) (we don't know what this is yet)
- \( \frac{d}{dx}[y^n] = ny^{n-1} \cdot \frac{dy}{dx} \) (Chain Rule!)
- \( \frac{d}{dx}[f(y)] = f'(y) \cdot \frac{dy}{dx} \) (Chain Rule applied)
Worked Examples
Example 1: Basic Circle
Fundamental ApplicationProblem: Find \( \frac{dy}{dx} \) for the circle \( x^2 + y^2 = 25 \)
Differentiate both sides with respect to x:
\[ \frac{d}{dx}[x^2 + y^2] = \frac{d}{dx}[25] \]
Apply derivatives term by term:
\[ \frac{d}{dx}[x^2] + \frac{d}{dx}[y^2] = 0 \]
\[ 2x + 2y \cdot \frac{dy}{dx} = 0 \]
Note: \( \frac{d}{dx}[y^2] = 2y \cdot \frac{dy}{dx} \) by Chain Rule
Solve for \( \frac{dy}{dx} \):
\[ 2y \cdot \frac{dy}{dx} = -2x \]
\[ \frac{dy}{dx} = -\frac{2x}{2y} = -\frac{x}{y} \]
Interpretation:
The slope of the tangent line to the circle \( x^2 + y^2 = 25 \) at any point (x,y) is \( -\frac{x}{y} \).
Example 2: Product with y
Product Rule + Chain RuleProblem: Find \( \frac{dy}{dx} \) for \( x^2y + y^3 = 6 \)
Differentiate both sides:
\[ \frac{d}{dx}[x^2y + y^3] = \frac{d}{dx}[6] \]
Apply Product Rule to \( x^2y \):
\[ \frac{d}{dx}[x^2y] = \frac{d}{dx}[x^2] \cdot y + x^2 \cdot \frac{d}{dx}[y] \]
\[ = 2x \cdot y + x^2 \cdot \frac{dy}{dx} \]
Apply Chain Rule to \( y^3 \):
\[ \frac{d}{dx}[y^3] = 3y^2 \cdot \frac{dy}{dx} \]
Combine and solve:
\[ 2xy + x^2\frac{dy}{dx} + 3y^2\frac{dy}{dx} = 0 \]
\[ (x^2 + 3y^2)\frac{dy}{dx} = -2xy \]
\[ \frac{dy}{dx} = -\frac{2xy}{x^2 + 3y^2} \]
Example 3: Finding Slope at a Point
Practical ApplicationProblem: Find the slope of the tangent line to \( x^3 + y^3 = 9 \) at the point (1,2)
Differentiate implicitly:
\[ \frac{d}{dx}[x^3 + y^3] = \frac{d}{dx}[9] \]
\[ 3x^2 + 3y^2 \cdot \frac{dy}{dx} = 0 \]
Solve for \( \frac{dy}{dx} \):
\[ 3y^2 \cdot \frac{dy}{dx} = -3x^2 \]
\[ \frac{dy}{dx} = -\frac{3x^2}{3y^2} = -\frac{x^2}{y^2} \]
Evaluate at (1,2):
\[ \frac{dy}{dx} = -\frac{(1)^2}{(2)^2} = -\frac{1}{4} \]
Interpretation:
The slope of the tangent line at (1,2) is \( -\frac{1}{4} \).
When to Use Implicit Differentiation
Use implicit differentiation when:
- The equation cannot be easily solved for y
- Solving for y would result in multiple functions
- You need the derivative at a specific point
- Working with related rates problems
- The relationship is naturally expressed implicitly
Common applications:
- Circles, ellipses, and other conic sections
- Curves defined by polynomials in both x and y
- Relationships in related rates problems
- Curves that aren't functions (fail vertical line test)
Common Mistakes to Avoid
| Mistake | Why It's Wrong | Correct Approach |
|---|---|---|
| Forgetting Chain Rule on y terms | Treats y as constant instead of function of x | Always multiply by \( \frac{dy}{dx} \) when differentiating y terms |
| Not solving for \( \frac{dy}{dx} \) completely | Leaves answer with \( \frac{dy}{dx} \) on both sides | Isolate \( \frac{dy}{dx} \) terms and factor them out |
| Algebra errors when solving | Makes mistakes in algebraic manipulation | Work carefully and check each step |
| Forgetting to apply other rules | Misses Product or Quotient Rule when needed | Identify the structure of each term before differentiating |
Quick Concept Check
Question 1: What is \( \frac{d}{dx}[y^3] \)?
