Rate of change with $a(t)=\frac{1}{t+4}$ on $[9,9+h]$ I am working on an exercise to find the rate of change between points $[9, 9+h]$ with the function $a(t)=\frac{1}{t+4}$.
The solution provided is $\frac{-1}{13(13+h)}$ whereas I arrive at $\frac{\frac{1}{h}}{h}$.
My working:
$a(t_1)$ = $\frac{1}{9+4}$ = $\frac{1}{13}.$
$x(t_2)$ = $\frac{1}{9+h+4}$ = $\frac{1}{13+h}.$
The rate of change is: $\frac{a(t_2)-a(t_1)}{t_2-t_1}.$ 
So: $\dfrac{\frac{1}{13+h}-\frac{1}{13}}{9+h-9}$ = $\dfrac{\frac{1}{13}+\frac{1}{h}-\frac{1}{13}}{h}$ = $\dfrac{\frac{1}{h}}{h}.$
Where did I go wrong and how can I arrive at $\frac{-1}{13(13+h)}$?

Because$$\frac1{13+h}\neq\frac1{13}+\frac1h.$$In fact\begin{align}\frac1{13+h}-\frac1{13}&=\frac{13}{(13+h)13}-\frac{13+h}{(13+h)13}\\&=\frac{13-13-h}{(13+h)13}\\&=-\frac h{(13+h)13}.\end{align}
