Solved Problems: Higher-Order Derivatives with Product, Quotient, and Chain Rules

📚 Higher-Order Derivatives: A Comprehensive Guide

Higher-order derivatives are simply derivatives of derivatives. If you differentiate a function once, you get its first derivative. Differentiating the first derivative gives you the second derivative, and so on. They tell us about the rate of change of the rate of change, and so on. Understanding them is crucial in fields like physics, engineering, and economics, where rates of change are constantly analyzed.

📜 History and Background

The concept of derivatives, including higher-order derivatives, was developed in the 17th century by Isaac Newton and Gottfried Wilhelm Leibniz independently. Their work laid the foundation for calculus and its applications in various scientific disciplines. The notation for derivatives evolved over time, with Leibniz's notation (e.g., $\frac{d^2y}{dx^2}$) being particularly useful for representing higher-order derivatives.

🔑 Key Principles

• 🔍 Notation: The $n$th derivative of a function $f(x)$ can be denoted as $f^{(n)}(x)$, $\frac{d^n}{dx^n}f(x)$, or $y^{(n)}$.
• 📏 Power Rule: If $f(x) = x^n$, then $f'(x) = nx^{n-1}$, $f''(x) = n(n-1)x^{n-2}$, and so on.
• ➕ Sum/Difference Rule: The derivative of a sum or difference of functions is the sum or difference of their derivatives. For example, $(f(x) + g(x))' = f'(x) + g'(x)$.
• ✖️ Product Rule: The derivative of the product of two functions $u(x)$ and $v(x)$ is given by $(uv)' = u'v + uv'$. For higher-order derivatives, you apply the product rule repeatedly.
• ➗ Quotient Rule: The derivative of the quotient of two functions $u(x)$ and $v(x)$ is given by $(\frac{u}{v})' = \frac{u'v - uv'}{v^2}$. Again, apply it repeatedly for higher-order derivatives.
• ⛓️ Chain Rule: The derivative of a composite function $f(g(x))$ is given by $(f(g(x)))' = f'(g(x))g'(x)$. This rule is particularly important and can become complex with higher-order derivatives.

💡 Real-World Examples

Example 1: Product Rule

Let $y = x^2 \sin(x)$. Find the second derivative, $y''$.

First derivative: $y' = 2x \sin(x) + x^2 \cos(x)$

Second derivative: $y'' = (2 \sin(x) + 2x \cos(x)) + (2x \cos(x) - x^2 \sin(x)) = 2 \sin(x) + 4x \cos(x) - x^2 \sin(x)$

Example 2: Quotient Rule

Let $y = \frac{e^x}{x}$. Find the second derivative, $y''$.

First derivative: $y' = \frac{e^x \cdot x - e^x \cdot 1}{x^2} = \frac{e^x(x - 1)}{x^2}$

Second derivative: $y'' = \frac{[e^x(x - 1) + e^x]x^2 - 2x[e^x(x - 1)]}{x^4} = \frac{e^x[x^2 - 2x + 2]}{x^3}$

Example 3: Chain Rule

Let $y = \cos(x^2)$. Find the second derivative, $y''$.

First derivative: $y' = -2x \sin(x^2)$

Second derivative: $y'' = -2 \sin(x^2) - 4x^2 \cos(x^2)$

📝 Practice Problems

Find the second derivative for each of the following functions:

1. 📐 $f(x) = x^3e^{2x}$
2. 📈 $g(x) = \frac{\ln(x)}{x}$
3. 🧮 $h(x) = \sin(3x^2)$

✅ Solutions to Practice Problems

1. ✨ $f'(x) = 3x^2e^{2x} + 2x^3e^{2x}$, $f''(x) = 6xe^{2x} + 6x^2e^{2x} + 6x^2e^{2x} + 4x^3e^{2x} = e^{2x}(4x^3 + 12x^2 + 6x)$
2. 🌟 $g'(x) = \frac{1 - \ln(x)}{x^2}$, $g''(x) = \frac{-3 + 2\ln(x)}{x^3}$
3. 💫 $h'(x) = 6x\cos(3x^2)$, $h''(x) = 6\cos(3x^2) - 36x^2\sin(3x^2)$

🎓 Conclusion

Mastering higher-order derivatives requires a solid understanding of the basic differentiation rules (product, quotient, and chain rules) and consistent practice. By working through examples and exercises, you can develop the skills needed to confidently tackle these types of problems. Good luck! 🎉