Answer
$$
f(x)=\frac{Kx}{A+x},
$$
$K, A$ are constants
Then, by the quotient rule,
$$
f^{\prime}(x)=\left(\frac{u(x)}{v(x)}\right)^{\prime}=\frac{v(x) \cdot u^{\prime}(x)-u(x) v^{\prime}(x)}{[v(x)]^{2}}
$$
where $u(x)=Kx$ and $v(x)=A+x$ , then
(a)
the rate of change of the growth rate with respect to the amount of food is given by
$$
\begin{aligned} f^{\prime}(x) &=\frac{(A+ x) K-Kx(1)}{(A+ x)^{2}} \\ &=\frac{A K}{(A+ x)^{2}} \\
\end{aligned}
$$
(b)
the rate of change of the growth rate when $x=A$ is given by
$$
\begin{aligned} f^{\prime}(x) &=\frac{A K}{(A+ A)^{2}} \\
&=\frac{A K}{(2A)^{2}} \\
&=\frac{A K}{4A^{2}} \\
&=\frac{ K}{4A}. \\
\end{aligned}
$$
Work Step by Step
$$
f(x)=\frac{Kx}{A+x},
$$
$K, A$ are constants
Then, by the quotient rule,
$$
f^{\prime}(x)=\left(\frac{u(x)}{v(x)}\right)^{\prime}=\frac{v(x) \cdot u^{\prime}(x)-u(x) v^{\prime}(x)}{[v(x)]^{2}}
$$
where $u(x)=Kx$ and $v(x)=A+x$ , then
(a)
the rate of change of the growth rate with respect to the amount of food is given by
$$
\begin{aligned} f^{\prime}(x) &=\frac{(A+ x) K-Kx(1)}{(A+ x)^{2}} \\ &=\frac{A K}{(A+ x)^{2}} \\
\end{aligned}
$$
(b)
the rate of change of the growth rate when $x=A$ is given by
$$
\begin{aligned} f^{\prime}(x) &=\frac{A K}{(A+ A)^{2}} \\
&=\frac{A K}{(2A)^{2}} \\
&=\frac{A K}{4A^{2}} \\
&=\frac{ K}{4A}. \\
\end{aligned}
$$
