We have
#1/(x^2-9) = 1/6 [1/(x-3)-1/(x+3)]#
Now, for #f(x) = 1/(x+a)#, we have
#d/dx f(x) = -1/(x+a)^2#
#d^2/dx^2 f(x) = ((-1)(-2))/(x+a)^3#
#d^3/dx^3 f(x) = ((-1)(-2)(-3))/(x+a)^4#
From these, the pattern
#d^n/dx^n f(x) = ((-1)(-2)...(-n))/(x+a)^(n+1)#
#qquad qquad =((-1)^n n!)/(x+a)^(n+1)#
Although the pattern should be pretty convincing, it is not a proof. The proof can easily be given by mathematical induction. Assuming that the formula works for #n# (we have already seen that it does work for #n=1,2#, and #3#), we can easily see that
#d^(n+1)/dx^(n+1) f(x) = d/dx ((-1)^n n!)/(x+a)^(n+1) #
#qquad = (-1)^n n! times (-(n+1))/(x+a)^(n+2) = ((-1)^(n+1) (n+1)!)/(x+a)^(n+2) #
So if the formula works for #n#, it works for #n+1#. Since it works for #n=1#, it works for all positive integers.
With this formula under our belt, it is easy to see that
#d^n/dx^n (1/(x^2-9)) = d^n/dx^n{1/6 [1/(x-3)-1/(x+3)]}#
#qquad = 1/6 [ d^n/dx^n(1/(x-3))- d^n/dx^n(1/(x+3))]#
#qquad = 1/6 [((-1)^n n!)/(x-3)^(n+1)-((-1)^n n!)/(x+3)^(n+1)]#
# qquad = ((-1)^n n!)/6[(x-3)^(-n-1)-(x+3)^(-n-1)]#
