differentiation

(a) f(x) = [1/(x^2+5x+6)] = 1/[(x+2)(x+3)]

Let

1/(x^2+5x+6) = A/(x+2) + B/(x+3)

1/(x^2+5x+6) = [A(x+3)+B(x+2)]/[(x+2)(x+3)]

1/(x^2+5x+6) = [(A+B)x+(3A+2B)]/[(x+2)(x+3)]

Comparing the coefficients,

We find that A = 1, B = -1

f(x) = 1/(x+2) - 1/(x+3)

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

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