$y = \frac{1}{x-3}$
$y '= (\frac{1}{x-3})'$
$y' = -\frac{(x-3)'}{(x-3)^2}=-\frac {1}{(x-3)^2}$
$y''=(-\frac {1}{(x-3)^2})'=\frac{((x-3)^2)'}{(x-3)^4}=\frac {2(x-3)}{(x-3)^4}=\frac {2}{(x-3)^3}$
$y^{(3)}=(\frac {2}{(x-3)^3})'=-\frac {2((x-3)^3)'}{(x-3)^6}=-\frac {2.3(x-3)^2}{(x-3)^6}=-\frac {2.3}{(x-3)^4}$
.
.
.
$y^{(n)}=\frac {(-1)^n.n!}{(x+3)^{n+1}} $. (n thuộc N)