#Second Derivatives of Implicit Functions

#Table of Contents

1. Introduction
2. Fundamentals of Implicit Differentiation
3. Finding the Second Derivative
4. Worked Example
5. Practice Questions
6. Glossary
7. Summary and Key Takeaways
8. Exam Strategy

#Introduction

Finding the second derivative of an implicit function involves differentiating the first derivative. This guide will walk you through the steps necessary to achieve this, using examples and providing tips to aid your understanding.

#Fundamentals of Implicit Differentiation

Before diving into second derivatives, ensure you're comfortable with implicit differentiation:

• Implicit Differentiation involves finding the derivative of a function defined implicitly by using the chain rule.

#Finding the Second Derivative

To find the second derivative of an implicit function, follow these steps:

1. Find the First Derivative:

   • Start with the implicit equation. For example, consider $x^2 + y^2 = 4x$.
   • Differentiate both sides with respect to $x$.

   $$\frac{d}{dx}(x^2 + y^2) = \frac{d}{dx}(4x)$$

   • Apply the chain rule to implicitly differentiate $y$:

   $$2x + 2y \frac{dy}{dx} = 4$$

   • Solve for $\frac{dy}{dx}$:

   $$\frac{dy}{dx} = \frac{2 - x}{y}$$

2. Find the Second Derivative:

   • Differentiate the first derivative with respect to $x$:

   $$\frac{d}{dx}\left( \frac{dy}{dx} \right) = \frac{d}{dx}\left( \frac{2 - x}{y} \right)$$

   • Use the quotient rule $\left( \frac{u}{v} \right)' = \frac{u'v - uv'}{v^2}$:

   $$u = 2 - x, \quad v = y$$ $$u' = -1, \quad v' = \frac{dy}{dx}$$

   • Substitute into the quotient rule:

   $$\frac{d^2y}{dx^2} = \frac{(-1) \cdot y - (2 - x) \cdot \frac{dy}{dx}}{y^2}$$

   • Substitute $\frac{dy}{dx} = \frac{2 - x}{y}$:

   $$\frac{d^2y}{dx^2} = \frac{-y - (2 - x) \cdot \frac{2 - x}{y}}{y^2}$$

   • Simplify:

   $$\frac{d^2y}{dx^2} = \frac{-y - \frac{(2 - x)^2}{y}}{y^2}$$ $$\frac{d^2y}{dx^2} = \frac{-y^2 - (2 - x)^2}{y^3}$$ $$\frac{d^2y}{dx^2} = -\frac{y^2 + (2 - x)^2}{y^3}$$

#Worked Example

Problem:

Show that the second derivative of $\sin x + \cos y = 0.8$ can be written as:

$$\frac{d^2y}{dx^2} = -\frac{\sin x \sin^2 y + \cos^2 x \cos y}{\sin^3 y}$$

Solution:

1. First Derivative:

   Differentiate both sides with respect to $x$:

   $$\frac{d}{dx}(\sin x + \cos y) = \frac{d}{dx}(0.8)$$ $$\cos x - \sin y \cdot \frac{dy}{dx} = 0$$ $$\frac{dy}{dx} = \frac{\cos x}{\sin y}$$

2. Second Derivative:

   Differentiate $\frac{dy}{dx}$ with respect to $x$:

   $$\frac{d}{dx} \left( \frac{dy}{dx} \right) = \frac{d}{dx} \left( \frac{\cos x}{\sin y} \right)$$ $$\frac{d^2y}{dx^2} = \frac{(-\sin x) \cdot \sin y - \cos x \cdot \cos y \cdot \frac{dy}{dx}}{\sin^2 y}$$

   Substitute $\frac{dy}{dx} = \frac{\cos x}{\sin y}$:

   $$\frac{d^2y}{dx^2} = \frac{-\sin x \sin y - \cos x \cos y \cdot \frac{\cos x}{\sin y}}{\sin^2 y}$$ $$\frac{d^2y}{dx^2} = \frac{-\sin x \sin y - \frac{\cos^2 x \cos y}{\sin y}}{\sin^2 y}$$ $$\frac{d^2y}{dx^2} = \frac{-\sin x \sin^2 y - \cos^2 x \cos y}{\sin^3 y}$$

   Therefore:

   $$\frac{d^2y}{dx^2} = -\frac{\sin x \sin^2 y + \cos^2 x \cos y}{\sin^3 y}$$

Illustrative Example: The above steps demonstrate how to find the second derivative of an implicit function involving trigonometric terms.

#Practice Questions

1. Question: Differentiate $x^3 + y^3 = 6xy$ to find $\frac{d^2y}{dx^2}$.

2. Question: Given $x^2 + y^2 = 1$, find $\frac{d^2y}{dx^2}$.

#Glossary

• Implicit Differentiation: A method to find the derivative of a function defined by an equation involving both $x$ and $y$.
• Quotient Rule: A rule for differentiating a function that is the ratio of two other functions.
• Chain Rule: A rule for differentiating compositions of functions.

#Summary and Key Takeaways

• First Derivative: Use implicit differentiation to find $\frac{dy}{dx}$.
• Second Derivative: Differentiate $\frac{dy}{dx}$ with respect to $x$ using the chain and quotient rules.
• Simplifying the result may involve substituting back the first derivative and combining like terms.

#Exam Strategy

• Understand Implicit Differentiation: Master the chain rule and implicit differentiation basics.
• Practice Quotient Rule: Be comfortable using the quotient rule for differentiation.
• Simplify Expressions: Always simplify your final answer and check for common factors.

By following these strategies and understanding the concepts, you will be well-prepared to tackle problems involving the second derivatives of implicit functions.