[MUSIC] Here is the so-called power rule for differentiating x^n.
Nevertheless, here we go. When n=1, the derivative of just x^1,
which is just x, is equal to 1. This should make sense because what's the
derivative measuring? The derivative is measuring output change compared to input
change. And, in this case, the function is just the function that sends x to x.
The input and the output are exactly the same.
So, the input and the output change is exactly the same,
their ratio is just 1. And consequently, the derivative of x,
the derivative of the identity function is 1.
For the time being, we're just going to think about this when n is a positive
whole number. But even there, it's pretty tricky.
Admittedly, when n=1, you're probably going to be pretty unimpressed.
The derivative of x^n is n*x^n-1. What's n? n can be any real number except
for zero. You should think about what you don't
want to plug in zero for n. When n=2, that means we're
differentiating x^2, which we studied a little bit ago.
Now, here is the power rule. If I plug in 2 for n, I've got the
derivative of x^2=2*x^2-1. Or a bit more nicely written, the
derivative of x^2=2x. I really remember, we really did study
this in quite some detail, you know, algebraically, numerically,
geometrically. When n=3, we can still study the
derivative of x^3 in a geometric way. So, here's the power rule.
You plug in n=3, and you get the derivative of x^3=3*x^3-1, 3*x^2.
We can see this geometrically. We start with a cube of side length x.
And we're going to glue on three green slabs of side length x, x, h.
Now, in order to actually thicken up the cube, we've got to glue on a few more
pieces, these blue pieces and this red corner piece.
But once we've done that, now we've built a cube of side length x+h.
How is the volume changed? Well, most of the change in volume happened in these
three green slabs, and those three green slabs have volume 3x^2h.
The change in the side length of cube is h.
Geometric argument is showing us that the derivative of x^3 is 3*x^2.
When n=4, we're trying to differentiate x^4.
But that would involve not a cube, but a hypercube.
[SOUND] It seems a bit ridiculous to try to gain intuition about the derivative of
x^3 by doing something as esoteric as studying 4-dimensional geometry.
So instead, let's differentiate x^3 directly by going back to the definition
of derivative. So, let's proceed directly. I want to
compute the limit as h approaches 0 of x+h^4-x^4/h.
What is this computing? This is the limit of the difference quotient.
This is the derivative of x^4 at the point x.
Now, to proceed, I'm going to make this a little bit smaller.
It's a bit too big to work with. This is the limit I'm trying to
calculate. The first step is to expand out x+h^4.
And if I expand x+h^4, this is what I get.
(h^4+4h^3x+6h^2x^2+4hx^3+x^4). And now, you'll notice something very
exciting. I've got an x^4-x^4 so I can cancel those two terms and I'll be left
with a limit of everything else. h^4+4h^3x+6h^2x^2+4hx^3/h.
But more good news, every single term up in the numerator
here, has an h in it. So, I can cancel those h's without
affecting the limit. And this limit is the same as the limit
of h^3+4h^2x+6hx^2+4x^3. Why? Well, look.
h^4/h gives me the h^3. 4h^3x/h gives me the 4h^x/h gives me the
4h^2x, and so forth. Now, we're practically there.
I want to evaluate this limit. Most of these terms here have got an h in
it, so when I take the limit, these terms are all 0.
The only term that survives is this one which as far as h is concerned is a
constant. It's the limit of 4x^3 as h approaches 0.
That's just 4x^3. And because this whole mess is
calculating the derivative of x^4, what I've really done here is shown, from the
definition of derivative, that the derivative of x^4 is 4x^3.
This limit calculation is perhaps complicated enough to give us a glimpse
into the whole story. What's the derivative of x^n? Trying to
show the derivative of x^n is nx^n-1. And to do that, we go back to the
definition of derivative and try to calculate this limit.
The limit is h goes to 0 of (x+h^n)-x^n/h.
Just like the case when n was 4, the first step is to expand this out.
But here, it's a bit trickier, right? To expand out x+h^n, I don't know exactly
what n is. n's just some positive whole number so I
can't write down exactly what it is. But I can write down enough of it to get
a sense of what's going on in the story. h^n+nh^n-1x+, and hidden in this dot,
dot, dot is all kinds of other terms that have h's in them,
plus nhx^n-1+x^n-x^n/h. Just like before, I've got an x^n and a
-x^n, so I can cancel those. And now, I'm left with just these terms,
still a bunch of terms with h's in them. And note that every single term in the
numerator here has an h, so I can then do the division just like before.
The h^n/h becomes h^n-1, and h^n-1x becomes nh^n-2x.
Everything in the dot, dot, dot here has at least an h^2 in it.
So, when I divide it by h, everything that's left over still has at least one h
in it. This last term nhx^n-1/h becomes nx^n-1
after I divide by h. And now, look.
This is a limit. As h approaches 0, this term dies, this
term dies, all of these terms with h's in them dies.
The only thing that's left is this term here, nx^n-1, and that means that this
entire limit is equal to nx^n-1. This limit is calculated in the
derivative of x^n. So, what we've really managed to do is
show that the derivative of x^n is nx^n-1.
[MUSIC]