"i already know that when there's a cusp on the graph, the derivative of that point is at zero"
Then you have a real problem. That's completely wrong. When there is a cusp on the graph, the derivative does not exist! It is when the graph is "level" (horizontal) that the derivative is 0.
ShawnD's advice is very good. From the left, the derivative starts out close to 0, increases and has an asymptote (goes to infinity) where the graph appears to be vertical. The slope is not negative after that (ShawnD was talking about the section after the first dotted line. I am still talking about the portion before it). It comes back down from "infinity" to some positive number at that dotted line. There is, of course, no derivative at the dotted line- the function shown is not even continuous there. On the right side of the dotted line, the derivative comes up from -infinity (the dotted line is a vertical asymptote), goes up to 0 where the curve "bottoms out" and the becomes positive. It has some positive value approaching that cusp, has no value at all (not zero!) at the cusp (show an "open" circle there) and the has some negative value immediately after. In fact, since the graph appears to be a straight line (with negative slope) the derivative graph should be a horizontal straight line (at some negative value). At the lower cusp, again an "open" circle showing no derivative and then a positive derivative. Almost horizontal- looks to me like the graph bends up just a little- suddenly increasing to a large positive value, then dropping back to 0 at the top of the "arch", then going to negative infinity as the graph approaches that second dotted line. An open circle at the dotted line (the function is not even defined there) and it looks to me like the function itself is a constant past that which means, of course, that the derivative is 0.