How to Calculate Second Derivatives Using the Product Rule Step-by-Step

📚 Understanding Second Derivatives and the Product Rule

In calculus, the second derivative of a function tells us about the rate of change of the rate of change – essentially, it describes the concavity of the function. When dealing with products of functions, the product rule becomes indispensable. Combining these concepts can seem daunting, but a step-by-step approach simplifies the process. This guide will walk you through calculating second derivatives using the product rule effectively.

📜 Historical Context

The development of calculus, largely attributed to Isaac Newton and Gottfried Wilhelm Leibniz in the late 17th century, provided the tools for analyzing rates of change and areas under curves. The product rule, a fundamental concept within differential calculus, emerged as a necessary tool for differentiating functions that are products of other functions. As calculus matured, its applications expanded into diverse fields, requiring more complex differentiation techniques, including the calculation of second derivatives of products.

🔑 Key Principles

• 🍎 Product Rule: The product rule states that the derivative of two functions, $u(x)$ and $v(x)$, is given by $(uv)' = u'v + uv'$.
• ⚙️ Second Derivative: The second derivative, denoted as $f''(x)$ or $\frac{d^2y}{dx^2}$, is the derivative of the first derivative, $f'(x)$.
• 💡 Applying the Product Rule Twice: When finding the second derivative of a product, you often need to apply the product rule twice – once for the first derivative, and again when differentiating the terms that arise from the first application.

🪜 Step-by-Step Calculation

Let's break down the process with a clear example. Suppose we want to find the second derivative of $y = x^2 \sin(x)$.

1. 🔍 Step 1: Find the First Derivative

   Using the product rule, where $u(x) = x^2$ and $v(x) = \sin(x)$:

   $u'(x) = 2x$

   $v'(x) = \cos(x)$

   Therefore, $y' = (x^2)'\sin(x) + x^2(\sin(x))' = 2x\sin(x) + x^2\cos(x)$.
2. 📝 Step 2: Find the Second Derivative

   Now we differentiate $y' = 2x\sin(x) + x^2\cos(x)$. Notice that we have two product terms: $2x\sin(x)$ and $x^2\cos(x)$. We need to apply the product rule to each of these terms.

   For $2x\sin(x)$: Let $u(x) = 2x$ and $v(x) = \sin(x)$. Then $u'(x) = 2$ and $v'(x) = \cos(x)$. The derivative is $2\sin(x) + 2x\cos(x)$.

   For $x^2\cos(x)$: Let $u(x) = x^2$ and $v(x) = \cos(x)$. Then $u'(x) = 2x$ and $v'(x) = -\sin(x)$. The derivative is $2x\cos(x) - x^2\sin(x)$.

   Combining these, we get: $y'' = (2\sin(x) + 2x\cos(x)) + (2x\cos(x) - x^2\sin(x)) = 2\sin(x) + 4x\cos(x) - x^2\sin(x)$.

✍️ Real-world Examples

• 📈 Physics: Analyzing the motion of a damped oscillator where the position is described by a product of a sinusoidal function and an exponential decay function.
• 💰 Economics: Modeling revenue as a product of price and quantity, where both price and quantity are functions of time.
• 🌡️ Engineering: Determining the rate of heat transfer in a system where temperature and area are both changing.

✅ Practice Problems

Let's solidify your understanding with some practice problems. Find the second derivative of the following functions:

1. ❓ $y = x e^x$
2. ❓ $y = x^3 \ln(x)$
3. ❓ $y = \cos(x) \sin(x)$

🔑 Solutions to Practice Problems

1. ✅ $y = x e^x$: $y'' = e^x + e^x + xe^x = 2e^x + xe^x $
2. ✅ $y = x^3 \ln(x)$: $y' = 3x^2 \ln(x) + x^2$, $y'' = 6x\ln(x) + 3x + 2x = 6x\ln(x) + 5x$
3. ✅ $y = \cos(x) \sin(x) = \frac{1}{2}\sin(2x)$: $y' = \cos(2x)$, $y'' = -2\sin(2x)$

📝 Conclusion

Calculating second derivatives using the product rule involves careful application of the rule and attention to detail. By breaking down the process into manageable steps and practicing with diverse examples, you can master this essential calculus skill. Remember, patience and persistence are key to success in calculus!