f(x) - 1/(x^2 - 9)

First lt us rewrite:

==> f(x) = 1/(x-3)(x+3)

We could write as follow:

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

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

==> 1 = Ax + 3A + Bx - 3B

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

==> A+B = 0 ==? A = -B

==> 3(A-B) = 1

==> 3(-2b) = 1

==> B = -1/6

==> A = 1/6

==>: f(x) = 1/6(x-3) - 1/6(x+3)

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

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

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

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

==> f''''(x) = 24/6(x-3)^5 - 24/6(x+3)^5

==> f^(5) = 24*5/(x-3)^6 - 24*5/6(x+3)^6\

==> f^(6) = 24*5*6/6(x-3)^7 - 24*5*6/6(x+3)^7

= 120/(x-3)^7 - 120/(x+3)^7

y = 1/(x^2-9).

To fibd the 6th derivative.

Solution:

y = 1/(x^2-9) = (1/6){1/(x-3) -1/(x+3)}

y = (1/6)(1/(x-3) ) - (1/6)(1/(x+3))

We know (1/(x+a))' = -1/(x+a)^2

(1/(x+a))" = (-1)(-2)/(x+a)^3.

D6 {1/(x+a)} = (-1)(-2)(-3)(-4)(-5)(-6)/ (x+a)^7 = 6!*(-1)^6/(x+a)^7 = 6!/(x+a)^7.

Using this we get:

Therefore D6 y = D6 (1/6){1/(x-3) - 2/(x+3)} = 6! (1/6) { 1/(x-3)^7 - 1/(x+3)^7} = 5!{1/(x-3)^7 - 5!/(x+3)^7}

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).