It’s Professor Dave, let’s define the derivative.
From our earlier investigations with slopes of tangent lines and instantaneous velocities
of falling objects, we gained a basic grasp of the concept of differentiation, so let’s
move forward and expand upon our understanding of this operation.
When we first introduced differentiation, we looked at a method for calculating the
slope of a tangent line. By using this method of exhaustion, we were performing differentiation,
and in doing so, we found the derivative of this function at this point. This derivative
was defined to be the slope of the tangent line at this point, and the slope of the line
was derived from the function. The derivative of a function tells us about the rate of change
in that function at a particular input value. If a function is increasing at a particular
point, its derivative is positive at that point, which is just a fancy way of saying
that the slope of its tangent line is positive at that point. If a function is decreasing
at a particular point, its derivative is negative at that point, and the slope of its tangent
line will be negative. Let’s take a moment to note something here.
It is crucial that we mentally separate the value of a function at some point from the
change in that function at that point. If examining position and speed, we could be
located at the origin, meaning our position is zero, but with non-zero speed, meaning
we aren’t standing there but just passing by. Or we could be at some other non-zero
location, but standing still, with zero speed. We can also think of stock prices. These can
be high or low, but their values are typically unrelated to their rate of change. They could
continue to climb, or plummet at any moment. Now looking at the tangent line again, let’s
formalize this a little bit more. Remember that our method involved choosing a second
point on the curve. We can give the first point the coordinates (a, f(a)), and the second
point can be (x, f(x)). As we said, the slope of this line, which we can call the secant
line, is rise over run, so f(x) minus f(a) over x minus A. Then we can allow point x
to approach point A, and we said that the slope of the tangent line will be the slope
of this secant line in the limit of x approaching A. We can express this another way, with the
second point in terms of the first, by saying that x is equal to A plus some other quantity,
h, so h is the distance in the x direction from the first point to the next. That means
this second point has the x coordinate A plus h, and the y coordinate f of (A plus h). Now
to get rise over run, we find f of (A plus h) minus f of A, over h. This time, the slope
of the tangent line is the limit of this expression as h approaches zero, meaning the distance
between the points becomes zero, and the points become one. What we must understand is that
we can just as easily refer to this value as the slope not just of the tangent line,
but the slope of the curve itself, at this precise point. If we were to zoom into the
curve at this point, closer and closer, it would eventually resemble a line with this
slope. And in fact, combining this knowledge with this expression will give us a better
understanding of the definition of a derivative. The derivative of some function, f, at some
input value, A, which we can represent as f prime of A, is precisely this expression
we just described. So to put it in words, the derivative of a function at some input
is equal to the slope of the tangent line at that point. Identically, the derivative
of a function at some input is equal to the rate of change of the function at that point.
These are two ways of saying precisely the same thing, and that thing is the derivative
of the function. Let’s use this expression to take the derivative
of some different functions. Let’s start with something easy. How about some constant
function, like f(x) = 5. If we take the derivative of this, when we go to plug in values for
f(x plus h) and f(x), these will both be 5, because no matter what point on the function
we are talking about, it’s equal to five. Five minus five is zero, so as h approaches
zero, f prime of x will become zero. This makes sense, as the rate of change anywhere
on this function is zero, because the value of the function never changes. So let’s
make a note that when we take the derivative of a constant, any constant, it will always
be zero. This should be reassuring, as it means that our precise math is aligning with
the intuitive idea we wanted it to reflect. Now let’s try the function f(x) = x. If
we try to find the derivative of this, for f(x plus h), we get (x plus h), and for f(x),
we get x. So (x plus h) minus x means the x’s cancel, which gives us h, and h over
h is one. So the derivative of x is one. Now let’s try x squared. Up top, remember that
we are plugging x plus h into the function first, so we have x plus h quantity squared,
minus the function itself, or x squared. Now we have to foil this first term on the top,
and we get x squared plus two xh plus h squared, minus the x squared, and that leaves us with
two xh plus h squared. Dividing both terms on top by the h on the bottom, we get two
x plus h, and when h goes to zero, that leaves us with two x. So the derivative of x squared
is two x. Let’s do one more, x cubed. Now we have x plus h quantity cubed minus x cubed.
We should know how to cube this binomial, by foiling two of them and multiplying the
result across the third binomial, but for the sake of time, let’s just show that this
will be x cubed plus (three x squared h) plus (three x h squared) plus h cubed. We subtract
x cubed, and we get this. Dividing through by the h on the bottom, we get (three x squared)
plus (three x h) plus h squared. As h approaches zero, we are left with three x squared, so
the derivative of x cubed is three x squared. By now we might be able to recognize a pattern.
It seems that every time we take a derivative, we are taking the exponent and bringing it
down in front to become a coefficient, and then reducing the exponent by one. This is
called the power rule, and it works for any function where the exponent, n, is a positive
integer. In this way, the derivative of x^n is equal to n times x^(n-1).
The power rule is something that we will be using day in and day out in order to take
the derivative of various functions, so the good news is that we don’t have to keep
using the other more tedious expression. But hopefully in deriving this new one, we have
developed an even better understanding of what it means to take the derivative of a
function. Just as before, it does mean to find the slope of the tangent line at a particular
point. But with this new power rule in our pockets, we can now safely learn how to find
derivatives without having to draw graphs or think about tangent lines at all. Instead,
we can simply apply the power rule, as long as we make sure to remember what’s really
going on under the hood of this fancy machinery. So let’s move forward and get some practice
with the power rule.