What we exactly do when we take derivative of any function? When we take differentiation of any function then what actually we do with that function? Ex.d/dx of x^2 is 2x. 
So what we have actually done with x^2.
 A: We take the limit
$$\frac{d}{dx}\ x^2 = \lim_{\Delta x\to 0}\frac{(x+\Delta x)^2 - x^2}{\Delta x}$$
or generally
$$\frac {d}{dx}\ f(x) = \lim_{\Delta x \to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$$
The right hand side is a function depending on $x$.
A: Historically it was determining at each point $x$ the slope for $y(x) = x^2$ by using the slope $m$ of a secant of width $\Delta x$, running through the points $(x, y)$ and $(x+\Delta x, y + \Delta y)$ and then calculating the limit value of that slope 
$$
m 
= \frac{\Delta y}{\Delta x}
= \frac{(x+\Delta x)^2-x^2}{\Delta x}
= \frac{2x \Delta x+ (\Delta x)^2}{\Delta x}
= 2x + \Delta x
$$
when $\Delta x$ vanishes thus the secant turns into a tangent.
A: The derivative of a continuous function $f(x)$ is defined as 
$$f'(x)=\lim_{h\rightarrow 0}\frac{f(x+h)-f(x)}{h}$$
This is interesting to calculate, because it describes the slope of the tangent line to $f(x)$ in any given point $x$ if $f'(x)$ exists as a finite value. It is often viewed as the rate of growth for the function $f$.
In your example, we get that $f(x)=x^2$ has the derivative
$$f'(x)=\lim_{h\rightarrow 0}\frac{(x+h)^2-x^2}{h}=\lim_{h\rightarrow 0}\frac{2xh+h^2}{h}=\lim_{h\rightarrow 0}(2x+h)=2x$$
which is the rate of growth for $f(x)=x^2$ in a given point $x$.
