We know definition of derivative is $\frac{\Delta f(x)}{\Delta x}$ when $\Delta x$ approaching 0. But what happens when we change $\Delta x$ by $(\Delta x)^2$.
$\lim\limits _{\Delta x\to0}\frac{f(x+\Delta x)-f(x)}{(\Delta x)^2}$.
Does it have any meaning of this rate of change formula? $(\Delta x)^2$ is the thing when we use it definition of second derivative.
P.s more from that could you recommend a book which plays this formulas?
Assume that the second derivative of $f$ exists at the point $x$. Recall that then we have the formula $$ f''(x) = \lim_{h \to 0} \frac{f(x + h) - 2 f(x) + f(x - h)}{h^2}. $$ for the second derivative. Now note that $$ \frac{f(x + h) - 2 f(x) + f(x - h)}{h^2} = \frac{f(x + h) - f(x)}{h^2} + \frac{f(x - h) - f(x)}{h^2}. $$ Since we are assuming that the limit $\lim_{h \to 0} \frac{f(x + h) - f(x)}{h^2}$ exists also note that $\lim_{h \to 0} \frac{f(x - h) - f(x)}{h^2}$ exists too, because it is obtained by the change of variables $h \mapsto -h$. Limit laws now mean that $$ f''(x) = 2 \lim_{h \to 0} \frac{f(x + h) - f(x)}{h^2}. $$ So your limit turns out to be twice the second derivative of $f$ at $x$.
On the other hand, existence of your limit is not equivalent to the second derivative of $f$ existing at the point $x$. For starters, recall that that the limit in the formula for $f''(x)$ above can exist even when the second derivative of $f$ does not exist. Next, your limit $ \frac{f(x - h) - f(x)}{h^2}$ exists less often than $\frac{f(x + h) - 2 f(x) + f(x - h)}{h^2}$, for the following reason: if your limit $ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h^2} $ exists, then by using limit laws we must have that the limit $$ \lim_{h \to 0} \frac{f(x + h) - f(x)}{h} = \left(\lim_{h \to 0} h\right) \cdot \left(\lim_{h \to 0} \frac{f(x + h) - f(x)}{h^2}\right) = 0 $$ exists, too. That means that the existence of your limit at $x$ implies that the derivative of $f$ at $x$ exists and is zero. Of course, there are plenty of functions for which the second derivative at a point exists, but nonetheless the derivative at the same point is nonzero.
An intuitive way to think about this when everything is smooth is to note that around the point $x$ the function $f$ is given by its Taylor series there: $$ f(x + h) = f(x) + f'(x) h + \frac{1}{2} f''(x) h^2 + O(h^3). $$ Your limit then simplifies into $$ \lim_{h \to 0} \left(\frac{f'(x)}{h} + \frac{1}{2} f''(x) + O(h) \right). $$ The only way this limit can exist is if $f'(x) = 0$, in which case it becomes $$ \lim_{h \to 0} \left(\frac{1}{2} f''(x) + O(h) \right) = \frac{1}{2} f''(x), $$ like we saw above.