Question

Develop a general rule for \([x f(x)]^{(n)}\) where \(f\) is a differentiable function of \(x\).

Short answer

The general rule for the n-th derivative of \([x f(x)]^{(n)}\) where \(f\) is a differentiable function of \(x\) is sum from \(i=0\) to \(n\) of \([nC_{i} x^{n-i} (f^{i}(x))]\).

Step-by-step solution

Prerequisites

Firstly, it's important to remember the product rule for derivatives: (uv)' = u'v + uv'. And Leibniz rule which says, the n-th derivative of a product of two functions is a sum of binomial coefficient multiplied terms involving derivatives of both functions.

Find First Few Derivatives

Let's start by finding first few derivatives to see if a pattern emerges. When \(n = 1\), \([x f(x)]^{(1)} = f(x) + x f'(x)\). When \(n = 2\), \([x f(x)]^{(2)} = f'(x) + f'(x) + x f''(x)\). When \(n = 3\), \([x f(x)]^{(3)} = f''(x) + 2f''(x) + 2f''(x) + x f'''(x)\). Note the pattern that emerges, in the nth derivative, the coefficient of \(f^{(i)}(x)\) is the number of ways to pick \(i\) items from \(n\) which is expressed as \(nC_{i}\).

Formulate General Rule

The nth derivative of \(x f(x)\) then becomes the sum from \(i=0\) to \(n\) of \([nC_{i} x^{n-i} (f^{i}(x))]\).