Step-by-Step Solution for Part (a):

We are given the function: $f(x) = \frac{Kx}{A + x}$ We need to find the rate of change of $f(x)$ with respect to $x$, i.e., $\frac{df}{dx}$.

Using the quotient rule for derivatives: $\frac{d}{dx}\left(\frac{u}{v}\right) = \frac{u'v - uv'}{v^2}$ where $u = Kx$ and $v = A + x$.

Step 1: Compute $u'$ and $v'$

• $u = Kx$, so $u' = K$.
• $v = A + x$, so $v' = 1$.

Step 2: Apply the quotient rule

$\frac{df}{dx} = \frac{(K)(A + x) - (Kx)(1)}{(A + x)^2}$

Simplify the numerator: $\frac{df}{dx} = \frac{KA + Kx - Kx}{(A + x)^2}$ $\frac{df}{dx} = \frac{KA}{(A + x)^2}$

So, the rate of change of the growth rate with respect to $x$ is: $\boxed{\frac{KA}{(A + x)^2}}$

Step-by-Step Solution for Part (b):

We are tasked to find the rate of change of the growth rate when $x = A$.

Substitute $x = A$ into $\frac{df}{dx}$:

$\frac{df}{dx} = \frac{KA}{(A + x)^2}$ When $x = A$: $\frac{df}{dx} = \frac{KA}{(A + A)^2}$ $\frac{df}{dx} = \frac{KA}{(2A)^2}$ $\frac{df}{dx} = \frac{KA}{4A^2}$ $\frac{df}{dx} = \frac{K}{4A}$

So, the rate of change of the growth rate when $x = A$ is: $\boxed{\frac{K}{4A}}$

Questions and Tip:

1. What is the interpretation of $f(x)$ in the context of population growth?
2. How does $A$ affect the maximum growth rate of the population?
3. What happens to $\frac{df}{dx}$ as $x \to 0$ or $x \to \infty$?
4. Can you explain why the rate of change decreases for large values of $x$?
5. How would you compute the second derivative of $f(x)$ for more detailed growth dynamics?

Tip: Always double-check calculations involving derivatives by simplifying each step systematically.