What does rate of change actually mean?

If we have a function, say, $y=x^2$, then $dy/dx=2x$.
Now $y=1,4,9,16...$ for $x=1,2,3,4...$
and $dy/dx=2,4,6,8...$ for $x=1,2,3,4...$
Now as $dy/dx$ represents rate of change of y w.r.t x but I can't understand how these particluar
values of $dy/dx$ represent change ?
I mean for x=3, the rate of change is 6. What does it mean ?

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Note: there are numbers other than integers... –  Arturo Magidin Jan 31 '11 at 14:36

You could think of it this way: taking the value of $\frac{dy}{dx}$ at some point $x_0$, tells you approximately how much the value of $y$ changes for values very close to $x_0$. So in your example, if $x_0 = 3$, the change is $6$: if you consider the value of $x^2$ at a point $x_0 + \epsilon$ for some small number $\epsilon$, the change in the values of the function from $x_0$ to $x_0 + \epsilon$ will increase by approximately $6\cdot \epsilon$.

You should try this with some numbers. Take $x_0 = 3$ and $\epsilon = 1$. Then $3^2 = 9$, while $(3+1)^2 = 16$. The change is $7$, which is close to $6\cdot \epsilon = 6$. If you take $\epsilon = \frac{1}{2}$, you can check that the change will be $3\frac{1}{4}$, which is close to $6\cdot \epsilon = 3$. Notice that as you make $\epsilon$ smaller, the error becomes smaller as well.

You should check wikipedia.

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Thanks for explaining with an example. –  Happy Mittal Jan 31 '11 at 7:28

When you let $x$ take the values 1, 2, 3, etc., you are taking fairly large steps, and then you don't see very well what the derivative $dy/dx$ means. Think of $dy/dx$ as measuring how much $y$ changes when $x$ changes by a very small amount, like when changing the value of $x$ from 3 to 3.00001; then $y$ changes by about 6 times as much, from 9 to approximately 9.00006 (the exaxt value is 9.0000600001).

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I find it helps to consider $x$ as a time variable. So let's write $t$ instead of $x$.

Suppose you've got a good racing car, and at time $t=0$ you're at the position $y=0$. You're planning to race along a straight road. Your car can accelerate in such a way that after $t$ seconds, the speed is $2t$ metres per second (that won't work very long ...). Then you can show that after $t$ seconds you'll be $y=t^2$ metres away from where you started.

Now let's check if that makes sense in terms of a rate of change: Your position at time $t$ is $y=t^2$, and the rate of change is $dy/dt = 2t$. OK, so the rate of change $dy/dt$ is just the current speed (wheras $y/t = t^2/t = t$ would be the avarage speed between time 0 and $t$).

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