To find $P'(t)$, the derivative of the given function with respect to $t$, let's first write the function:

$P(t) = 400 \left( 4 + \frac{4t}{45 + t^2} \right)$

We will differentiate this function using the chain rule and the quotient rule where necessary.

Step 1: Differentiate the Constant Multiple

The function $P(t)$ has a constant multiple of 400. The derivative of a constant times a function is the constant times the derivative of the function. So we focus on differentiating the inner function:

$P(t) = 400 \left( 4 + \frac{4t}{45 + t^2} \right)$

The derivative of 4 is 0, so we only need to differentiate the second term $\frac{4t}{45 + t^2}$.

Step 2: Apply the Quotient Rule

The second term is a quotient, so we apply the quotient rule:

$\frac{d}{dt} \left( \frac{4t}{45 + t^2} \right) = \frac{(45 + t^2)(4) - 4t(2t)}{(45 + t^2)^2}$

Simplifying this:

$\frac{(45 + t^2)(4) - 4t(2t)}{(45 + t^2)^2} = \frac{180 + 4t^2 - 8t^2}{(45 + t^2)^2}$

$= \frac{180 - 4t^2}{(45 + t^2)^2}$

Step 3: Combine the Results

Now, multiplying this result by the constant 400, we get:

$P'(t) = 400 \times \frac{180 - 4t^2}{(45 + t^2)^2}$

This simplifies to:

$P'(t) = \frac{400(180 - 4t^2)}{(45 + t^2)^2}$

Thus, the derivative $P'(t)$ is:

$P'(t) = \frac{72000 - 1600t^2}{(45 + t^2)^2}$

This is the rate of change of the population $P(t)$ in bacteria per hour.

Would you like a detailed breakdown of any part of this solution or further clarification?

Related Questions:

1. How do we apply the quotient rule for more complex functions?
2. What is the biological significance of the derivative in this context?
3. How does the population change over time for larger values of $t$?
4. Can we find the critical points of $P'(t)$ and what do they represent?
5. How would you approximate $P'(t)$ for specific values of $t$?

Tip:

The derivative $P'(t)$ can help determine when the population growth rate is at its maximum or when it starts to decrease, which is useful in modeling real-world biological processes.