I'd like to add to HallsofIvy's comment a link to concepts you are probably familiar with from analytical geometry:
When your teacher introduced you to "the derivative", I would think that he/she first talked about secants/secant lines and then about tangents/tangent lines. (?)
In particular, you may have learned that the slope of the tangent line at some point of a curve may be found as the limit of the slopes of secant lines associated with that point when the distance between the two points (on the curve) defining a secant line goes to zero.
To make maths out of this:
1.
Let the two points on the curve be:
P_{1}=(x,y(x)), P_{2}=(x+\bigtriangleup{x},y(x+\bigtriangleup{x}))
2.
Since you've got two points, P_{1},P_{2}, you can evidently draw a straight line between them!
This straight line is called the secant line S with respect to the points P_{1},P_{2}.
3.
Now, I would think that you know that a straight line L in the plane usually can be represented as a function Y(X)=Ax+B, where Y(X) is the vertical coordinate Y at a point at L, while X is the horizontal coordinate of the same point at L.
A and B are constants for L (equal values for all choices of X!); B is called the Y-intercept (lies on the Y-axis, X=0), while A is called the slope of L.
4.
Going back to our secant line S, how can I find its slope?
We know 2 things about S:
a) If X=x, then Y(X)=y(x) (i.e, we're at the point P_{1})
b) If X=x+\bigtriangleup{x}, then Y(X)=y(x+\bigtriangleup{x}) (i.e, we're at the point P_{2})
Going back to the general equation for a line L, we must therefore have for S:
A*x+B=y(x), A*(x+\bigtriangleup{x})+B=y(x+\bigtriangleup{x})
Solving these equations for A and B, we find:
A=\frac{y(x+\bigtriangleup{x})-y(x)}{\bigtriangleup{x}}
B=y(x)-A*x
Hence, we may represent the Y-coordinate of a point on S, S_{Y}, as a function of the horizontal coordinate, X, like this:
S_{Y}(X)= \frac{y(x+\bigtriangleup{x})-y(x)}{\bigtriangleup{x}}X+y(x)-A*x
This is the way in which the secant line S can be represented in the usual manner of a line L.
5.
We are interested in the slope of S, \frac{y(x+\bigtriangleup{x})-y(x)}{\bigtriangleup{x}}
This is called the quotient of differences, as HallsofIvy says.
6.
In order to find the slope of the tangent line at point P_{1}, we evaluate the slope expression from S as we let the difference between the values of horizontal coordinates of P_{1},P_{2} shrink to zero.
(That difference is \bigtriangleup{x}).
Geometrically, this limiting process has the interpration that we evaluate the slopes of different secant lines which have P_{1} in common, but where each secant line's P_{2} is chosen to be progressively closer to P_{1}.
The tangent line's slope is found when P_{2} becomes P_{1}.
The derivative of y at x, \frac{dy}{dx}, is the name of the slope of the tangent line at P_{1}.
