# Difference between function, derivative and second derivative?

If you are looking at three graphs: one is the original function, one is the derivative and the other is the second derivative, what is the accepted way of determining which is which?

For example this input.

Is it a simple as counting the number of maximums/minimums?

-

Not really; those examples are unrealistically simple. One of the most useful techniques is to match up zeroes of one graph $-$ places where it crosses the $x$-axis $-$ with ‘flat spots’ $-$ horizontal tangents $-$ of another. If graph $A$ has a zero everywhere that graph $B$ has a horizontal tangent, there’s a good chance that $A$ represents the derivative of the function whose graph is $B$. Of course you should then check further: is $A$ positive $-$ above the $x$-axis $-$ where $B$ is increasing, and negative where $B$ is decreasing? If so you’ve almost certainly found a match: either $A$ is the graph of $f\;'$ and $B$ that of $f$, or $A$ is the graph of $f\;''$ and $B$ that of $f\;'$.
Once you’ve matched up one pair like this, the third graph is generally pretty easy to identify. In the example just described, in which $A$ shows the derivative of the $B$ function, there are just two possibilities: either $C$ shows the derivative of the $A$ function, or $B$ shows the derivative of the $C$ function. It shouldn’t be hard to choose between the two by using the same ideas that I mentioned in the first paragraph.
@ParachutingPanda: It’s pretty unsafe to do so: the graphs might be of functions with a great many local extrema outside the region pictured. Even if that’s not the case, you can easily go astray: $y=x^3$ has no local extrema, but its derivative has an absolute minimum at the origin. You might count extrema to get an idea of where to start, provided that you remember that it can be very misleading, but in the end you’re going to have use the kind of analysis that I described: I simple count of extrema will give you the right answer only if you’re very lucky. – Brian M. Scott Nov 6 '11 at 11:49