Professor Dave here, let’s talk about differentiation.
If we are going to learn calculus, we are going to have to understand the concept of
differentiation.
To introduce this concept, let’s examine a particular problem.
Say we want to find the equation of a line that is tangent to a curve at some point.
With the word tangent, we mean that the line has precisely one point in common with this
section of the curve, so there can be only one tangent line, as if the slope were any
bigger or smaller, it would necessarily intersect the curve at a second point.
Say the curve we are looking at is our good old friend, y equals x squared, the basic
parabola.
Let’s choose a point on this curve, like (1, 1).
We know this point is on the curve, because one squared equals one.
What is the equation of the tangent line at this point?
If we recall our algebra, we know that to get the equation of a line in slope-intercept
form, we either need any two points on the line, or we need one point and the slope of the line.
But alas, we don’t have this information.
We just have the one point.
Well one thing we can do is pick some other point on the curve, like this one, and we
know that any point on the curve must have the coordinates (x comma x squared), because
the y value is always the square of the x value.
If we make a different line between these two points, which we call the secant line,
we can write an expression to get the slope of this line, because we now have two points,
those being the new one, and the point (1, 1) that we started with.
Remember that slope is rise over run, so it’s y2 minus y1 over x2 minus x1.
That will be x^2 minus one, over x minus one.
If we move this second point around, the line changes, as does its slope, but this expression
will stay the same.
Now let’s plug in some values for this second point and see what happens to the slope of the line.
Say we start with x equals two.
X squared is four, so we get three over one, or a slope of three for the secant line.
Now let’s try plugging in 1.5.
X squared will be 2.25, so we get 1.25 over 0.5, which gives us 2.5 as the slope.
Now let’s keep pushing this second point closer and closer to our initial point.
We can try X equals 1.1, then 1.01, then 1.001.
The resulting line gets closer and closer to being equivalent to the tangent line we
are looking for, and its slope gets closer and closer to two.
We can even do this from the other direction, starting with x equals zero, going up to x
equals one half, 0.9, 0.99, and we see the same effect.
The line gets closer and closer to becoming the tangent line, and we again approach a
slope of two, just from the other direction.
This would seem to suggest, or even demand, that the slope of the tangent line must be
equal to two.
If the slope is equal to two, then we just do some algebra, by plugging in the coordinates
of the point as well as the slope into the equation for a line.
We get one equals two times one, plus B, and the intercept must be negative one.
Therefore, the equation of this tangent line is y = 2x – 1.
In finding the slope of this tangent line, we have performed differentiation, and we
have thereby discovered something about the rate of change in this function at this point.
Speaking of rates of change, this is another lens we can use to comprehend the concept
of differentiation, so let’s move forward and examine things from another point of view.