Step 1 :The problem is asking for the change in population of bacteria from $x=1$ to $x=1.5$. This can be approximated using the differential of the function $P(x)$, which represents the rate of change of the population with respect to time. The differential of a function is given by its derivative. So, first we need to find the derivative of $P(x)$, then we can use this to approximate the change in population.
Step 2 :Let's find the derivative of $P(x)$. Given $P(x)=\frac{20x}{5+x^{2}}$, the derivative $P'(x)$ is calculated as $P'(x) = -\frac{40x^{2}}{(x^{2} + 5)^{2}} + \frac{20}{x^{2} + 5}$.
Step 3 :Next, we calculate the change in $x$, which is $\Delta x = 1.5 - 1 = 0.5$.
Step 4 :Then, we use the derivative to approximate the change in population, $\Delta P = P'(1) \cdot \Delta x$.
Step 5 :By substituting the values into the equation, we get $\Delta P = 1.111$ million.
Step 6 :Final Answer: The approximate change in population of bacteria from $x=1$ to $x=1.5$ is \(\boxed{1.111}\) million.