Understanding rate of change If we are given the function $y=3x+5$, and were asked to obtain its derivative we would get $y'=3$, which means that as $x$ increases, the rate at which $y$ changes is always $3$.. so if $x=1$, then $y=8$ and if $x=2$ then $y=8+3=11$ and so on.. I was trying to test this concept in a different function, say $y=x^{2}+x$, the derivative is $y'=2x+1$, does that mean that the rate at which $y$ changes is always $2x+1$? If $x$ increases by $1$, then we would have $y=1+1=2$, and when $x$ increases even more by $1$, we will get $y=2^{2}+2=6$.. how can I relate $2$ and $6$ to $y'=2x+1$ like how I did with $y=3x+5$?
 A: As mentioned in the comments, the derivative is a localized change at $x$. For example, in you case you have $y'=2x+1$. That means that if you are sitting at $x=1$ and you change $x$ by a very small amount $\varepsilon$, your rate of change $$\frac{\Delta y}{\varepsilon}=2\cdot1+1=3$$
But if you are $x=2$, the same change $\varepsilon$ in $x$ would change $y$ by a different amount
$$\frac{\Delta y}{\varepsilon}=2\cdot 2+1=5$$
A: I believe the best way to explain this is to take the motivation from physics - which was the original motivation for Newton to invent the differential calculus.
Let $x$ denote time, and let $y$ be the position of a particle at time $x$ along the number line. So, if $y=3x+5$, this means that the movement of the particle followed the following table:
$$\begin{array}{r|r}\text{Time }(x)&\text{Position }(y)\\\hline0&5\\1&8\\2&11\\3&14\end{array}$$
You can make up a similar table for $y=x^2+x$:
$$\begin{array}{r|r}\text{Time }(x)&\text{Position }(y)\\\hline0&0\\1&2\\2&6\\3&12\end{array}$$
Now, a proper way to understand the derivative $y^\prime$ is that it is the velocity of the particle at the time $x$.
In the first case, the particle moves at constant velocity of $3$ units of position per $1$ unit of time. This is why $y^\prime$ always comes up as $3$.
In the second case, the velocity changes with time: it is $y^\prime=2x+1$. This means that the velocity follows the following table:
$$\begin{array}{r|r}\text{Time }(x)&\text{Velocity }(y^\prime)\\\hline0&1\\1&3\\2&5\\3&7\end{array}$$
Obviously, as the velocity changed (in this case increased), you cannot expect it to match the average velocities in certain time periods: for example, between time $x=1$ and $x=2$ the particle moved from $y=2$ to $y=6$, so the average velocity in that interval was $\frac{6-2}{2-1}=4$. This is somewhere between the velocity at the beginning of the time interval ($3$) and at the end ($5$).
Now, you may ask yourselves - what is the velocity at a given time point $x$ in the first place? The average velocity calculation requires two time points. How do you do with just one? The answer is:


*

*First take your point $x$ and another time point $z\ne x$ and calculate the average velocity beteween $x$ and $z$ (or between $z$ and $x$, if $z<x$).

*Then see how this average velocity changes when $z$ gets closer to $x$.


In our case, with $y=x^2+x$, the average velocity on the time interval between $x$ and $z$ is:
$$\begin{array}{rl}\frac{y(z)-y(x)}{z-x}&=\frac{z^2+z-x^2-x}{z-x}\\&=\frac{(z-x)(z+x+1)}{z-x}\\&=z+x+1\end{array}$$
Now, as $z$ gets closer to $x$, $z+x+1$ gets closer to $2x+1$ and that is why, in the "limit case" we get the velocity at the point $x$ to be $2x+1$. The role of calculus is to formalise what it means when we say "limit case", so you have a precise definition of a limit, and because $\lim_{z\to x}(z+x+1)=2x+1$ you can justify the previous (slightly loose) observation.
Bottom line: Derivative at the point is not very useful, out of the box*, to calculate average rates of change, but the average rate of change and the derivative are otherwise very closely related, and the latter is a limit of the former when the $x$ variable changes only slightly. I advise you to get back to your problem, but instead of looking at $y(1)$ vs. $y(2)$ you try to look at $y(1)$ vs. $y(1.0001)$ and you will see a lot better correspondence with the value $y^\prime(1)$.
*Barring further knowledge, e.g. of integral calculus, that is. Knowing that, you might know that the change of $y(2)$ vs. $y(1)$ is given by $\int_1^2 y^\prime(x)dx=\int_1^2 (2x+1)dx$ - but this is better left for a separate question!
A: Understanding derivative is a crucial thing. Among many definitions, "the derivative of a function $f(x)$ at a point $a$ is the best linear approximation" is an important one.
Let us take your second function. $f(x)=x^2+x$. So the derivative of $f$ at a point $x_0$ is $3x_0+1$. What do we mean by this is the best linear approximation?
you can draw many lines passing through $(x_0,f(x_0)) $ right. take any such line and let $m$ be the slope of that line. then what is the point on that line correspond to $x$? That is $f(x_0)+m(x-x_0)$. Now, what is the error if we think this line is the linear approximation of $f(x)$ at $x_0$.
That is $f(x)-(f(x_0)+m(x-x_0))$. If we  set $x=x_0+\Delta x$ we choose the $m$ s.t 
$\lim\limits_{\Delta x \to 0}\frac{f(x_0+\Delta x)-(f(x_0)+m(x-(x_0+\Delta x)))}{\Delta x}=\lim\limits_{\Delta x \to 0}\frac{f(x_0+\Delta x)-f(x_0)-m\Delta x}{\Delta x}=0$ That $m$ is the derivative of $f(x)$ at $x_0$.

