AP Calculus AB 3.6 The Second Derivative (Example 4 using Implicit Differentiation)

### AP Calculus AB 3.6: The Second Derivative (Example 4 Using Implicit Differentiation)

**Objective:**
This section focuses on computing the second derivative of a function using implicit differentiation. Students will learn how to apply implicit differentiation to find the second derivative without explicitly solving for \( y \) in terms of \( x \).

### Key Concepts

1. **Definition of the Second Derivative:**
The second derivative of a function \( f(x) \) is defined as the derivative of the first derivative \( f'(x) \). It is denoted as \( f''(x) \) and is given by:
\[
f''(x) = \frac{d^2}{dx^2} f(x)
\]
The second derivative provides insights into the curvature and concavity of the graph of the function.

2. **Implicit Differentiation:**
Implicit differentiation is used when a function is not explicitly solved for one variable in terms of another. If you have an equation involving both \( x \) and \( y \), you differentiate both sides with respect to \( x \), applying the chain rule to \( y \) terms. For example:
\[
\frac{dy}{dx} = \frac{dy}{dx} \cdot \frac{dy}{dx} + \text{other terms}
\]

3. **Geometric Interpretation:**
- The first derivative \( f'(x) \) indicates the slope of the tangent line to the graph of \( f(x) \).
- The second derivative \( f''(x) \) reveals the concavity of the function:
- If \( f''(x) \) is greater than zero, the graph is concave up.
- If \( f''(x) \) is less than zero, the graph is concave down.

### Example Problem

**Problem:** Find the second derivative of the equation \( x^2 + y^2 = 25 \).

1. **First Derivative Using Implicit Differentiation:**
Start by differentiating both sides of the equation with respect to \( x \):
\[
\frac{d}{dx}(x^2) + \frac{d}{dx}(y^2) = \frac{d}{dx}(25)
\]

This results in:
\[
2x + 2y \frac{dy}{dx} = 0
\]

Now solve for \( \frac{dy}{dx} \):
\[
2y \frac{dy}{dx} = -2x
\]
\[
\frac{dy}{dx} = -\frac{x}{y}
\]

2. **Second Derivative Using Implicit Differentiation:**
Differentiate \( \frac{dy}{dx} \) to find \( \frac{d^2y}{dx^2} \):
\[
\frac{d}{dx}\left(\frac{dy}{dx}\right) = \frac{d}{dx}\left(-\frac{x}{y}\right)
\]

Apply the quotient rule:
\[
\frac{d}{dx}\left(-\frac{x}{y}\right) = \frac{-y \cdot \frac{d}{dx}(x) - x \cdot \frac{d}{dx}(y)}{y^2}
\]
Substitute \( \frac{d}{dx}(y) = \frac{dy}{dx} \):
\[
\frac{d}{dx}\left(-\frac{x}{y}\right) = \frac{-y(1) - x\left(-\frac{x}{y}\right)}{y^2}
\]
This simplifies to:
\[
\frac{d^2y}{dx^2} = \frac{-y + \frac{x^2}{y}}{y^2}
\]
Combine the terms:
\[
\frac{d^2y}{dx^2} = \frac{-y^2 + x^2}{y^3}
\]

3. **Interpretation:**
- The second derivative \( \frac{d^2y}{dx^2} = \frac{-y^2 + x^2}{y^3} \) can be analyzed to determine the concavity of the curve defined by the equation \( x^2 + y^2 = 25 \), which represents a circle.
- Students can investigate the values of \( x \) and \( y \) to find points of inflection, indicating changes in concavity.

### Key Takeaways

- **Understanding the Second Derivative:** Students should recognize the importance of the second derivative in determining concavity and analyzing function behavior without explicitly solving for \( y \).
- **Application of Implicit Differentiation:** Mastery of implicit differentiation is crucial for handling equations involving both \( x \) and \( y \).
- **Critical Thinking and Problem Solving:** Students are encouraged to think critically about function behavior and apply their knowledge of differentiation to analyze and interpret results.

### Conclusion

In AP Calculus AB 3.6, students learn to compute the second derivative of functions using implicit differentiation. This skill is essential for understanding the behavior of curves defined by implicit equations. By working through examples involving the second derivative and the implicit differentiation process, students develop their differentiation skills and deepen their understanding of calculus concepts.

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

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