In this problem, we're told that the population of a species of bird is given by the function \( p(t) = \frac{6t+5}{0.3t^2+2} \), where \( t \) is specifically time measured in years. Now we are asked here to find \( p'(t) \), the derivative, and also interpret its meaning. So we are using our math skills and our derivative skills here and combining that with some critical thinking in order to determine what this math we're doing means in the context of this problem. So let's go ahead and get started in first finding our derivative here, \( p'(t) \). Now looking at my function here, since I have one function on the top \( 6t+5 \) divided by another function on the bottom \( 0.3t^2+2 \), I need to use the quotient rule.
Remember, the quotient rule tells us \( \text{low} \cdot \text{d high} - \text{high} \cdot \text{d low} \) over the square of what's below is our derivative. So let's go ahead and apply that quotient rule here. We're starting with that low function, which is \( 0.3t^2 + 2 \) times the derivative of that high function. The derivative of \( 6t+5 \) is just 6. And then here, we're subtracting high \( d \) low.
That is our high function, \( 6t+5 \) times the derivative of that bottom function \( 0.3t^2+2 \). The derivative of this function is just going to be \( 0.6t \) using the power rule. Now that's our full numerator. Now for our denominator, it's over the square of what's below, so I'm taking that function on the bottom, \( 0.3t^2 + 2 \), and I'm going to fully square that because remember it's the square of what's below, not just what's below. So from here, we can do some simplification.
We want to go ahead and distribute this 6 in here to see if we can end up canceling anything out and then also this \( 0.6t \). So in the top, I have \( 6 \times 0.3 \). That's going to give me \( 1.8t^2 + 12 \) for that first term and then I am subtracting this \( 0.6t \) distributed throughout. \( 6 \times 0.6 \) is going to give me \( 3.6t^2 \) because I have a \( t \) in both of those terms and then a plus \( 5 \times 0.6t \) which is going to give me \( 3t \). Now from here I want to distribute this negative in here just to go ahead and do that all in one step.
So just know that both of these terms are going to be negative because we have distributed that. Then my denominator stays the same at \( (0.3t^2+2)^2 \). And I can go ahead and do some further simplification here. I have some like terms. I have this \( 1.8t^2 \).
I have a minus \( 3.6t^2 \). So this becomes a negative \( 1.8t^2 \). And then a minus \( 3t \), just writing this in standard form of a polynomial, plus \( 12 \). Then my denominator stays the same at \( (0.3t^2+2)^2 \). All of that is squared and this is my final derivative here \( p'(t) \).
Now we have answered the first part of our question which is to find \( p'(t) \), our derivative, but we want to answer that second part too. We want to interpret what this actually means. Now this is a population function. We're told that this is a function that represents the population of a species of bird where \( t \) is measured in years. So in finding the derivative here, remember that the derivative is a rate of change, so we found the rate of change of the population of this bird.
So if we were to plug in a value and find, say, \( p'(2) \), we would be finding the rate of change of this bird population at the time of 2 years. Now if you have any questions here, feel free to let us know and I will see you in the next video.
