Problem 40 - Calculus - AP Edition

Question

Suppose that the function \(f\) is differentiable everywhere and \(F(x)=x f(x)\). (a) Express \(F^{\prime \prime \prime}(x)\) in terms of \(x\) and derivatives of \(f\). (b) For \(n \geq 2,\) conjecture a formula for \(F^{(n)}(x)\).

Step-by-Step Solution

Answer

(a) \(F'''(x) = 3f''(x) + x f'''(x)\); (b) \(F^{(n)}(x) = n f^{(n-1)}(x) + x f^{(n)}(x)\)."

Step 1: Apply Product Rule for First Derivative

To find the first derivative, use the product rule on the expression \(F(x) = x f(x)\). The product rule states \((uv)' = u'v + uv'\). Here, \(u = x\) and \(v = f(x)\) which gives us, \(F'(x) = 1 \cdot f(x) + x \cdot f'(x) = f(x) + x f'(x)\).

Step 2: Calculate Second Derivative

Differentiate \(F'(x) = f(x) + x f'(x)\) again using the product rule. The derivative of \(f(x)\) is \(f'(x)\), and for \(x f'(x)\) apply the product rule again: \((x f'(x))' = 1 \cdot f'(x) + x f''(x)\). Therefore, \(F''(x) = f'(x) + f'(x) + x f''(x) = 2f'(x) + x f''(x)\).

Step 3: Calculate Third Derivative

Differentiate \(F''(x) = 2f'(x) + x f''(x)\) using the product rule. The derivative of \(2f'(x)\) is \(2f''(x)\), and for \(x f''(x)\) apply the product rule: \((x f''(x))' = 1 \cdot f''(x) + x f'''(x)\). Thus, \(F'''(x) = 2f''(x) + f''(x) + x f'''(x) = 3f''(x) + x f'''(x)\).

Step 4: Conjecture a General Formula for \(F^{(n)}(x)\)

Observe the patterns in the derivatives calculated: \(F'(x) = f(x) + x f'(x)\), \(F''(x) = 2f'(x) + x f''(x)\), \(F'''(x) = 3f''(x) + x f'''(x)\). Notice each derivative has a pattern: \(F^{(n)}(x) = n f^{(n-1)}(x) + x f^{(n)}(x)\). Therefore, the general case can be conjectured as this formula.

Key Concepts

Product Rule

The product rule is a critical tool in calculus for finding the derivative of products of two functions. If we have two functions, say \( u(x) \) and \( v(x) \), the product rule helps us find the derivative of their product. The rule states that if \( F(x) = u(x) \cdot v(x) \), then the derivative, \( F'(x) \), is given by:\[F'(x) = u'(x) \cdot v(x) + u(x) \cdot v'(x)\]
This means we differentiate one function while keeping the other as it is, then sum the results after switching which function is differentiated.

• In our example, \(F(x) = x \cdot f(x)\), we have \( u(x) = x \) and \( v(x) = f(x) \).
• Applying the product rule, \(F'(x) = 1 \cdot f(x) + x \cdot f'(x) = f(x) + x \cdot f'(x)\).

This process is essential in calculus because many real-world problems involve functions that are products of simpler functions. Understanding the product rule can help untangle complex relationships in functions by breaking them into manageable parts.

Higher-order Derivatives

Higher-order derivatives refer to the derivatives of a function taken multiple times. The first derivative, \(F'(x)\), provides the slope of the function or how fast \(F(x)\) changes with respect to \(x\). Successive derivatives, like \(F''(x)\), \(F'''(x)\), and so forth, give us even deeper insights into the function's behavior.
In the context of our exercise, we start with \(F(x) = x \cdot f(x)\) and use the product rule to find the first derivative, \(F'(x)\). To find the second derivative, \(F''(x)\), you again employ the product rule on \(F'(x)\):

• The first part of \(F'(x) = f(x) + x \cdot f'(x)\) is \(f(x)\), leading to \(f'(x)\) as its derivative.
• For the second part, \(x \cdot f'(x)\), using the product rule gives \(1 \cdot f'(x) + x \cdot f''(x)\).
• Hence, \(F''(x) = 2f'(x) + x \cdot f''(x)\).

Higher-order derivatives like \(F'''(x) = 3f''(x) + x \cdot f'''(x)\) help identify patterns and general behaviors, crucial for understanding advanced calculus problems. Incorporating these insights helps predict and explain the function's long-term behavior and changes.

Differentiable Functions

A function is called differentiable if it has a derivative at all points within its domain. Differentiability implies the function is smooth and continuous, meaning there are no sharp edges or jumps.
In calculus, when we deal with differentiable functions, we are assured of a couple of essential features:

• The function is continuous over its domain, which means there are no breaks or gaps.
• The function has a tangent at every point within the domain, indicative of a well-defined slope.

The function \(f(x)\) in our exercise is differentiable everywhere, allowing us to calculate not just the first derivative, but also higher-order derivatives like the second, third, and so on. This property simplifies complex calculus operations because it establishes a baseline level of smoothness and predictability. Such characteristics make it feasible to use calculus rules safely and expect consistent behavior across the function's domain.