Determine the sixth derivative of y=1/x^2-9 .

Because of the fact that the denominator of the function
is a difference of squares, we'll re-write it as:

 y =
1/(x^2-9)

y =
1/(x-3)(x+3)

Now, we'll write the function as a sum or
difference of 2 irreducible ratios.

1/(x-3)(x+3) = A/(x-3)
+ B/(x+3)

We'll calculate the least common denominator of
the 2 ratios and we'll multiply each ratio with the proper
value.

1 = Ax+Bx + 3A -
3B

After factorization we'll
get:

1 = x(A+B) + 3(A-B)

The
expression from the left side could be written as:

1 = 0*x
+ 1

So, for the identity to hold, the both expressions from
both sides have to correspond.

A+B =
0

3(A-B) = 1

We'll divide by 3
both sides:

A-B = 1/3

 but A =
-B

2A = 1/3

We'll divide by
2:

A=1/6

B=-1/6

1/(x-3)(x+3)
= 1/6(x-3) - 1/6(x+3)

Now, we'll calculate the first
derivative:

f'(x) = [1/6(x-3) -
1/6(x+3)]'

f'(x) = 6/36(x+3)^2 -
6/36(x-3)^2

f'(x) = 1/6(x+3)^2 -
1/6(x-3)^2

We'll calculate the second
derivative:

f''(x) = 12(x-3)/36(x-3)^4 -
12(x+2)/36(x+3)^4

After reducing the
terms:

f''(x) = 2/6(x-3)^3 -
2/6(x+3)^3

And to establish a final form, we'll calculate
the third derivative:

f'''(x) = 2*3*6(x+3)^2/36(x+3)^6 -
2*3*6(x-3)^2/36(x-3)^6

After reducing the
terms:

f'''(x) = 2*3/6(x+2)^4 -
2*3/6(x-2)^4

We can conclude that the 6-th derivative
is:

f(x)^(VI) = 6!/6(x-3)^7 -
6!/6(x+3)^7

Note: We'll write the 6-th
derivative as f(x)^(VI).