Well, if you plug the explicit function s(t) in the definition of derivative what do you get? That is,
lim (h goes to 0) of [s(t+h)-s(t)]/h.
This will give you the answer.
I have a function s=4.4t^2 where t is time and s is a distance traveled in feet. I understand that to find out how many feet per second the car is going after t seconds I need to find the derivative. I have read through the chapter in my textbook on derivatives and I have studied the formula of the difference quotient f(t+h) - f(t) / h. This is the very first derivative I have ever calculated. I'm finding that I'm stuck and I don't see how the derivative of this function is calculated. I'm hoping someone will enlighten me as to what the steps to calculating this derivative look like.
Thanks in advance...
I know the derivative of 4.4t^2 is 8.8t because there is a rule that says the derivative of 4.4^2 the power goes down by one and 4.4 is multiplied by the power however what I am being asked to do is plug the equation into the difference quotient which I have done a couple of times so far and not gotten the correct answer yet. I'm hoping someone might show me the steps. In your reply above I am having trouble determining what you mean by s because in the equation s is f(t). Are you saying s is 4.4?
There's no need to be rude Plato. The OP SAID he hasn't done derivatives before.
What you need to understand, sepeto, is the idea of an "average" rate of change being an approximation to a rate of change at a point, and then the "instantaneous" rate of change being the exact rate of change at that point, which you get by improving on the approximation.
Think of a car's speedometer. It gives us the speed at which a car is travelling at any point in time. So it's an instantaneous rate.
But is it really? How do you think the speedometer goes about working out the speed at any point in time? Well, we know that speed = distance/time, but it's not like it can draw a graph, rule in a tangent and evaluate its gradient. What it does is it calculates the distance travelled over a very small period of time (like a second). This gives the speedometer a pretty good approximation for the speed at any time. How could it improve on the approximation? Why, by evaluating the distance travelled over an even smaller amount of time. The smaller the change in time, the better the approximation will be.
This is exactly what's happening with the derivative. Say you think of the distance "s" as being some function of time "t". Then at ANY point of time "t" we have a corresponding distance travelled "s(t)". In order to get an approximation for the speed, we need to evaluate the distance travelled over a small change in time, which we'll call "h". Then the corresponding distance at "t + h" is "s(t + h)". Then the speed is .
This is a good approximation as long as the change in time is small, so in order to make it exact, we make this change in time SMALLER, i.e. make it approach 0. Therefore, the EXACT speed at any point in time is , which means after you get an expression for the approximate speed, it will be in terms of h, and you need to turn the h into a 0. This is what Plato showed you.
The general formula for the derivative is given in terms of a general function f(t). In your case s(t)=4.4t^2 =f(t).
Hence S(t+h)= 4.4(t+h)^2. If you expand the binomial e plug it into the derivative formula, then you get the answer.