Derivatives of Curves. How do they work? How is the second derivative of a quadratic equation a constant? Is the second derivative the slope of the slope?

In: Mathematics

A derivative of a curve, y, evaluated at a point along the x-axis, x, is the slope of the line tangent to y at x.

When you calculate the derivative as a function it is just a quick way to find the slope of a tangent line at any point.

The second derivative is, indeed, the slope of lines tangent to points along the first derivative. Since the derivative of x^2 is just the line 2x passing through the origin, a tangent line at any point will just be the same line with a slope of 2.

Since this second derivative is just a constant, a horizontal line on the x-y plane, its tangent is just 0. It has no slope at any point.

A fourth derivative of a quadratic and beyond would all just be taking the slope of the horizontal line following the x axis and are mutually 0.

> Derivatives of Curves. How do they work?

Basically, the slope of a curve at each point.

> How is the second derivative of a quadratic equation a constant?

Well the slope of quadratic curve is increasing at a constant rate.

The quadratic curve is increasing at a quadratic rate (by definition).

But the slope is increasing at a constant rate, which is why the 1st derivative is a straight diagonal line.

So the second derivative is simply the slope of the 1st derivative, which is constant since it is a straight line.

> How is the second derivative of a quadratic …

The highest power of x in a quadratic equation is 2, the square term. When you differentiate a variable raised to a power, the index decreases by one. So differentiating twice reduces that index to zero, a constant therefore no longer dependent on x.

> Is the second derivative ….

Yes the slope of the slope, or more succinctly, the curvature.

By taking the derivative of a function, we are essentially looking at the rate of change. Let’s take a simple example: f(x) = x^(2) – x + 1

If we take the first derivative, we end up with f'(x) = 2x + 1

If we plot both functions you’ll see that when f'(x) is equal to 0. The change in f(x) is also 0. Because the parabola has a minimum, at this point, it doesn’t go lower or higher. You can see this by looking at the green arrow in my [plot](http://prntscr.com/10n6ihw).

Before the minimum, if we move along the x-axis (left to right), we see that the parabola is decreasing. This is also indicated by f'(x) because the linear equation is negative at this point. Vice versa, if we move beyond the parabola’s minimum, it will start to increase. Again, here its derivative is positive.

Now for simplicity we will make the first derivative it’s own function! so f'(x) = g(x). Taking the derivative of this will indeed result in a constant. Why? Because the rate of change of g(x) is also constant. If we look at a new [plot](http://prntscr.com/10n6n1l) we see that that g(x) = 2x -1 increases by 2 y units, for every whole unit of x. This is also supported by its derivative g'(x) = 2

So when we take the second derivative of a function, we first look at the rate of change, and then we look at the rate of change OF the rate of change.

It’s more difficult to explain by words than to look at the graphs, I hope this helped you visualize it some more.

To add to other explanations: think of it in terms of physics. Start with the position. Take the time derivative, you get velocity which is the rate of change of position with time. Take the time derivative of velocity and you get acceleration, which is the rate of change of velocity with time. When, say, a car is accelerating at a constant rate, the acceleration is constant and thus the position function will be quadratic.

The derivate of a function is the rate of change of that function (with respect to a variable).

So if the second derivative is constant that means the rate of change of the rate of change is fixed.

As an example, you can think of the function as a constant acceleration. That implies that the velocity (first derivative) is changing linearly which implies that the distant (the function itself) increases quadratically.

>Is the second derivative the slope of the slope?

That’s exactly right.

>How is the second derivative of a quadratic equation a constant?

It just is. That’s more or less how a quadratic is defined, its a shape whose second derivative is constant, a shape whose slope increases proportionally to the x or whatever input you are using.

The derivative definition of the limit as h approaches x of (f(x) – f(h))/(x – h) is pretty self explanatory imo. This is just the slope formula, rise over run, but taken to an infinitesimally small rise over an infinitesimally small run.