Before Calculus, the only functional that an individual can find the slope of is a straight line.  What is the slope of the line below?  It's trivial to determine that the rise/run slope of this function is 1/2.

Now, what's the slope of the graph below?

It changes.  As the x-coordinate increases, the slope increases.  Using differential Calculus we can analyze this function's rate of change to find, at any given point, the exact slope.  Voila! At x=1 the slope is exactly 2, a straight line with the slope of 2 and the correct y-intercept is perfectly tangent to the function!

Knowing that the slope is 2 at x=1, a line can be generated using the following equation, commonly referred to as the point-slope form:

Substituting in the x and y values (x1 = 1, y1 = 1), and m = slope = 2, the equation for our perfectly tangent line is:

Plotted together with our original function, we have the following:

But how was this accomplished? First imagine the case where we attempted to get the slope of the function. It doesn't work because the slope clearly changes at every point.  Let's try to get the slope at the coordinate x=1.  We can start by using the rise/run formula between x=[0, 1].  We let our first coordinate be the origin, (0,0) and our second coordinate be (1,1).  Running our calculations through, our graph looks something like the following:

This isn't correct by a rather large margin.  Let's try moving our origin coordinate a little closer, to (0.5, 0.25).  Our slope is now, and our graph is now:

That's a lot better, but we can still see that there's some difference between the function and our line.  Let's try it again, moving our origin coordinate much closer, to (0.99, 0.9801).  Our graph is now:

It should now be very clear how differential Calculus works.  Finding the rate of change of a function is accomplished by using infinitely small units anywhere in the function.  The equation for accomplishing this is below:

The lim (limit) means that, as the delta-x becomes infinitely small, we approach the point where the rise/run approaches the perfectly correct amount.  An example:

Let:

If we substitute f(x) with x^2 within the limit, then we have the following:

Expanding this equation within the limit and simplifying, the following can be obtained:

Finally, if we do the limit and substitute in the fact that then we have the final result:

This process can be applied to ANY equation. However, some equations are impossible to differentiate. There are many tricks and accepted general formulas to differentiating many equations

Below are some graphs that I think are cool.  The derivative of the sine() is cosine().  The derivative of cosine() is -sine().  The derivative of -sine() is -cosine().  And finally, the derivative of -cosine() is sine().  These four graphs are all plotted together.

In addition to making pretty graphs, Calculus has applications in nearly every field of physical study.  One application is maximization and minimization problems.

Suppose a person wishes to maximize their viewing angle at a movie theatre, as shown below.  The bottom of the screen is ten feet from the ground, and screen is fifty feet tall.  The seat is oriented so that a person's eyes are 3 feet above the ground, and the floor is inclined at 15 degrees. How far back should a person sit to maximize their viewing angle?

We want to create an equation that tells exactly what the angle of viewing is as a function of the distance from the screen.  First it is important to note that the offset height due to the floor's incline, y, can be expressed as a function of the distance from the screen, x, with the following equation:

The function can be described as the top angle subtracted by the bottom angle. The top angle can be computed by noting that the tangent of the top angle is equal the difference of the movie screen's top height minus the viewing height, divided by the screen distance, (basically rise/run).  The same can be done for the bottom angle, and the final equation is below.

How can this equation be useful? To find the maxima or minima of an equation, simply set the derivative of the equation equal to zero. How does this work? Think about it.  When does a mountain peek occur? When the change in the mountain height switches from positive to negative. When dealing with infinitely small units, a maximum point or minimum point can be found by searching for when the change switches from positive to negative (maximum) or negative to positive (minimum).

What a monster! Set it equal to zero and solve for x, thank goodness for math programs, and we have...

Alright, so now we know that the best seat in the house is 19.29 feet horizontally from the screen.  If we plot the graph of the viewing angle, we can see clearly that this makes perfect sense. Note that the viewing angle is in terms of degrees, while in the equations above it was in respect to radians.

There are many ways to prove that a point is either a maximum or a minimum.  My favorite method is the second derivative test. Take the derivative of the first derivative to find if the slope itself is increasing or decreasing, implying that the whole thing is is either a minimum or maximum respectively.

• Differentiation - University of Texas Mathematics Department
• Derivative - MathWorld (a good listing of the many tricks of derivatives)