Question

What is the relationship between the first and second derivatives?

Answer 1

What I'm trying to ask is, what is implied if the second derivative is negative? What about if it is positive?

Answer 2

A negative second derivative means that the first derivative is decreasing on that interval.

Answer 3

Okay I'm completely lost. CliffSedge, could you possibly elaborate?

Answer 4

Derivatives are rates of change. The first derivative tells you how the original function is changing; the second derivative tells you how the first derivative is changing.

Answer 5

As an example, let's say you have a function \(\large y=x^3\),
at x=2, y=8.

The first derivative of y w.r.t. x is \(\large y'=3x^2\),
at (2,8), the slope of the tangent line at that point is 12. That is the instantaneous rate of change. It tells that \(\large y=x^3\) is increasing at that point.

The second derivative of \(\large y=x^3\) , which is the first derivative of \(\large y'=3x^2\), is \(\large y''=6x\).

Ath the point (2,8), the second derivative is also 12; this means that the first derivative is increasing. The slope is getting bigger.

Answer 6

If you think of the original function as telling you about position, the first derivative represents velocity, and the second derivative represents acceleration.

Answer 7

For that example, wouldn't the slope remain the same if the derivatives are both the same? I'm horribly sorry for sounding so dumb but this is just not sticking in my brain.

Answer 8

The values of the first and second derivatives being the same is just a coincidence here.
Since one is a rate of change of the other, the rate-of-change of the rate-of-change is increasing, so the slope cannot stay the same.

Answer 9

Try graphing \(\large y=x^3\) and trace along it to see that the slope is increasing as the function itself also increases.

Answer 10

Okay. That makes some sense. I get the velocity/acceleration now.

Answer 11

But not everywhere. The slope is practically (if not) 0 near the origin. That's the part I don't understand...doesn't that mean that the rate of change isn't constant everywhere?

Answer 12

That is true. The rate of change is not constant.
The first derivative, \(3x^2\) is a function of x, so depends on what x is.
The second derivative, \(6x\) is also a function of x, and it depends on what x is too.

Answer 13

The slope, and the rate of change of slope is zero at the origin (that makes it an inflection point). The first derivative is going from an interval of decrease to an interval of increase at that point.

Answer 14

That sort of makes sense! Calculus is frustrating me beyond belief right now.
Thanks for the help! :)

Answer 15

I recommend paying a lot of attention to the shapes of the graphs of functions.
When you learn about critical points and concavity it should make more sense.

Answer 16

I tried when we first learned, and it just didn't stick. I guess I'll have to try again...thanks!