help understand derivative in this example the derivative is a measure of how a function changes as its input changes.(from Wikipedia)
for instance, for function $y = x^2$ derivative $y'= 2x$, hence
$$\begin{align*}
x&= 6,&\quad y &=36,&\quad y'&= 12;\\ 
x&= 7,& y &=49,& y'&= 14;\\ 
x&= 8,& y &=64,& y'&= 16;\\
x&= 9,& y &=81,& y'&= 18;\\
x&= 10,& y &=100,& y'&= 20.
\end{align*}$$
In this case, intervals between  two consecutive output values are roughly equal to corresponding $y'$.
But for function $y= x^3$, derivative $y'= 3x^2$, hence 
$$\begin{align*}
x&= 1,&\quad y &=1,&\quad y'&= 3;\\ 
x&= 2,& y &=8,& y'&= 12; \\
x&= 3,& y &=27,& y'&= 27;\\
x&= 4,& y &=64,& y'&= 48;\\
x&= 5,& y &=125,& y'&= 75;
\end{align*}$$
But here  intervals between  two consecutive output values are not at all equal to corresponding $y'$.
What's wrong?
Did I understand "change in $y$ with respect to $x$" incorrectly?
What is the best way to understand "measure of how a function changes as its input changes"?
 A: The best way to understand "measure of how a function changes as its input changes" is to visualize it.
When you are given any function $f(x)$, you can look at the derivative in some point $a$ as the slope of the tangent line in that point. Take a look at this picture for an illustrative example.
When we take a look at linear functions ($f(x) = ax+b$), we see that the slope is the same for every point. We can calculate it by dividing the change in y by the change in x ( $\frac{y_2-y_1}{x_2-x_1}$ for any two points $(x_1,y_1) (x_2,y_2)$). 
When we take a look at functions of a higher degree ($a_nx^n + b_{n-1}x^{n-1} + \ldots + a_{1}x + a_0$), we notice that the slope is not necessarily the same for two points. We cannot use our general change in y over change in x formula anymore. This is where derivatives come in handy. So picture them as a way of finding the slope in any point for any given function of any degree (technically not every function, but you shouldn't worry about that for now). They also work for linear ones by the way.
To get back to the original example, if you would calculate the derivative of that function and would fill in the black dot's x-value, you would get the slope of the function in that point!
I hope it's a bit more clear for you now.
A: That is a very reasonable question. Really - I wish the students I'd taught had as much insight to ask such things.
Some people refer to the derivative as the best linear approximation. This, in a sense, is what you ask. It is closely related to Euler's Method, pictured below

The distinction here is that for higher degree polynomials, the derivative itself changes very rapidly (i.e. the concavity can really have large effects). How to compensate? You could calculate the derivative at the midpoints, such as at 1.5, 2.5, etc. instead. That often reduces some error. Or you could use smaller step sizes, i.e. to estimate the function at 2 you could find the derivative at 1 and use that to approximate the function at 1.2 - with that derivative you could approximate the function at 1.4, and so on until you get to 2. Then the error is much less! In fact, with polynomials, you can get arbitrarily small error by reducing the step size enough (very, very small).
Soon, perhaps, you will learn Taylor's Theorem. This answers the question: why should I only approximate with lines? What if I want to incorporate the changing slope, i.e. the concavity too? The jerk? The whatever-you-call-the-change-in-the-change-in-the-change-in-the-change of y over the change in x? etc.
A: The trouble is, your steps are too big. With $y=x^3$, if you go from $x=4$ to $x=5$ then $y$ changes by 61 which, as you note, isn't very close to $y'(4)=48$. But if you only go from 4 to 4.1, then $y$ changes by 4.921, which is pretty close to $(48)(.1)$, the derivative times the step size. Take a step of size .01, from 4 to 4.01, and $y$ changes by .4812, even closer to $(48)(.01)$. 
The derivative is a measure of instantaneous change. It's an exact measure in the limit, that is, as the change in the input goes to zero; it's an approximate measure when the change is not zero, and gets worse as the change gets bigger. For $y=x^3$, an input change of 1 is too big. 
A: They're not equal because $f(x) = x^2$ changes at a constant rate $f'(x) = 2x$, but as you pointed out $g(x) = x^3$ doesn't change linearly but rather as a quadratic $g'(x) = 3x^2$... The rate of change of the rate of change is different, if that makes sense. If you make that table for the first derivative of $g$ and the second derivative of $g$, you'd see what you saw in $f$ and $f'$.
