Answer
If $f(x) = xe^x,~~$ then $~~f^{(n)}(x) = (x+n)e^x$
Work Step by Step
$f(x) = xe^x$
$f'(x) = e^x+xe^x = (x+1)e^x$
The statement is true for $n=1$
Suppose that $f^{(k)} = (x+k)e^x$
We can find $f^{(k+1)}$:
$f^{(k+1)} = e^x+(x+k)e^x = (x+k+1)e^x$
The statement is true for $n = k+1$
Therefore, by induction, the statement is true for all positive integers.
If $f(x) = xe^x,~~$ then $~~f^{(n)}(x) = (x+n)e^x$
