Second Derivative Question -- Help Understanding the Importance Please

[NoahsArk]

TL;DR Summary

Help Understanding the Importance of Second Derivatives

Hello. My understanding of the importance of second derivatives is that they help us to know whether the graph of a function is concave upward or concave downward. In the equation $f(x) = x^2 + 2x$ we already know from the first derivative, $f\prime (x) = 2x + 2$, that the graph is concave upward because the 2 in front of the x is positive. Why do we need to even figure out the second derivative, which is also a positive number, 2, to determine that? Seems like we are taking an unnecessary step. Thanks.

 

[andrewkirk]

The method is unnecessary for a quadratic like the one above, because that is a very simple function.
But your short cut won't work for more complex functions, like $f(x)=x^3-3x^2+2x + 10$. That function is concave up in some places and concave down in others, and you need to use second derivatives to find out where.

 

[WWGD]

Second derivatives are also used to determine maxima and minima of functions of two variables.

 

[NoahsArk]

I appreciate the help.

 

[NoahsArk]

But your short cut won't work for more complex functions, like $f(x)=x^3−3x^2+2x+10$. That function is concave up in some places and concave down in others, and you need to use second derivatives to find out where.

The part about being concave upwards and concave downwards in some places resolves part of the question for me. I still don't understand why we need the second derivative to find out where. Why not use the first derivative or the third derivative, for example? Also, although I know how to find the derivative of a derivative, I am not sure what it really means. A derivative tells us how to find the slope of a curve at a given point which you can use to draw a tangent line. What is the derivative of the derivative telling us?

Thanks

 

[PeroK]

The part about being concave upwards and concave downwards in some places resolves part of the question for me. I still don't understand why we need the second derivative to find out where. Why not use the first derivative or the third derivative, for example? Also, although I know how to find the derivative of a derivative, I am not sure what it really means. A derivative tells us how to find the slope of a curve at a given point which you can use to draw a tangent line. What is the derivative of the derivative telling us?

Thanks

The second derivative is important because it tells you the nature of a turning point. You can't get that from the first or third derivatives.

If you take a function of time, then the second derivative is the acceleration. Just as the first derivative of displacement is the velocity, the second derivative is the rate of change of velocity, which is acceleration.

Higher derivatives are not used as much. The first and second are by far the most important.

 

[etotheipi]

What is the derivative of the derivative telling us?

It's telling you how the derivative is changing. For example if the second derivative in a certain neighbourhood is greater than zero, $\frac{d^2 y}{dx^2} = \frac{d}{dx} \frac{dy}{dx} > 0$, that means that the gradient is increasing. At a local minimum of a function, the gradient starts off negative, then increases to zero, and then increases to positive values.

 

[NoahsArk]

Thank you for the responses.

If you take a function of time, then the second derivative is the acceleration. Just as the first derivative of displacement is the velocity, the second derivative is the rate of change of velocity, which is acceleration.

Does this mean that the second derivative tell us how much of an additional increase there is in velocity for each unit of x? For example, in $D = t^2$ the derivative is 2t and the second derivative is 2. Does that mean that the speed is increasing by 2 for each extra unit of time. Say at t = 3 the speed is 6kph and at t = 4, it is 8kph. In that example the change in velocity of 2 makes sense. In the equation $d = t^3$ the derivative would be $3t^2$ and the second would be 6t. Does that mean that the speed is increasing 6t for each extra unit of time? At t = 3 the speed would be 27kph, and at t = 4 it would be 64. The increase in speed is 37 which does not = 6t, so I think I am misunderstanding how the second derivative = change in velocity.

@etotheipi does that mean whenever the second derivative is greater than zero, then the gradient is increasing? I am assuming there are cases when the first derivative is less than zero but the second isn't and vice versa.

If there are good exercises for this online please let me know. I started to look on Khan Academy and haven't found any yet, but I bet they are there.

 

[etotheipi]

@etotheipi does that mean whenever the second derivative is greater than zero, then the gradient is increasing? I am assuming there are cases when the first derivative is less than zero but the second isn't and vice versa.