Now if we come to problem you are facing, Now your $x_0=1$ and $f(1)=2$. Now what is $m$ of $f(x)$ at $1$? It is $4$. Now, what is the point on the best approximation line correspond to $2$? It is $f(x_0)+m(x-x_0)=5$. But the actual point $f(2)=6$, So as we seen simbolicly above, here we get an error of $1$.
Now can you understand why you didn't get an error when you trying to workout $f(x)=3x+5$?
A: In the first case $y=3x+5$ is a linear function that is just characterized by the fact that the rate of increment is constant at any point (and in any direction for functions of more variables).
In the second case the function is not linear and the derivative is just the best linear approximation of the function near  a point. This is the best and more general definition of derivative and make clear why we cannot use the derivative to find the values of the function for distant points.
A: For a linear function, such as $y=3x+5$, the rate of change is a constant everywhere, which is $y'=3$. 
In contrast, for a non-linear function, such as $y=x^2+x$, its rate of change $y=2x+1$ varies with the location of $x$. For $x=1$, it is 3, while for $x=2$, it is 5. The rate of change increase as $x$ becomes larger.
A: This is why I've been inclined - esp. since it was brought up in this post:
https://mathoverflow.net/a/40136/11576
to question the utility of the notion of a "rate of change" as a good way to introduce the derivative.
What the derivative, $f'(x_0)$, of a function $f$, at the point $x_0$, "is", intuitively, is it is the "sensitivity of the output value of the function to 'small' changes in the input value", when viewed as how strongly the output value "responds" as compared, proportionately, to the size of the small change in input value. That is, given that the input value is $x_0$, and the output value is $f(x_0)$, suppose I perturb this input by a "tiny" amount $dx$, i.e. modify it to
$$x_0 \pm dx$$
. Then I observe the change in the value of $f$, i.e. how
$$f(x_0 - dx)$$
and
$$f(x_0 + dx)$$
compare against the original value, i.e.
$$f(x_0)$$
. Namely, I then find how the change, i.e.
$$df = f(x_0 + dx) - f(x_0)$$
compares with $dx$, i.e.
$$\frac{df}{dx} = \frac{f(x_0 + dx) - f(x_0)}{dx}$$
.
You can think about this like a machine with a knob on it, and a needle display. At the beginning, the knob (reprsenting the input to the function, here $x_0$) is set at some particular setting, and the needle on the display is also pointing at some value (i.e. $f(x_0)$). If you now nudge or "wiggle" the knob just a little bit one or the other way, the derivative is telling how much bigger the needle "nudges" or "wiggles" in response to your small nudge. If it wiggles a lot, the derivative is big, if it wiggles only a little, the derivative is small, if it wiggles in the same direction as you wiggle the knob, the derivative is positive, and if it wiggles in the contrary direction, the derivative is negative.
Now, the tricky part which pertains to your question are both the notion that this is a "sensitivity to changes" and, moreover, that the changes involved must be "small". Because changes are what count, the derivative need not be the same at each point $x_0$ - and moreover, because the changes must be small, one should not expect that a large change will follow a naive "slope" calculation read off the value of the derivative evaluated at any given point. In terms of our machine example, suppose the knob is set to a considerably different setting. It may be that we may observe that the needle on the dial may shift around more or less depending on how we wiggle the knob a little bit around that other setting, than around the first setting - hence, the derivative varies from point to point.
Thus, when you have
$$f(x) := x^2 + x$$
so that $f'(x_0) = 2x_0 + 1$, and we evaluate the derivative at, say, $9$, to get $f'(9) = 19$, what this value means is that if I "wiggle" the input $9$ plus or minus only by a tiny enough amount, will the similarly-tiny change in $f(x)$ be about 19 times larger than the tiny change I made to x (here, $9$).
And this is easily demonstrated: suppose I change $9$ by $0.00001$. We have
$$f(9) = 90$$
but
$$f(9.00001) = 90.0001900001$$
(exactly!), which is a change of ever so slightly more than 0.00019 in the output, versus a change of 0.00001 in the input, so the ratio is about 19, as we should expect. If we make the change even smaller, it will be even closer to 19: thus, the "limit" concept - the "proper" definition deals with an "evanescent" impulse to the input value.
Yet, quite clearly, this doesn't work if I make a huge change: consider $f(1)$ - that's a change to the input by a whole factor of $9$, so not small at all. Then $f(1) = 2$, or a ratio of $45$, despite that $19 \cdot 9 = 171$. 
Putting this all together, the answer to your question, using your terminology, is that near any given point $x$, a suitably-small change applied to $x$ will cause $y$ to change by a factor of approximately $2x + 1$ more than that small change, with that approximation being better the smaller the change we use. The reason we can use a small change only is because the function is non-linear: only a linear function has a constant sensitivity to changes regardless of at which point we take it, hence the response of a non-linear function cannot be fully characterized by a single number alone.
