Answer
$f^{n}(x)$ = $[x^{2}+2nx+n(n-1)]e^{x}$
Work Step by Step
$f(x)$ = $x^{2}e^{x}$
$f'(x)$ = $x^{2}e^{x}+2xe^{x}$ = $(x^{2}+2x)e^{x}$
$f''(x)$ = $(x^{2}+2x)e^{x}+(2x+2)e^{x}$ = $(x^{2}+4x+2)e^{x}$
$f''(x)$ = $(x^{2}+4x+2)e^{x}+(2x+4)e^{x}$ = $(x^{2}+6x+6)e^{x}$
$f^{4}(x)$ = $(x^{2}+6x+6)e^{x}+(2x+6)e^{x}$ = $(x^{2}+8x+12)e^{x}$
$f^{n}(x)$ = $[x^{2}+2nx+n(n-1)]e^{x}$
