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This might be an unusual question but I was wondering why the 2nd derivative of this function is a straight line? I kind of have the feeling this is not that easy to answer. But it kind of struck me that it is exactly linear.

Here's a picture:

enter image description here

I mean, yes mathematically you can say that it just is as it is, but is there also an intuitive answer to it? Thank you for answering!

Regards!

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    $\begingroup$ If you start with a polynomial of degree (at most) $3$, it will be. Otherwise... it has no reason to be. (And the idea is that differentiating a polynomial $k$ times makes the degree "drop" by $k$.) $\endgroup$ – Clement C. Jan 16 '17 at 22:54
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    $\begingroup$ $(\sin x)'' = -\sin x.$ Oops ... $\endgroup$ – zhw. Jan 16 '17 at 23:44
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    $\begingroup$ Everyone seems to interpret this question as "why is the 2nd derivative of any function a straight line?" Maybe what is meant is "why is the 2nd derivative of a function I'm just about to show you a straight line?", i.e., as only relating to the function whose graph is shown in the picture? @VeraMarya? $\endgroup$ – Casper Jan 17 '17 at 0:25
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    $\begingroup$ Simply put: Each derivative shows you the gradient of the tangent of the curve derived as a function of x. So the second derivative shows that the gradient of the first derivative starts negative, and gradually and linearly changes to a positive value as x increases. $\endgroup$ – Shiri Jan 17 '17 at 11:19
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    $\begingroup$ At first it's decelerating, and then it's at constant velocity (at the green line), and then it's accelerating $\endgroup$ – Akiva Weinberger Jan 17 '17 at 16:57
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Suppose the second derivative of a function is a straight line:

$$f''(x)=ax+b$$

It follows that

$$f'(x)=\frac12ax^2+bx+c$$

$$f(x)=\frac16ax^3+\frac12bx^2+cx+d$$

So the only functions that have their second derivatives as straight lines are polynomials degree $3$ or lower.


Now, for some insight, let's just worry about the first derivative of $x^2$, which is obviously $2x$, but I wish to make it more intuitive:

enter image description here

Rather than letting $h\to0$, we fix $h=1$ and just take finite differences. It is easy enough to see then that $(x+1)^2-x^2=2x+1=1+\frac d{dx}x^2$.

Similarly, if we apply a finite difference on $x^3$ 2 times, which is analogous to the second derivative, we get

$$(x+1)^3-x^3=3x^2+3x+1$$

$$3(x+1)^2+3(x+1)+1-3x^2-3x-1=6x+6=6+\frac{d^2}{dx^2}x^3$$

Indeed, you should be able to prove by induction that the $n$th derivative of a polynomial degree $n+1$ is linear.


And as a last comment, usually, when we have $h\to0$, all the left-over constants and such go to $0$.

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    $\begingroup$ I'm not sure how intuitive this may be for the OP... $\endgroup$ – user347489 Jan 17 '17 at 5:28
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    $\begingroup$ Yeap, this answer has no value whatsoever for OP, or that's what I gather from his question. $\endgroup$ – YoTengoUnLCD Jan 17 '17 at 8:23
  • $\begingroup$ @user347489 oh, I think I see what you mean... You think the OP wants an intuitive answer? Hm... I'll see what I can do. $\endgroup$ – Simply Beautiful Art Jan 17 '17 at 11:57
  • $\begingroup$ @user347489 So, do you think this is any better? $\endgroup$ – Simply Beautiful Art Jan 17 '17 at 15:25
  • $\begingroup$ @YoTengoUnLCD If you cared, how about now? $\endgroup$ – Simply Beautiful Art Aug 10 '17 at 11:49
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I think you are confused. The second derivative of any arbitrary function will not necessarily be a straight line. For instance, the exponential $f(x) = e^x$ has second derivative $f''(x) = e^x$ which is clearly not linear. It is possible to come up with linear approximations in a neighborhood of a point, and 3rd-degree polynomials of the form $g(x) = ax^3 + bx^2 + cx + d$ will always have second derivatives that are straight lines. But for arbitrary functions this is not the case in general.

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    $\begingroup$ The question is “why is this function's second derivative straight”, not about second derivatives in general. $\endgroup$ – Anton Sherwood May 30 '17 at 6:50
  • $\begingroup$ @AntonSherwood: Yeah, but the guy has got a total of $15 \cdot 10 - 2 \cdot 2 = 146$ points for this useless answer, so how do you expect him to be honest and delete it? Part of the trouble is ignorant users who vote like sheep, without bothering to think. (Now let's think for a moment about the fundamentals of democracy... umm, no, better not, or else we'll become depressed.) $\endgroup$ – Alex M. Aug 16 '17 at 9:53
  • $\begingroup$ @AlexM.— (I'm an anarchist …) $\endgroup$ – Anton Sherwood Aug 19 '17 at 14:39
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I understand this is math.stackexchange, so the answers that have already been put up are certainly sound. However, you said you were looking for something a little more intuitive.

