Basically, I just want to know the significance of the 2nd derivative of a function, or what does it tell us.
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1This looks like an extremely broad question. Even within physics, there are so many things for which you'd need to calculate the second derivative: an easy example is acceleration. It's the second derivative of position with respect to time. – Sep 10 '18 at 11:21
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1Would [math.se] be a better home for this question? – Qmechanic Sep 10 '18 at 11:21
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1@Qmechanic Probably, although here we can discuss why the 2nd derivative is so often important in physics. Eg, stuff about simple harmonic motion, and even the basic connection between force & acceleration. – PM 2Ring Sep 10 '18 at 11:30
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Not clear what you are asking. What is your difficulty? What is the context? – sammy gerbil Sep 10 '18 at 11:49
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Possible duplicate: https://physics.stackexchange.com/q/20714/2451. Related: https://physics.stackexchange.com/q/45517/2451 , – Qmechanic Sep 10 '18 at 11:50
3 Answers
In broadest terms, it tells you “how quickly the change is changing”.
Specific example: position $x$. Unless an object is stationary, its position $x$ changes with time. We call that change (the first derivative of $x$ with time, sometimes denoted as $\dot x$) the velocity.
But that velocity may be changing too. The change in velocity (which we call acceleration) is the time derivative of $v$, or the second derivative of $x$, $a=\ddot x$. So the second derivative, in this instance, describes the acceleration.
Other examples have a different meaning. For example, take a deflected string under tension. If the string is in the X direction and deflection in Y, then $\frac{dy}{dx}$ is the slope of the string - the extent to which it doesn’t point along X. The second derivative in that case, $\frac{d^2y}{dx^2}$ describes the rate of change of the slope which is the curvature of the string. It so happens that the curvature determines the local force on an infinitesimal element of the string, and can be used to compute the over all shape and its time evolution.
So different physical quantities will have different meaning of their second derivative - but it always means the change in the rate of change.

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It tells you the (instantaneous) change rate of (instantaneous) change rate of $f$. Let $f(t)$ be the travelled distance for example. Then the first derivative gives you the velocity and the second derivative gives you the change rate of velocity, namely the acceleration.

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The derivative of a function tells us the rate of change of that function with respect to whatever variable we're interested in. The 2nd derivative would then just be the rate of change of that 1st derivative.
The simplest physical example is probably distance, speed and acceleration...
$x$ is a particle's position in space. Differentiate this with respect to time, and you get the velocity, $v = \frac{\text{d}x}{\text{d}t}$ which is the rate of change of the particle's position in space. Differentiating this again gives the acceleration, $a = \frac{\text{d}v}{\text{d}t} = \frac{\text{d}^2x}{\text{d}t^2}$ which is the rate of change of the particle's velocity.
Any derivative is just a "rate of change" of something - the fact that it's the second, third etc... very much depends on the context (and doesn't matter as much).

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1if you also add a mathematical description as in for example http://mathworld.wolfram.com/FirstDerivativeTest.html and http://mathworld.wolfram.com/SecondDerivativeTest.html it would be a better answer to the OP. – physicopath Sep 10 '18 at 11:52