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[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.

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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.

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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.

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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.

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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.

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