The derivative of a function just describes the slope of that function. When the function is increasing, its slope (derivative) will be positive. When it is increasing "faster", its derivative will be more positive. Similarly - when the function is decreasing, its derivative will be negative.

The diagram you posted has dotted lines drawn from the top curve at inflection points, where the slope changes from positive to negative. Notice how they're drawn downwards to the middle curve (the parabola) exactly where it crosses the x-axis.

The derivative of the parabola is negative while the parabola is decreasing, and positive while increasing. That alone doesn't guarantee a straight line:

$$f(x) = x^4$$ is very "parabola" shaped, and from initial inspection you may expect it to be a parabola, but the derivative ends up being cubic: $$f'(x) = 4x^3.$$

The second derivative is not guaranteed to be a straight line for any arbitrary function $f(x)$; it just describes the "slope of the slope" of that function.

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I will make an attempt to respond to this part:

is there also an intuitive answer to it?

Recall that the derivative of a function [is a function that] takes an $x$-value as its input, considers the tangent line drawn at that $x$-value, and gives the slope of that tangent line as its output.

Looking at the first picture, you have two turning points, so you will need a derivative that has two zeroes. You can also see that the function, while initially increasing (so: positive slopes of the tangent lines), is doing so at a decreasing rate till it reaches its vertex (so: tangent lines get less steep till their slope reaches zero). Afterwards, the function is decreasing (so: negative slopes of the tangent lines) pretty fast until your middle dotted line, at which point it continues to decrease (so: tangent line slopes are still negative) but not quite as quickly as it approaches the next vertex (where the tangent line's slope is again zero).

That is all a bit wordy, but the same sort of idea can be carried out in looking at the second image. It has one vertex (so: its derivative should be equal to zero only once) and goes from very negative tangent line slopes to very large positive tangent line slopes (so: a straight line seems to be a good candidate).

I happen to think that considering a quadratic function written in standard form, $ax^2 + bx + c$, and recalling that its vertex occurs when $x = -b/2a$, is enough to guess (using one's intuition) not only that the derivative would correspond to a linear function, but that, in particular, the derivative is $2ax + b$. I wrote up this idea and put it as the second example in the following article, published in a journal of mathematics education, which you may (or may not) find of interest:

Dickman, B. (2016). Looking Back to Support Problem Solving. Mathematics Teacher, 110(1), 54-58. Link (no paywall).

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Let's talk only about intuition.

If you look at your function $f(x)$, you'll notice that there are regions where it's value increases and other where the value decreases. Since the function is continuous, at some point it should stop increasing. That's what the first derivative tells you: if $f'(x)$ is positive, then the value of $f(x)$ increases as $x$ increases, if $f'(x)$ is negative, then the value of $f(x)$ decreases as $x$ increases, and if $f'(x)=0$ then the function is not increasing not decreasing; i.e. the slope of $f(x)$ is zero.

Now, if your function is continuous and has intervals where it's value increases and intervals where it's value decreases, it's reasonable to think that the rate at which it's value increases/decreases changes, and that's given by to second derivative. In particular, when $f''(x)>0$ and $f'(x)>0$, the rate at which the value increases is faster, and if $f''(x)>0$ and $f'(x)>0$, the rate at which the value increases is slower (see for yourself what happens when $f'(x)<0)$. And when $f''(x)=0$, you have an inflection point, where the rate of increase/decrease of the function value changes from faster to slower or vice-versa.

Now, your particular example:

Your $f(x)$ plot shows a function with one local maximum and one local minimum, and you can expect that at such points $f'(x)=0$. And it turns out that your $f'(x)$ plot has one (absolute) minimum, so $f(x)$ should have an inflection point. It makes sense: your function reaches a local maximum and starts decreasing and then reaches a local minimum and starts increasing; at some point between maximum and minimum the "speed" at which the function decreases has to slow down: that point is the inflection point, where $f''(x)=0$.

