Here is the calculus that you can make. I note $\tau_{GDPc} $ the rate of change of the per capita GDP, $\tau_{pop}$ the one for the population and $\tau_{GDP}$ for the GDP rate of change.
A rate of change for any variable A is
$$ \tau _A = \frac{A(t)-A(t-1)}{A(t-1)}$$
t being the moment considered, $(t-1)$ the previous moment for the calculus of the rate of change. So knowing $GDPc = \frac{GDP}{pop}$ you can write:
$$ \tau_{GDPc} = \frac{\frac{GDP(t)}{pop(t)} - \frac{GDP(t-1)}{pop(t-1)}}{ \frac{GDP(t-1)}{pop(t-1)}}$$
With the definition of the rate of change, you write
$$pop(t)= pop(t-1)*(\tau_{pop}+1) $$
You can then simplify the $pop(t-1)$ term in the previous equation and you get
$$\tau_{GDPc} = \frac{1}{\tau_{pop}+1}\left[\frac{GDP(t)-GDP(t-1)}{GDP(t-1)}-\tau_{pop}\right] $$
$$\tau_{GDPc} = \frac{1}{\tau_{pop}+1}\left[\tau_{GDP}-\tau_{pop}\right]$$
It is indeed different than Brandon marcus answer.
We can make a try with simple numbers:
- let's consider at t-1 a population of 100 for a GDP of 100. The per capita income is then 1.
- At t, you will have a population of 102,5 fo a GDP of 101,5, that is a per capita of 0,99024.
The rate of chage of the per capita income will be $\frac{0,99024-1}{1} \simeq -0,975 \%$.
With the formula I gave you:
$$\tau_{GDPc} = \frac{1}{1,025}(0,015 - 0,025) = -0,975 \% $$
Seems to work.