For f(t) equal to (6t+6)/(t+9), f'(t) is given by [(t+9) times derivative of (6t+6) minus (6t+6) times derivative of (t+9)] divided by square of (t+9).
That is, f'(t) = [(t+9)(6) − (6t+6)(1)] ÷ (t+9)2 which simplifies to [6t + 54 − 6t − 6]/(t+9)2 or 48/(t+9)2.
Then f'(a) is equal to 48/(a+9)2.