Finally, the second derivative being a straight line requires to see all plots together:

  • $f'(x)<0$ and $f''(x)>0$: slope is positive and rate of change is negative: speed of change is slowing
  • $f'(x)=0$ and $f''(x)<0$: local maximum
  • $f'(x)<0$ and $f''(x)<0$: slope is negative and rate of change is negative: speed of change is accelerating
  • $f'(x)<0$ and $f''(x)=0$: slope is negative and rate of change is zero: speed of change starts to slow down
  • $f'(x)<0$ and $f''(x)>0$: slope is negative and rate of change is positive: speed of change is accelerating
  • $f'(x)=0$ and $f''(x)>0$: local minimum
  • $f'(x)>0$ and $f''(x)>0$: slope is positive and rate of change is positive: speed of change is accelerating

As noted in other answers, the only functions that have a straight line as a second derivatives are polynomials of degree 3 or less. Let me add that a non-zero slope for a straight line second derivative implies that the degree of the polynomial must be exactly 3.

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The relationship between a curve and its derivative is mutual, and looking at it the other way round will explain why the derivative of a parabola is a straight line: Because the integral of a straight line is a parabola.


Others have mentioned what the intuitive meaning of a derivative is: How steep the curve is at that point, i.e. how steep a line would be that touches the curve only at that point: a tangent.

So why is the derivative of a parabola a straight line? The first thing to note is how the derivative line crosses the x axis precisely where the slope of the parabola is horizontal, i.e. its "steepness" is 0. Before that the derivative was negative. This means that the parabola curve falls with increasing x, which is exactly what we see. Afterwards the slope is positive; the y value of the parabola grows with growing x.

So we have established that the derivative line is qualitatively correct, but it could still be a curved line. Why is it straight? This picture may explain it:

Line with triangular areas under it, and a parabola indicating the area size

We see a parabola in red and its derivative in green. Because the reversal of deriving is integrating, the parabola gives us the value for the area under the green curve for each x. The areas are triangles which I highlighted in shades of yellow and orange. Remember that the area of a triangle is half the area of a corresponding rectangle, marked in a very light shade above the green line. Therefore, the triangles' area — which is the area under the green line — is growing with the square of x! A function whose value grows with the square of its argument is a parabola. I have marked the triangles' areas and the corresponding value of the parabola with arrows. It's actually easy to just count the areas with the aid of the grid.

Soooo — if integrating a straight line results in a parabola, deriving a parabola must result in a straight line!

A nice classical real world application of such a relationship is speed and distance during steady acceleration. The speed indicates how fast we move, and would be the green line in my image. The distance covered so far is the red parabola. We are accelerating, so the speed grows over time (the green line points up). "Steady acceleration" means here that the speed graph is a straight line.

If we have the speed of a vehicle at each point after it started, we can compute the distance covered: It is the area under the speed graph, in our diagram the triangles. The red graph shows that distance. It becomes steeper and steeper, because the vehicle accelerates. It is a parabola because the triangles' area grows quadratically.

A nice example with more graphics for the speed/distance relationship can be found here.

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Differentiation is an invertible operation.

Given an arbitrary function $f(x)$ (with some restrictions that are irrelevant here) you can find a function $g(x)$ such that the derivative of $g(x)$ is $f(x)$.

By iterating, you can find a function $h(x)$ such that its derivative is $g(x)$, thus its second derivative is $f(x)$.

You can find a function such that its second derivative has any shape.


In the case at hand, if

$$f(x)=ax+b,$$

$$g(x):=\frac{ax^2}2+bx+c$$ will ensure $g'(x)=f(x)$ (for any constant $c$).

And

$$h(x):=\frac{ax^3}6+\frac{bx^2}2+cx+d$$ is such that $h''(x)=g'(x)=f(x)$.

In fact taking the antiderivative of any polynomial is straightforward.

For other functions $f$, quite often you won't be able to find an analytical expression that works, but this doesn't mean that the functions $g$ and $h$ do not exist. In this case, it is not possible to "write it down", but you can plot it anyway.


It turns out that the next question, "is there a function which is the same as its own derivative" is an extremely important one, which has innumerable applications in the study of physical systems.

As you can check (if you learnt the properties of the exponential), $f(x)=e^x$ is such that

$$f(x)=f'(x)=f''(x)=\cdots.$$

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What derivative does is it shows you relative change. Let me explain:

If you have a function $$f_{(x)} = x$$ it will give You a straight rising line. A derivative of this function is $$f'_{(x)} = 1$$ This means that for each 1 of x, f(x) rises by 1 as well.

For the second power You'd have respectively: $$g_{(x)} = x^2$$ And $$g'_{(x)} = 2x$$

Now if you look at the chart of the g(x) function, You'll see that it changes a lot on the left side, then almost no change near 0 and then it starts changing more and more.

Now on the left side, the value goes down. And the derivative is negative. the bigger the change, the lower the value of g'(x).

Near zero, g'(x) reaches zero, so the change is very small.

On the right side, the value goes up. The derivative is positive. The bigger the change, the higher the value of g'(x).

Hope this helps to get an intuitive feel of derivatives.

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