Question 2: For \( x^2 + y^2 = 25 \), what is \( \frac{dy}{dx} \)?
Step-by-Step Strategy for Implicit Differentiation
- Differentiate both sides of the equation with respect to x
- Apply Chain Rule to all terms containing y (multiply by \( \frac{dy}{dx} \))
- Apply other rules (Product, Quotient) as needed
- Collect all \( \frac{dy}{dx} \) terms on one side of the equation
- Factor out \( \frac{dy}{dx} \) from those terms
- Solve for \( \frac{dy}{dx} \)** by dividing both sides
- Simplify the final expression
AP Exam Tip
Implicit differentiation frequently appears on AP Calculus exams. Remember to:
- Show all steps clearly, especially the Chain Rule applications
- Clearly indicate when you're multiplying by \( \frac{dy}{dx} \)
- Box your final answer for \( \frac{dy}{dx} \)
- If finding slope at a point, substitute coordinates after finding the general derivative
- Practice with various equation types (polynomials, trig functions, etc.)
Practice Problems
Problem 1: Basic Implicit
Find \( \frac{dy}{dx} \) for \( x^3 + y^3 = 8 \)
Differentiate both sides:
\[ \frac{d}{dx}[x^3 + y^3] = \frac{d}{dx}[8] \]
\[ 3x^2 + 3y^2 \cdot \frac{dy}{dx} = 0 \]
\[ 3y^2 \cdot \frac{dy}{dx} = -3x^2 \]
\[ \frac{dy}{dx} = -\frac{3x^2}{3y^2} = -\frac{x^2}{y^2} \]
Problem 2: Product with y
Find \( \frac{dy}{dx} \) for \( xy + x^2y^2 = 6 \)
Differentiate both sides:
\[ \frac{d}{dx}[xy + x^2y^2] = \frac{d}{dx}[6] \]
For \( \frac{d}{dx}[xy] \), use Product Rule:
\[ \frac{d}{dx}[x] \cdot y + x \cdot \frac{d}{dx}[y] = 1 \cdot y + x \cdot \frac{dy}{dx} \]
For \( \frac{d}{dx}[x^2y^2] \), use Product Rule + Chain Rule:
\[ \frac{d}{dx}[x^2] \cdot y^2 + x^2 \cdot \frac{d}{dx}[y^2] = 2x \cdot y^2 + x^2 \cdot 2y \cdot \frac{dy}{dx} \]
Combine: \( y + x\frac{dy}{dx} + 2xy^2 + 2x^2y\frac{dy}{dx} = 0 \)
\[ (x + 2x^2y)\frac{dy}{dx} = -y - 2xy^2 \]
\[ \frac{dy}{dx} = -\frac{y + 2xy^2}{x + 2x^2y} = -\frac{y(1 + 2xy)}{x(1 + 2xy)} = -\frac{y}{x} \]
Problem 3: Trigonometric Implicit
Find \( \frac{dy}{dx} \) for \( \sin(x) + \cos(y) = 1 \)
Differentiate both sides:
\[ \frac{d}{dx}[\sin(x) + \cos(y)] = \frac{d}{dx}[1] \]
\[ \cos(x) + (-\sin(y)) \cdot \frac{dy}{dx} = 0 \]
\[ \cos(x) - \sin(y) \cdot \frac{dy}{dx} = 0 \]
\[ -\sin(y) \cdot \frac{dy}{dx} = -\cos(x) \]
\[ \frac{dy}{dx} = \frac{\cos(x)}{\sin(y)} \]