AP Calculus AB 3.2 Implicit Differentiation (Example 5 with Quotient Rule)

### AP Calculus AB 3.2: Implicit Differentiation (Example 5: Using the Quotient Rule)

**Objective:**
In this section, students will learn how to use **implicit differentiation** to find the derivative of a function defined implicitly, particularly when the function involves a quotient. This example will demonstrate how to differentiate equations using the **Quotient Rule**.

### What is Implicit Differentiation?

Implicit differentiation is a technique that allows us to differentiate equations where the dependent variable \( y \) is not explicitly expressed as a function of \( x \). This method is especially useful for equations that involve fractions or ratios.

#### Example: Implicit Differentiation Involving a Quotient

Consider the following implicit equation:

\[
\frac{y^2 + 1}{x + y} = 3
\]

Our goal is to differentiate this equation with respect to \( x \) and find \( \frac{dy}{dx} \).

### Step-by-Step Process:

1. **Differentiate Both Sides of the Equation**:
We will differentiate both sides with respect to \( x \).

- Differentiate the left side using the **Quotient Rule**:
The Quotient Rule states that if you have a function \( \frac{u}{v} \), then its derivative is given by:
\[
\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{v \frac{du}{dx} - u \frac{dv}{dx}}{v^2}
\]

For our function:
- Let \( u = y^2 + 1 \) and \( v = x + y \).

- Differentiate \( u \):
\[
\frac{du}{dx} = 2y \frac{dy}{dx}
\]

- Differentiate \( v \):
\[
\frac{dv}{dx} = 1 + \frac{dy}{dx}
\]

Applying the Quotient Rule, we get:
\[
\frac{d}{dx}\left(\frac{y^2 + 1}{x + y}\right) = \frac{(x + y)(2y \frac{dy}{dx}) - (y^2 + 1)(1 + \frac{dy}{dx})}{(x + y)^2}
\]

- Differentiate the right side:
\[
\frac{d}{dx}(3) = 0
\]

Thus, we can set up the equation:
\[
\frac{(x + y)(2y \frac{dy}{dx}) - (y^2 + 1)(1 + \frac{dy}{dx})}{(x + y)^2} = 0
\]

2. **Simplify the Equation**:
Since the fraction equals zero, the numerator must also equal zero:
\[
(x + y)(2y \frac{dy}{dx}) - (y^2 + 1)(1 + \frac{dy}{dx}) = 0
\]

Distributing gives:
\[
(x + y)(2y \frac{dy}{dx}) - (y^2 + 1) - (y^2 + 1) \frac{dy}{dx} = 0
\]

Rearranging terms leads to:
\[
(x + y)(2y \frac{dy}{dx}) - (y^2 + 1) = (y^2 + 1) \frac{dy}{dx}
\]

3. **Isolate \( \frac{dy}{dx} \)**:
Combine all terms involving \( \frac{dy}{dx} \):
\[
(x + y)(2y \frac{dy}{dx}) - (y^2 + 1) \frac{dy}{dx} = (y^2 + 1)
\]
Factor out \( \frac{dy}{dx} \):
\[
\frac{dy}{dx} \left((x + y) \cdot 2y - (y^2 + 1)\right) = y^2 + 1
\]

Therefore, we have:
\[
\frac{dy}{dx} = \frac{y^2 + 1}{(x + y) \cdot 2y - (y^2 + 1)}
\]

### Final Answer:

The derivative \( \frac{dy}{dx} \) for the equation \( \frac{y^2 + 1}{x + y} = 3 \) is:

\[
\frac{dy}{dx} = \frac{y^2 + 1}{(x + y) \cdot 2y - (y^2 + 1)}
\]

### Key Takeaways:

- **Implicit differentiation** is a valuable technique for finding derivatives of equations where the dependent variable is not isolated.
- The **Quotient Rule** is essential when differentiating functions that are expressed as ratios.
- Remember to isolate \( \frac{dy}{dx} \) clearly to express the derivative.

### Importance of Implicit Differentiation with the Quotient Rule:

Understanding how to apply implicit differentiation, particularly when using the Quotient Rule, is critical for solving complex calculus problems. This method equips students with the necessary skills to analyze various mathematical relationships, especially in real-world applications where functions are often interdependent.

By mastering this technique, students enhance their problem-solving abilities and prepare for more advanced calculus topics, such as parametric equations and higher-dimensional analysis. This foundational knowledge strengthens their overall understanding of calculus and its applications in different fields.

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Nick Perich
Norristown Area High School
Norristown Area School District
Norristown, Pa

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