Answer
(a) $S = \frac{-54.4x^{-0.6}}{(1+4x^{0.4})^2}$
(b) We can see a sketch of the graphs of $R$ and $S$ below.
Work Step by Step
(a) $R = \frac{40+24x^{0.4}}{1+4x^{0.4}}$
$S = \frac{dR}{dx} = \frac{(0.4)(24x^{-0.6})(1+4x^{0.4})-(40+24x^{0.4})(0.4)(4x^{-0.6})}{(1+4x^{0.4})^2}$
$S = \frac{9.6x^{-0.6}+38.4x^{-0.2}-38.4x^{-0.2}-64x^{-0.6}}{(1+4x^{0.4})^2}$
$S = \frac{-54.4x^{-0.6}}{(1+4x^{0.4})^2}$
(b) We can see a sketch of the graphs of $R$ and $S$ below.
At low levels of brightness $x$, we would expect the area of the pupil $R$ to be at a maximum, but to decrease quickly as the brightness $x$ increases.
At low levels of brightness, since the area of the pupil decreases quickly as the brightness $x$ increases, we would expect $S$ to be a large magnitude negative value, and then to decrease in magnitude as the brightness $x$ increases.
