Find the point of inflection for f(x)=(x-2)/(x^2-3x-4)I need help finding the second derivative

Indeed, to determine the inflection point of the function,
we'll have to find out the 2nd derivative.

First, we need
to find out the 1st derivative and for this reason, we'll use the quotient
rule:

(u/v)' = (u'*v -
u*v')/v^2

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

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

f'(x) = (x^2 - 3x - 4 - 2x^2
+ 7x - 6)/(x^2 - 3x - 4)^2

f'(x) = (-x^2 + 4x - 10)/(x^2 -
3x - 4)^2

We'll differentiate now
f'(x):

f"(x) = {(-x^2 + 4x - 10)'*(x^2 - 3x - 4)^2 - (-x^2
+ 4x - 10)*[(x^2 - 3x - 4)^2]'}/(x^2 - 3x - 4)^4

f"(x)
=[(-2x + 4)*(x^2 - 3x - 4)^2 - 2(x^2 - 3x - 4)*(2x-3)*(-x^2 + 4x - 10)]/(x^2 - 3x -
4)^4

f"(x) = [(-2x + 4)*(x^2 - 3x - 4) - 2*(2x-3)*(-x^2 +
4x - 10)]/(x^2 - 3x - 4)^3

f"(x) = (-2x^3 + 6x^2 + 8x +
4x^2 - 12x - 16 + 4x^3 - 16x^2 + 40x - 6x^2 + 24x - 60)/(x^2 - 3x -
4)^3

f"(x) = (2x^3 - 12x^2 + 60x - 76)/(x^2 - 3x -
4)^3

f"(x) = 2(x^3 - 6x^2 + 30x - 38)/(x^2 - 3x -
4)^3

The second derivative of the function is
f"(x) = 2(x^3 - 6x^2 + 30x - 38)/(x^2 - 3x -
4)^3.