I used an image because I already had the format done:
$\displaystyle P = \frac{130R}{(R+0.8)^2}$
$\displaystyle \frac{dP}{dR} = \frac{(R+0.8)^2 \cdot 130 - 130R \cdot 2(R+0.8)}{(R+0.8)^4}$
$\displaystyle \frac{dP}{dR} = \frac{130(R+0.8)[(R+0.8)-2R]}{(R+0.8)^4}$
$\displaystyle \frac{dP}{dR} = \frac{130(0.8-R)}{(R+0.8)^3}$