Answer
\[
\frac{0.0021}{(0.015+[S])^{2}}=\frac{d v}{d[S]}
\]
Work Step by Step
$\frac{0.14[S]}{0.015+[S]}=v$
To find $\frac{d v}{d[S]}$ use $\left(\frac{f}{g}\right)^{\prime}=\frac{g f^{\prime}-f g^{\prime}}{g^{2}}$
$\begin{aligned} &\frac{(0.015+[S]) \cdot 0.14-0.14[S] \cdot 1}{(0.015+[S])^{2}}= \frac{d v}{d[S]} \\ &=\frac{0.0021}{(0.015+[S])^{2}} \end{aligned}$
$d v / d[S]$ will tell you how $\mathrm{v}$ will change if $[\mathrm{S}]$ (the concentration of the substrate) changes.
