What is actual result of derivative? I'm pretty new to derivatives and I'm not sure if I understand the concept well.
Let's say we have function $f(x) = x^2$.If we set $x=3$, our example point is set at coordinates $(3,9)$.Using formula we can find derivative of $f(x)$:  $f'(x)=6$.
My question is: what does this number $6$ mean? From what I've read, it should be slope of the chosen point. Is the slope defined by one number? What is actual use of such slope?
Thanks!
 A: It's the slope of the line tangent to the graph of $f(x)=x^2$ at the point $(3,9)$.

Say you have a differentiable function (that is a function with a derivative) $f$. Now consider any point $(x,f(x))$ on the curve $f$. No matter how curvy the function is at that point, you can always zoom in enough so that the function looks pretty much like a straight line in a very small interval around that point.  The derivative $f'(x)$ is the slope of the line most similar to the curve at that point.

Notice how in the left picture, if we zoom in enough the graph looks like a line.  The derivative is the slope of that line.  The right picture on the other hand doesn't look like a line when you zoom in really close.  That tells us that that function doesn't have a derivative at the point right where it makes a 'v'.
A: If you have $f(x) = x^2$, then $f'(3) = 6$ is the slope of the tangent line to the graph of $f$ at the point $(3,f(3)) = (3,9)$.
In general, given a differentiable function $f: \Bbb R \to \Bbb R$, and $a \in \Bbb R$, $f'(a)$ is the slope of the tangent line to the graph of $f$ at the point $(a,f(a))$.
Armed with this interpretation, try to convince yourself that $f(x) = x^2$ has $f'(0) = 0$ just by looking at the graph, for example.
A: Notice, for a given function $y=f(x)$ the derivative $f'(x)$ shows the slope of the tangent line to the curve $y=f(x)$ at a given point say $(3, 9)$.
The slope is very important to find out the equation of the tangent at a given point on the curve.  
