Question

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

Short answer

The general rule for \((xf(x))^{(n)}\) is \((xf(x))^{(n)} = nf'^{(n-1)}(x) + xf'^{(n)}(x)\)

Step-by-step solution

Understand the basic first derivative

The function being considered is \(xf(x)\), for which the first derivative can be calculated using the product rule of differentiation which is \((uv)'=u'v+uv'\). Here \(u=x\) and \(v=f(x)\), upon applying the rule, the first derivative becomes: \(f(x) + xf'(x)\).

Understand the second derivative

Now, let's consider the second derivative of the function `\(xf(x)\)`. For that purpose, we have to differentiate the result from Step 1: \(f(x) + xf'(x)\). Applying the product rule on the second term, the result becomes: \(f'(x) + f'(x) + xf''(x)\), which further simplifies to \(2f'(x) + xf''(x)\).

Recognise the pattern

By comparing the results from Step 1 and Step 2, a pattern can be recognised. The number of \(f'(x)\) is equal to the derivative number (n = 1, 2). Therefore, we can infer that the nth derivative will consist of n \(f'^{(n-1)}(x)\). The next step is to formulate this pattern into a general rule.

Formulate the rule

Applying the observed pattern, the nth derivative of \(xf(x)\) can be expressed as: \((Xf(x))^{(n)} = nf'^{(n-1)}(x) + xf'^{(n)}(x)\). This rule gives the nth derivative of the function \(xf(x)\).