The function $f(x) = Ae^{kx}$, where $A$ and $k$ are positive constants is said to have "constant, relative" rate of change.
How does it differ from the statement that it has a constant rate of change?
Thanks,
[Math] What does it mean to for a function to have a “constant, relative” rate of change
calculus
Related Solutions
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.
Let's say the function is $f(t)$ which gives distance, so that the derivative is $f'(t)$ or instantaneous speed. You can think of it as follows: what speed is the car (for example) travelling at that precise moment in time. Or: if the car was to continue at the same speed, what would it be going. $f'(t)$ gives the speed at time t exactly. That is the intuition behind what a rate of change at one point means.
EDIT: So say we want to know the speed right at time $t$. We could go a little ways out to time $t+\Delta t$ where $\Delta t$ is a very small number, and see how much our distance has changed over that time interval: so take $f(t+\Delta t)-f(t)$. We have $$\text{speed}=\frac{\text{change in distance}}{\text{change in time}}$$ so the speed is $$\text{speed}=\frac{f(t+\Delta t)-f(t)}{\Delta t}$$ approximately. We can closer and closer to the exact speed at $t$ itself by making the change $\Delta t$ smaller and smaller. So the real speed is the limiting value $$\text{speed}=\lim_{\Delta t\to 0}{\frac{f(t+\Delta t)-f(t)}{\Delta t}}=f'(t)$$ Remember that limits don't tell us what the function actually reaches when $\Delta t=0$; we can make $\Delta t$ as small as we like, but it can never equal zero. The limit tells us what the function seems to be approaching as we make $\Delta t$ tend to $0$ - we infer that the speed at the point itself, if we could find it, would be that limiting value (which is, as you should notice, the derivative by definition).
EDIT2: How can the derivative be exact?
Consider the following graph: 
You are looking at the function $$f(x)=\frac{\sin x}{x}$$ This function approaches $1$ as $x$ goes to zero because $$\lim_{x\to 0}{\sin x\over x}=1$$ However, the function is undefined at $x=0$ itself.
Finding a derivative is a little like asking: if there was a point at $x=0$, what would it be? We can make $x$ get closer and closer to $0$, and the function $f$ will get closer and closer to $1$. But it will never get there exactly.
However, if there was, hypothetically speaking, a point at $x=0$ like the blue one I have drawn in, what would its y-coordinate be? Saying that "the limit of f is 1" does not mean that "the function equals 1". It means that "the function never equals 1, BUT... it gets closer and closer, and if there was a value there, it would have to be 1".
So when taking a derivative, you are working your way around a value that "could" exist, but that just doesn't work at the point itself. However, you can infer what the value must be by looking at what the function gets closer and closer to - in other words, by taking the limit. The exact limit.
Best Answer
It means that the rate of change, relative to the actual value, is constant. It always increases (or decreases) by a fixed percentage in a given time interval.