The Difference Quotient: The Bridge between Algebra (Slope) and Calculus (the Derivative)One of the cornerstones of calculus is the difference quotient. The difference quotient — along with limits — allows you to take the regular old slope formula that you used to compute the slope of lines in algebra class and use it for the calculus task of calculating the slope (or derivative) of a curve. Here’s how it works.

In the following example, you want to find the slope at a point on the parabola.

To compute the slope, you need two points to plug into this formula. For a line, this is easy. You just pick any two points on the line and plug them in.

You can see the line drawn tangent to the curve at (2, 4), and because the slope of the tangent line is the same as the slope of the parabola at (2, 4), all you need is the slope of the tangent line. But you don’t know the equation of the tangent line, so you can’t get the second point — in addition to (2, 4) — that you need for the slope formula.

Here’s how the inventors of calculus got around this roadblock.

The above figure is the graph of y = x2 with a tangent line and a secant line. It shows the tangent line again and a secant line intersecting the parabola at (2, 4) and at (10, 100).    

A secant line is a line that intersects a curve at two points. This is a bit oversimplified, but it’ll do.

The slope of this secant line is given by the slope formula:

You can see that this secant line is quite a bit steeper than the tangent line, and thus the slope of the secant, 12, is higher than the slope you’re looking for.

Now add one more point at (6, 36) and draw another secant using that point and (2, 4) again. See the above figure.

Calculate the slope of this second secant:

You can see that the slope of this secant line is a better approximation of the slope of the tangent line than the slope of the first secant was.

Now, imagine what would happen if you grabbed the point at (6, 36) and slid it down the parabola toward (2, 4), dragging the secant line along with it. Can you see that as the point gets closer and closer to (2, 4), the secant line gets closer and closer to the tangent line, and that the slope of this secant thus gets closer and closer to the slope of the tangent?

So, you can get the slope of the tangent if you take the limit of the slope of this moving secant.

So here’s the limit you need:

Watch what happens to this limit when you plug in three more points on the parabola that are closer and closer to (2, 4):

When the point slides to (2.01, 4.0401), the slope is 4.01

When the point slides to (2.001, 4.004001), the slope is 4.001

Sure looks like the slope is headed toward 4.

As with all limit problems, the variable in this problem, the run, approaches but never actually gets to zero. If it got to zero — which would happen if you slid the point you grabbed along the parabola until it was actually on top of (2, 4) — you’d have a slope of 0/0, which is undefined. But, of course, that’s precisely the slope you want — the slope of the line when the point does land on top of (2, 4). Herein lies the beauty of the limit process.

And the slope of the tangent line is — you guessed it — the derivative.

The derivative of a function, f(x), at some number x = c, written as f’(c), is the slope of the tangent line to f drawn at c.

Okay, here’s the most common way of writing the difference quotient (you may run across other, equivalent ways).

Take a look at the following figure, which shows how a limit produces the slope of the tangent line at (2, 4).

Doing the math gives you, at last, the slope of the tangent line at (2, 4):

So the slope is 4. (By the way, it’s a meaningless coincidence that the slope at (2, 4) happens to be the same as the y-coordinate of the point.)

How to Determine Maximum and Minimum Speeds of Moving ObjectsBy Mark Ryan from Calculus For Dummies, 2nd Edition

One of the most practical uses of differentiation is finding the maximum or minimum values of a real-world function, for example, the maximum and minimum speeds of a moving object.

You can think of velocity as the more technical version of speed.

Here's an example. A yo-yo moves straight up and down. Its height above the ground, as a function of time, is given by the function H(t) = t3 − 6t2 + 5t + 30, where t is in seconds and H(t) is in inches. At t = 0, the yo-yo is 30 inches above the ground, and after 4 seconds, it's at a height of 18 inches, as shown in this figure.

The yo-yo's height, from 0 to 4 seconds.

To determine the total distance the yo-yo travels, you need to add up the distances traveled on each leg of the yo-yo's trip: the up leg, the down leg, and the second up leg.

First, the yo-yo goes up from a height of 30 inches to about 31.1 inches (where the first turn-around point is). That's a distance of about 1.1 inches. Next, it goes down from about 31.1 to about 16.9 (the height of the second turn-around point). That's a distance of 31.1 minus 16.9, or about 14.2 inches. Finally, the yo-yo goes up again from about 16.9 inches to its final height of 18 inches. That's another 1.1 inches. Add these three distances to obtain the total distance traveled:

Note: Compare this answer to the total displacement of  –12, which you get from subtracting the final height of the yo-yo, 18 inches, from its initial height of 30 inches. The displacement is negative because the net movement is downward. And the positive amount of the displacement (namely 12) is less than the distance traveled of 16.4 because with displacement the up legs of the yo-yo's trip cancel out part of the down leg distance. Check out the math:

The yo-yo's average speed is given by the total distance traveled divided by the elapsed time. Thus,

Say you determine that the yo-yo's maximum velocity is 5 inches per second, and its minimum velocity is –7 inches per second. A velocity of –7 is a speed of 7, so that's the yo-yo's maximum speed. Its minimum speed of zero occurs at the two turnaround points.

A good way to analyze maximum and minimum speed is to consider the speed function and its graph. (Or, if you're a glutton for punishment, check out the following mumbo jumbo.) Speed equals the absolute value of velocity.

Velocity, V(t), is the derivative of position (height, in this problem). Thus:

So, for the yo-yo problem, the speed function,

Check out the graph of S(t) in the following figure.

The yo-yo's speed function S(t)

Looking at this graph, it's easy to see that the yo-yo's maximum speed occurs at t = 2

and that the minimum speed is zero at the two x-intercepts.

Minimum and maximum speed: For a continuous velocity function, the minimum speed is zero whenever the maximum and minimum velocities are of opposite signs or when one of them is zero. When the maximum and minimum velocities are both positive or both negative, then the minimum speed is the lesser of the absolute values of the maximum and minimum velocities. In all cases, the maximum speed is the greater of the absolute values of the maximum and minimum velocities. Is that a mouthful or what?

How to Differentiate the Trigonometric FunctionsYou should memorize the derivatives of the six trig functions. Make sure you memorize the first two in the following list — they’re a snap. If you’re good at rote memorization, memorize the last four as well.

Here they are:

You might enjoy the following mnemonic trick for the last four trig derivatives. (This might seem a bit convoluted, but it works). Imagine you’re taking a test and can’t remember these derivatives. You lean over to the guy next to you and whisper, “Psst, hey, what’s the derivative of cscx?” Now, take the last three letters of psst (sst) — those are the initial letters of sec, sec, tan. Write these three functions down, and below them, write their cofunctions: csc, csc, cot. Put a negative sign on the csc in the middle. Finally, add arrows:

Using this diagram, the trig derivatives are very easy to remember. Look at the top row. The sec on the left has an arrow pointing to sec tan — so the derivative of secx is secx tanx.

The bottom row works the same way, except that both derivatives are negative.