Find the 2009-th derivative of the function f(x)=1/(x^2-4). Calculate for x=0.

First, we notice that the denominator of the function is a difference of squares, so we'll re-write it as:

 f(x)=1/(x^2-4) = 1/(x-2)(x+2)

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

1/(x-2)(x+2) = A/(x-2) + B/(x+2)

We'll calculate the common denominator, to the right side and we'll multiply each ratio with the corresponding amount.

1 = Ax+Bx + 2A - 2B

After factorization we'll get:

1 = x(A+B) + 2(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

2(A-B) = 1

A-B = 1/2, but A=-B

2A = 1/2

A=1/4

B=-1/4

1/(x-2)(x+2) = 1/4(x-2) - 1/4(x+2)

Now, we'll calculate the first derivative:

f'(x) = [1/4(x-2) - 1/4(x+2)]'

f'(x) = 4/16(x+2)^2 - 4/16(x-2)^2

f'(x) = 1/4(x+2)^2 - 1/4(x-2)^2

We'll calculate the second derivative:

f''(x) = 8(x-2)/16(x-2)^4 - 8(x+2)/16(x+2)^4

After reducing the terms:

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

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

f'''(x) = 2*3*4(x+2)^2/16(x+2)^6 - 2*3*4(x-2)^2/16(x-2)^6

After reducing the terms:

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

It is obvious that:

f(x)^(2009) = 2009!/4(x+2)^2010 - 2009!/4(x-2)^2010

Now, we'll calculate for x=0

We'll factorize:

f(0)^(2009) = 2009!/4(1/2^2010 - 1/2^2010)

f(0)^(2009) = 0