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Currently learning about Concavity and using the second derivative to measure the concavity of a function. However, I don't quite understand what the second derivative is telling me. I know that the first derivative tells us the rate at which the function is changing or the slope at any point. However, I can't quite conceptualize what the second derivative is telling me. When someone tells me it's the "rate at which the rate of change is changing" my mind goes blank. Could someone explain this to me please.

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    The first derivative of position is speed, the second acceleration. – 2017-01-05

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$f'$ tells you the slope of $f$. With this, you know that the function is increasing/decreasing or none of those. $f''$ gives you the possible inflection points of $f$ (and also information about critical points), which are points in which the curve "bends to the side contrary to the side it was going originally". I prefer not to understand in terms of "rate at which the rate of change is changing" because it seems cumbersome, take a look at the article I mentioned and see if it fits better.

For example:

$$\bbox[border:2px solid lightblue]{f(x)=x^3}\quad\quad\quad\quad\quad\quad \bbox[border:2px solid orange]{f(x)=3x^2}\quad\quad\quad\quad\quad\quad \bbox[border:2px solid lightgreen]{f(x)=x^3}$$

enter image description here

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The sign of second derivative show the concavity. That is if $y'' >0 $ then the function is convex or concave upward and if $y''<0$ then the function is concave.

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    Just to add to the answer - convex or concave upwards means the tangent line will lie below the graph of the function and the opposite for concave down. – 2017-01-05
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Short answer:

It seems you've conceptualized 'the rate of change' with slope, if so, then consider that a function's slope may change for different X values. Thus its slope will have a rate of change with respect to X. The 2nd derivative conveys this 'rate of change of slope'. If a function's slope is changing than its graph will curve, thus the 2nd derivative can express how a function is curving with respect to the X axis, which we call concavity.

Long answer:

The phase "rate at which the rate of change is changing" may be conceptualized by considering f(x)=3x. Its rate of change with respect to x is 3, f '(x)=3 aka slope=3. Looking at its graph one would see a line that increases 3 in the Y direction for an increase of 1 in the X direction.

And since each time you increase X by 1 Y will increase by 3, the 'rate of change' is not changing, it's always 3. So tying f(x)=3x back to the original phase, the "rate at which the rate of change is changing" is zero, because the rate of change is always 3. And since the slope is not changing, the graph f(x)=3x is a straight line and its 2nd derivative is zero, f "(x)=0

Now consider the graph of f(x)=3x^2, the first time X is increased by 1, Y increases by 3; (X=1,Y=3), but the next time X is increased by 1, Y increases by 9 (X=2, Y=12). So the "rate at which the rate of change is changing" is no longer zero. Thus the slope of f(x) is changing with x and its graph is no longer straight line. Instead it curves upward and its 2nd derivative is no longer zero, f "(x) =6.

Hopefully this helps connect the 2nd derivative with how a function is 'curving' which defines its concavity.

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    It would be very helpful if you used MathJaX formatting to make your answer more reader-friendly. This looks good! – 2017-03-30