Having difficulties understanding the concept of the slope:
Suppose we have $f(x)=x^2$, its derivative $f'(x)=2x$
At $x=10$, $f(x)=100$ and $f'(x)=20$. So the rate of change of the function at $(x,y)=(10,100)$ is $20$, but what does that mean? If we take $x+1=11$, we get: $f(11)=121$ and not $120$.
So what was the point of calculating the slope at (10,100)? What information did it provide us?
On
The line through the two points $(10,100)$ and $(11,121)$ on the graph is called a secant line, and it's slope is $21$.
If you pick a point closer to $(10,100)$ than the point $(11,121)$ you will find that the resulting secant line will have slope closer to $20$.
In a way, you are "sneaking up" on the slope of the tangent line by approaching it with a sequence of secant lines.
The slope of the tangent line is the limit of the slopes of the secant lines.
Well, it provided us with the slope at that exact point on the graph. Of course in this case the slope will change as soon as we move away from this point.
What can we do with the slope? We can for example see if that point on the graph is a possible local minima/maxima. If the slope is not equal to 0, there is no possibility for this to be the case, since we could move a bit up/down to find a higher/lower point. If we examine all cases where the slope is 0, we have a chance of finding our targets.
Another example would be the slope being acceleration of a particle in a physics model. Finding the acceleration at a point in time can allow us to derive other properties of the particle, such as force.