[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]