I have done some calculations recently on squares and cubes and $4^{th}$ powers, and so on, and I found out something. It goes like this:
$$ \begin{align} f(x)&=x^2\\ f'(x)&=2x^1=2x\\ f"(x)&=2 \end{align} $$
This is only for the rate of change in the squares. For the rate of increase, you get:
$$\begin{align} f(x)&=x^2+(x^2+x+1)+(x^2+x)\\ f'(x)&=2x+(2x+1)+(2x+1)\\ f''(x)&=2+(2)+(2)\\ &=6 \end{align} $$
To make any noticeable pattern appear, we have to add another term or two. the cubes' differentiation would be this:
$$ \begin{align} f(x)&=x^3+(x^3+x^2+x+1)+(x^3+x^2)\\ f'(x)&=3x^2+(3x^2+2x+1)+(3x^2+2x)\\ f''(x)&=6x+(6x+2)+(6x+2)\\ f'''(x)&=6+(6)+(6)\\ &=18 \end{align} $$
Notice the pattern yet? No? Here's one more:
$$ \begin{align} f(x)&=x^4+(x^4+x^3+x^2+x+1)+(x^4+x^3)\\ f'(x)&=4x^3+(4x^3+3x^2+2x+1)+(4x^3+3x^2)\\ f''(x)&=12x^2+(12x^2+6x+2)+(12x^2+6x)\\ f'''(x)&=24x+(24x+6)+(24x+6)\\ f''''(x)&=24+(24)+(24)\\ &=72 \end{align} $$
Do you notice the pattern yet? It is this:
$$\text{if} f(x)=x^z+(x^z+x^{z-1}+x^{z-2}+\cdots+x^0)+(x^z+x^{z-1}), \\ \text{then} f''{^{(\text{z times})}}''(x)=3(z!)$$
Basicially, in the equation $x^z+(x^z+x^{z-1}+x^{z-2}+\cdots+x^0)+(x^z+x^{z-1})$, which is the rate of increase in a cube or polytope, the $z^{th}$ derivative (or, in calculus terms, you could put it as $\frac{d^zf}{dx^2}$)is always $3\times z!$
My question is this: is there any practical use for this? My friend said it may have something to do with the rate that the exponential function increases (as in $3z! $ added to a number added to a number and so on z times until you get $x^z$), but I am not sure. Is it for that? If not, where is it useful? Thanks in advance.
Let $f(x) = x^z + (x^z + x^{z-1} + \cdot\cdot\cdot + x^0) + (x^z + x^{z-1})$.
First, rewrite this as $f(x) = 3x^z + (x^{z-1} + x^{z-2} + \cdot\cdot\cdot + x^0) + (x^{z-1}) $ Now we know that the $n$-th derivative of $x^n$ is $n!$ and it's trivial that the $n$-th derivative of $x^k$ is $0$ for any integer $0 < k < n$.
Since all the terms after $3x^z$ become $0$ as we differentiate, we only have to concern ourselves with the first term.
$$ \frac{d^z}{dx^z}\left[ 3x^z \right] = 3 \cdot \frac{d^z}{dx^z}\left[ x^z \right] = 3z!$$
Addendum: Why is $\frac{d^n}{dx^n} [x^n] = n! \,$ ? As you did in your post, write out the first couple derivatives to notice the pattern.
$$ \begin{align*} f(x) &= x^n \\ f'(x) &= n x^{n-1} \\ f''(x) &= n (n-1) x^{n-2} \\ f'''(x) &= n (n-1) (n-2) x^{n-3} \\ f^{(k)}(x) &= \frac{n!}{(n-k)!}x^{n-k} \\ f^{(n)}(x) &= \frac{n!}{(n-n)!} x^{n-n} = \frac{n!}{1} x^0 = n! \end{align*} $$