Answer
$$
s(x)=\frac{x}{m+nx},
$$
$m, n$ are constants
Then, by the quotient rule,
$$
s^{\prime}(x)=\left(\frac{u(x)}{v(x)}\right)^{\prime}=\frac{v(x) \cdot u^{\prime}(x)-u(x) v^{\prime}(x)}{[v(x)]^{2}}
$$
where $u(x)=x$ and $v(x)=m+nx$ , then we have
$$
\begin{aligned} s^{\prime}(x) &=\frac{(m+n x) 1-x(n)}{(m+n x)^{2}} \\ &=\frac{m+n x-n x}{(m+n x)^{2}} \\ &=\frac{m}{(m+n x)^{2}} \end{aligned}
$$
(b)
The rate of contraction when the concentration of the
drug is $x=50 , m=10 , n=3$ is given by
$$
\begin{aligned} s^{\prime}(50) &=\frac{10}{(10+50 (3) )^{2}}\\
&=\frac{1}{2560} \\
&\approx 0.000391 \text { mm. per ml.}.
\end{aligned}
$$
Work Step by Step
$$
s(x)=\frac{x}{m+nx},
$$
$m, n$ are constants
Then, by the quotient rule,
$$
s^{\prime}(x)=\left(\frac{u(x)}{v(x)}\right)^{\prime}=\frac{v(x) \cdot u^{\prime}(x)-u(x) v^{\prime}(x)}{[v(x)]^{2}}
$$
where $u(x)=x$ and $v(x)=m+nx$ , then we have
$$
\begin{aligned} s^{\prime}(x) &=\frac{(m+n x) 1-x(n)}{(m+n x)^{2}} \\ &=\frac{m+n x-n x}{(m+n x)^{2}} \\ &=\frac{m}{(m+n x)^{2}} \end{aligned}
$$
(b)
The rate of contraction when the concentration of the
drug is $x=50 , m=10 , n=3$ is given by
$$
\begin{aligned} s^{\prime}(50) &=\frac{10}{(10+50 (3) )^{2}}\\
&=\frac{1}{2560} \\
&\approx 0.000391 \text { mm. per ml.}.
\end{aligned}
$$
