Step-by-Step Implicit Differentiation Problems: Product & Quotient Rules

📚 Understanding Implicit Differentiation

Implicit differentiation is a technique used to find the derivative of a function where $y$ is not explicitly defined in terms of $x$. Instead, you have an equation relating $x$ and $y$. It's super useful in related rates problems and various applications in physics and engineering.

• 🔍 Definition: Implicit differentiation involves differentiating both sides of an equation with respect to $x$, treating $y$ as a function of $x$.
• 🕰️ History: The method evolved alongside the development of calculus in the 17th century, with contributions from mathematicians like Isaac Newton and Gottfried Wilhelm Leibniz.
• 🔑 Key Principle: Remember to apply the chain rule when differentiating terms involving $y$. That is, $\frac{d}{dx}[f(y)] = f'(y) \cdot \frac{dy}{dx}$.

➕ Product Rule in Implicit Differentiation

When dealing with products of $x$ and $y$, the product rule becomes essential. The product rule states that the derivative of $uv$ with respect to $x$ is $u'v + uv'$, where $u$ and $v$ are functions of $x$.

• 🧱 Formula: If you have a term like $x \cdot y$, then $\frac{d}{dx}(xy) = 1 \cdot y + x \cdot \frac{dy}{dx}$.
• 💡 Tip: Always remember to write $\frac{dy}{dx}$ when differentiating $y$ with respect to $x$.
• ✍️ Example: Consider $x^2y + y^2 = 5$. Differentiating with respect to $x$, we get $2xy + x^2\frac{dy}{dx} + 2y\frac{dy}{dx} = 0$.

➗ Quotient Rule in Implicit Differentiation

When you have a fraction involving $x$ and $y$, the quotient rule comes into play. The quotient rule states that the derivative of $\frac{u}{v}$ with respect to $x$ is $\frac{u'v - uv'}{v^2}$, where $u$ and $v$ are functions of $x$.

• ➗ Formula: If you have a term like $\frac{y}{x}$, then $\frac{d}{dx}(\frac{y}{x}) = \frac{\frac{dy}{dx} \cdot x - y \cdot 1}{x^2}$.
• 🧪 Application: This is commonly used in equations involving rational functions of $x$ and $y$.
• 📈 Example: Consider $\frac{y}{x} + x^3 = 10$. Differentiating with respect to $x$, we get $\frac{\frac{dy}{dx}x - y}{x^2} + 3x^2 = 0$.

⚙️ Step-by-Step Examples

Let's walk through some examples to solidify your understanding.

Example 1: Product Rule

Given: $x^2y + xy^2 = 6$

1. ✏️ Differentiate both sides with respect to $x$: $\frac{d}{dx}(x^2y) + \frac{d}{dx}(xy^2) = \frac{d}{dx}(6)$.
2. ➕ Apply the product rule: $(2x \cdot y + x^2 \cdot \frac{dy}{dx}) + (1 \cdot y^2 + x \cdot 2y \cdot \frac{dy}{dx}) = 0$.
3. 🧮 Simplify: $2xy + x^2\frac{dy}{dx} + y^2 + 2xy\frac{dy}{dx} = 0$.
4. 🤝 Collect terms with $\frac{dy}{dx}$: $(x^2 + 2xy)\frac{dy}{dx} = -2xy - y^2$.
5. ➗ Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{-2xy - y^2}{x^2 + 2xy}$.

Example 2: Quotient Rule

Given: $\frac{x}{y} + y^2 = 5$

1. ✏️ Differentiate both sides with respect to $x$: $\frac{d}{dx}(\frac{x}{y}) + \frac{d}{dx}(y^2) = \frac{d}{dx}(5)$.
2. ➗ Apply the quotient rule: $\frac{1 \cdot y - x \cdot \frac{dy}{dx}}{y^2} + 2y\frac{dy}{dx} = 0$.
3. 🧮 Simplify: $\frac{y - x\frac{dy}{dx}}{y^2} + 2y\frac{dy}{dx} = 0$.
4. ✔️ Multiply through by $y^2$ to eliminate the fraction: $y - x\frac{dy}{dx} + 2y^3\frac{dy}{dx} = 0$.
5. 🤝 Collect terms with $\frac{dy}{dx}$: $(2y^3 - x)\frac{dy}{dx} = -y$.
6. ➗ Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = \frac{-y}{2y^3 - x} = \frac{y}{x - 2y^3}$.

Example 3: Combining Product and Quotient Rules

Given: $\frac{xy}{x+y} = 1$

1. ✏️ Differentiate both sides with respect to $x$: $\frac{d}{dx}(\frac{xy}{x+y}) = \frac{d}{dx}(1)$.
2. ➗ Apply the quotient rule: $\frac{\frac{d}{dx}(xy)(x+y) - xy\frac{d}{dx}(x+y)}{(x+y)^2} = 0$.
3. ➕ Apply the product rule and simplify: $\frac{(y + x\frac{dy}{dx})(x+y) - xy(1+\frac{dy}{dx})}{(x+y)^2} = 0$.
4. ✔️ Multiply through by $(x+y)^2$: $(y + x\frac{dy}{dx})(x+y) - xy(1+\frac{dy}{dx}) = 0$.
5. 🧮 Expand: $xy + y^2 + x^2\frac{dy}{dx} + xy\frac{dy}{dx} - xy - xy\frac{dy}{dx} = 0$.
6. 🤝 Simplify and collect terms with $\frac{dy}{dx}$: $x^2\frac{dy}{dx} + y^2 = 0$.
7. ➗ Solve for $\frac{dy}{dx}$: $\frac{dy}{dx} = -\frac{y^2}{x^2}$.

📝 Practice Quiz

Test your understanding with these practice problems!

1. ❓ Find $\frac{dy}{dx}$ for $x^3 + y^3 = 8$.
2. ❓ Find $\frac{dy}{dx}$ for $x^2y - xy^2 = 2$.
3. ❓ Find $\frac{dy}{dx}$ for $\frac{x^2}{y} + y^3 = 1$.
4. ❓ Find $\frac{dy}{dx}$ for $xy = \sin(y)$.
5. ❓ Find $\frac{dy}{dx}$ for $\cos(xy) = x$.
6. ❓ Find $\frac{dy}{dx}$ for $x = \tan(y)$.
7. ❓ Find $\frac{dy}{dx}$ for $\sqrt{x} + \sqrt{y} = 4$.

💡 Conclusion

Implicit differentiation, especially when involving product and quotient rules, might seem daunting initially. However, with practice and a systematic approach, it becomes a manageable and valuable tool in calculus. Keep practicing, and you'll master it in no time!