Re: HSC 2013 2U Marathon
y=uv \ \ y'=u'v +v'u
y''=u''v +2 u'v' +v''u \ \ y'''= u'''v +3u''v' + 3 u'v'' +uv''' \ \ etc
(1+x)^2 = 1+2x+x^2 \ \ (1+x)^3= 1+ 3x +3x^2 +x^3 \ \ etc
$pattern is pascals triangle$
\therefore \frac{d^5}{dx^5}(e^{-x} (1+x^2)) = -e^{-x} (x^2 +10x +21)