I was just wondering what a double derivative of function implies. Is it the rate of change of rate of change of the function, or something else. And if that is the case then help me understand, why (d^n/dx^n) x^n is a constant i.e. n!.
For some reason, I am only able to visualise it graphically and I am trying to draw tangent lines over tangent lines and it's not helping. So please help me visualise a higher order derivative of a function.
To explain what double derivative is, we shall better transfer to Physics.
As you may know, and as you pointed out, the derivative of a function with respect to a variable (the variable of the function) represents indeed the rate of change of the function in terms of the variable.
Physics time: suppose the variable is time, $t$, and your function is $s(t)$, that is a distance walked in a time $t$.
Now, the derivative of $s(t)$ with respect to $t$ represents the rate of change of $s(t)$ in time, which in physics defines the velocity $v(t)$:
$$\dfrac{d s(t)}{dt} = v(t)$$
Geometrically, we say the function $v(t)$ is the tangent curve to the function $s(t)$ at the point $t$ (to be chosen).
If we now consider the double derivative of $s(t)$, we sure can interpret it as the rate of change of the rate of change of $s(t)$, but this nested rate of change can make you lose the sense of what you're doing.
A simpler approach: the double derivative of $s(t)$ is the rate of change of $v(t)$ in time! Indeed
$$\dfrac{d^2 s(t)}{dt^2} \equiv \dfrac{d}{dt}\dfrac{ds(t)}{dt} = \dfrac{d}{dt} v(t) = \dfrac{d v(t)}{dt} = a(t)$$
Physically speaking, this defines the acceleration of a body, which is indeed the rate of change of the velocity in time.
In mechanics we can define further derivatives, for example the third derivative in time of $s(t)$ which would mean the rate of change of the acceleration in time.
But we can also have space derivatives, for example in Thermodynamics we can call the "temperature gradient", that is the rate of change of Temperature in space (we use here partial derivatives often, but let's just focus on the $1$D case).
Eample: consider the heating of a long metallic rod, which we consider to be unidimensional (just very long and thin). So we can study how Temperature spreads and varies along the rod, that is along the $x$ direction. In this case the Temperature Gradient reads
$$\dfrac{dT}{dx}$$
and it does represent the rate of change of the Temperature along the rod.
Similar stidues are done in Atmospheric Phsyics (to speak rather simply) to study the Temperature Gradiend in the atmosphere. In this case we need three-dimensional analysis because we need the rate of chance of the Temperature in the whole space.
Now to answer to the mathematical question, just look:
$$\dfrac{d}{dx} x^n = n\cdot x^{n-1}$$
I assume you know this. I mean I am assuming that you know the power-rule differentiation I wrote above.
Now, let's take the second derivative:
$$\dfrac{d^2}{dx^2} x^n \equiv \dfrac{d}{dx} \left(\dfrac{d}{dx} x^n\right) = \dfrac{d}{dx} \left(n\cdot x^{n-1}\right) = n \dfrac{d}{dx} x^{n-1} = n\cdot (n-1) x^{n-2}$$
I'm assuming you're starting to understand how it goes on: every derivative you compute, lows down of one unit the exponent of the $x$ (at the beginning thou hast $x^n$ then after a derivative you have $x^{n-1}$ then after the second derivative you get $x^{n-2}$ and so on.)
Every derivative, in addition, brings down the exponent term, producing it to the previous. As you can see you nave
$$x^n$$
then you derive getting
$$\color{red}{n}\cdot x^{n-1}$$
then you derive getting
$$\color{red}{n}\cdot (\color{blue}{n-1}) x^{n-2}$$
suppose you derive again:
$$\color{red}{n}\cdot (\color{blue}{n-1}) \cdot (\color{green}{n-3}) x^{n-4}$$
If you go on, you will arrive at a point where the exponent will be $1$ and the terms brought down read
$$n\cdot (n-1)\cdot (n-2)\cdot (n-3)\ldots$$
which is nothing but $n!$ (I assume you know what "!" means).
How this clarifies you a bit what you asked for.