Answer
$\frac{dS}{dw}=\frac{\sqrt 5}{20\sqrt {w}}$
$S$ will increase more rapidly with respect to weight at lower body weights.
Work Step by Step
Step 1. Given the equation $S=\frac{1}{60}\sqrt {wh}$ and $h=180$ cm, we have $S=\frac{1}{60}\sqrt {180w}=\frac{\sqrt 5}{10}\sqrt {w}$
Step 2. The rate of change of body surface area with respect to weight is $\frac{dS}{dw}=\frac{\sqrt 5}{20\sqrt {w}}$
Step 3. We can see that for higher weight $w$, the rate of increase will be smaller; thus $S$ will increase more rapidly with respect to weight at lower body weights.
