Let f(x) = xe^x

Question

Find f '(x), f "(x), and f "'(x). Based on your results, find a general formula for f^n (x).

Answer 1

This can be done by doing the first few derivatives and discerning a pattern. We begin with f'(x).

f(x) = xe^x

Using the Product Rule [(uv)' = uv' + vu'], we can set u = x and v = e^x, and therefore, u' = 1 and v' = e^x.

This makes the derivative f'(x) = xe^x + e^x = e^x(x + 1).

Next we find f''(x). f''(x) = e^x + e^x(x + 1) = e^x(1 + x + 1) = e^x(x + 2).

We'll do one more, since that is what the question calls for.

f'''(x) = e^x + e^x(x + 2) = e^x(1 + x + 2) = e^x(x + 3).

The pattern that is emerging is that, with each new derivative, a new e^x gets added, or the constant inside the binomial (which multiplies by e^x) increases by 1. In fact, the constant inside the parenthesis is the same number as the number of derivatives that you've done. So the general formula would be f⁽ⁿ⁾(x) = e^x(x + n).