Problem 39 - Calculus - AP Edition

Question

Find a general formula for \(F^{\prime \prime}(x)\) if \(F(x)=x f(x)\) and \(f\) and \(f^{\prime}\) are differentiable at \(x\).

Step-by-Step Solution

Answer

The second derivative is \( F''(x) = 2 f'(x) + x f''(x) \).

Step 1: Differentiate the First Time

Given the function \( F(x) = x f(x) \), we need to find its second derivative \( F''(x) \). To do this, we start by calculating the first derivative \( F'(x) \). We apply the product rule, which states that \( (uv)' = u'v + uv' \). Here, let \( u = x \) and \( v = f(x) \). Then \( u' = 1 \) and \( v' = f'(x) \). Therefore, \( F'(x) = 1 \cdot f(x) + x \cdot f'(x) = f(x) + x f'(x) \).

Step 2: Differentiate the Second Time

Now we will differentiate \( F'(x) = f(x) + x f'(x) \) to find \( F''(x) \). The derivative of \( f(x) \) is \( f'(x) \). For the second term \( x f'(x) \), we again use the product rule: \( (x f'(x))' = (x)' f'(x) + x (f'(x))' = f'(x) + x f''(x) \). Thus, \( F''(x) = f'(x) + f'(x) + x f''(x) = 2 f'(x) + x f''(x) \).

Key Concepts

Understanding the Product Rule

The product rule is a crucial tool in calculus, especially when dealing with the differentiation of products of two functions. It's like having a reliable method to tackle those situations where functions multiply each other.
The rule states that if you have two differentiable functions, say \( u(x) \) and \( v(x) \), their derivative when multiplied is given by:

• \((uv)' = u'v + uv'\)

In simple terms, you differentiate the first function and multiply by the second, then add the product of the first with the derivative of the second.
This method allows us to methodically break down complex problems, as seen in the example where we needed to find the derivative of \( F(x) = x \, f(x) \). Here, we let \( u = x \) and \( v = f(x) \).
The derivatives became \( u' = 1 \) and \( v' = f'(x) \), which, when applied to the rule, give us a straightforward path to the solution.

Concept of Differentiation

Differentiation is the backbone of calculus. It's all about finding how things change; in math terms, it's figuring out the rate at which one quantity changes relative to another.
When you differentiate a function, you're looking to find something called the derivative, which provides the rate of change at any point. This is tremendously useful in analyzing how functions behave.
In our example, we needed to differentiate \( F(x) = x f(x) \), which led us to first calculate the first derivative \( F'(x) \) using the product rule. Once we found \( F'(x) = f(x) + x f'(x) \), we dug deeper to get the second derivative, showing us how the rate of change itself changes.
Differentiation allows us to understand not just how functions grow or shrink but also how their behavior curves and oscillates over intervals. Mastering this concept means gaining insights into predicting behavior in fields ranging from physics to economics.

Effective Calculus Problem Solving

Solving calculus problems can seem daunting at first, but following systematic rules and understanding principles makes it manageable. Here, conceptual knowledge blends with procedural skills to tackle problems turn by turn.
When faced with the problem of finding \( F''(x) \) for \( F(x) = x f(x) \), we approached it step-by-step:

• First, apply the product rule to find the first derivative \( F'(x) \).
• Second, differentiate \( F'(x) \) again, using the rules of differentiation, such as the product rule and knowing the derivatives of simpler functions.

Remember, calculus isn't just about memorizing formulas, but about understanding the logic behind the steps. Practicing these steps in multiple scenarios strengthens your calculus skills.
Breaking down problems, identifying which rules apply, and executing the operations can make calculus a powerful tool in your academic toolkit. As demonstrated, correctly applying the product rule twice helped us find the sought second derivative, \( F''(x) = 2f'(x) + x f''(x) \), a solution that tells us even more about how the function behaves beyond just its surface changes.