Yes, that's right. An example where the first derivative < 0 whilst the second derivative > 0 would be an object moving in the negative direction, but accelerating in the positive direction. E.g. something slowing down.

 

[PeroK]

Does this mean that the second derivative tell us how much of an additional increase there is in velocity for each unit of x? For example, in $D = t^2$ the derivative is 2t and the second derivative is 2. Does that mean that the speed is increasing by 2 for each extra unit of time. Say at t = 3 the speed is 6kph and at t = 4, it is 8kph. In that example the change in velocity of 2 makes sense. In the equation $d = t^3$ the derivative would be $3t^2$ and the second would be 6t. Does that mean that the speed is increasing 6t for each extra unit of time? At t = 3 the speed would be 27kph, and at t = 4 it would be 64. The increase in speed is 37 which does not = 6t, so I think I am misunderstanding how the second derivative = change in velocity.

The kinematic equation for constant acceleration is:
$$s(t) = ut + \frac 1 2 a t^2$$
Where $s(t)$ is the displacement at time $t$, $u$ is the initial velocity and $a$ the constant acceleration.

If you differentiate this equation (with respect to time) you get the velocity (at any time):
$$v(t) = u + at$$
And, if you differentiate again you get the acceleration:
$$a(t) = a$$

 

[Adesh]

Second order derivatives are one of the most important thing in advanced physics. Most famous equations like- Schrodinger's Equation, Navier-Stokes equation, Electromagnetic-Wave equations and more.

Second order derivatives are not something only used for determination of concavity and convexity. Consider the scenario of spring, experiments show that the restoring force is proportional to the distance we have stretched the spring from its equilibrium, that is $$ F = -kx $$ by Newton's law we know that the force is mass times acceleration and as @PeroK sir has told you that acceleration is the second derivative of position, we have
$$ ma = -kx \\
m \frac{d^2x}{dt^2} = -kx \\
\frac{d^2x}{dt^2} = \frac{-k}{m} x$$

There we again got a second order derivative in our simplest of physical situation. They have shown up here because acceleration is something that we can measure and position is something that we can measure and hence the equation, but to solve it mathematically we had to convert acceleration into something which involved position.

Side Note: You can know the importance of second order derivatives by the fact that PeroK sir has written an article on How to Solve Second-Order Partial Derivatives.

 

[Ssnow]

In addition to previous information I want to underline the relation between of the second derivative and the "signed curvature" of the graph of a real function $y=f(x)$ (at the point $x$):

$$\mathcal{k}(x)=\frac{y''(x)}{(1+{y'(x)}^2)^{\frac{3}{2}}}.$$

In particular when the slope of the graph is small the signed curvature is well approximated by $y''$. This relation give us a geometrical interpretation of the second derivative that is usually used in the geometric study of curves and surfaces.

Ssnow

 

[FactChecker]

Summary:: Help Understanding the Importance of Second Derivatives

Hello. My understanding of the importance of second derivatives is that they help us to know whether the graph of a function is concave upward or concave downward. In the equation $f(x) = x^2 + 2x$ we already know from the first derivative, $f\prime (x) = 2x + 2$, that the graph is concave upward because the 2 in front of the x is positive.

But that 2 IS the second derivative. So you are answering your own question. Just because this example is so easy that you know the answer by sight does not invalidate the method.

Why do we need to even figure out the second derivative, which is also a positive number, 2

"also" is the wrong word. It is the same '2' as the one you pointed out earlier. There are many examples that are much harder to see without formally doing the calculation of the second derivative and where the second derivative will be positive in some places and negative in other places.

 

[NoahsArk]

I think I'm starting to get a much better idea of what the second derivative is with the help of the responses here. First derivatives show velocity at a certain point, and second derivatives show how the velocity is changing for each unit of time. That's why in a concave downward graph, as we approach a zero slope, although the slope is positive until we get to the top, the slope is decreasing and that's why the second derivative is negative.