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[MUSIC] Here is the so-called power rule for differentiating x^n.

Â Nevertheless, here we go. When n=1, the derivative of just x^1,

Â which is just x, is equal to 1. This should make sense because what's the

Â derivative measuring? The derivative is measuring output change compared to input

Â change. And, in this case, the function is just the function that sends x to x.

Â The input and the output are exactly the same.

Â So, the input and the output change is exactly the same,

Â their ratio is just 1. And consequently, the derivative of x,

Â the derivative of the identity function is 1.

Â For the time being, we're just going to think about this when n is a positive

Â whole number. But even there, it's pretty tricky.

Â Admittedly, when n=1, you're probably going to be pretty unimpressed.

Â The derivative of x^n is n*x^n-1. What's n? n can be any real number except

Â for zero. You should think about what you don't

Â want to plug in zero for n. When n=2, that means we're

Â differentiating x^2, which we studied a little bit ago.

Â Now, here is the power rule. If I plug in 2 for n, I've got the

Â derivative of x^2=2*x^2-1. Or a bit more nicely written, the

Â derivative of x^2=2x. I really remember, we really did study

Â this in quite some detail, you know, algebraically, numerically,

Â geometrically. When n=3, we can still study the

Â derivative of x^3 in a geometric way. So, here's the power rule.

Â You plug in n=3, and you get the derivative of x^3=3*x^3-1, 3*x^2.

Â We can see this geometrically. We start with a cube of side length x.

Â And we're going to glue on three green slabs of side length x, x, h.

Â Now, in order to actually thicken up the cube, we've got to glue on a few more

Â pieces, these blue pieces and this red corner piece.

Â But once we've done that, now we've built a cube of side length x+h.

Â How is the volume changed? Well, most of the change in volume happened in these

Â three green slabs, and those three green slabs have volume 3x^2h.

Â The change in the side length of cube is h.

Â Geometric argument is showing us that the derivative of x^3 is 3*x^2.

Â When n=4, we're trying to differentiate x^4.

Â But that would involve not a cube, but a hypercube.

Â [SOUND] It seems a bit ridiculous to try to gain intuition about the derivative of

Â x^3 by doing something as esoteric as studying 4-dimensional geometry.

Â So instead, let's differentiate x^3 directly by going back to the definition

Â of derivative. So, let's proceed directly. I want to

Â compute the limit as h approaches 0 of x+h^4-x^4/h.

Â 3:18

What is this computing? This is the limit of the difference quotient.

Â This is the derivative of x^4 at the point x.

Â Now, to proceed, I'm going to make this a little bit smaller.

Â It's a bit too big to work with. This is the limit I'm trying to

Â calculate. The first step is to expand out x+h^4.

Â And if I expand x+h^4, this is what I get.

Â (h^4+4h^3x+6h^2x^2+4hx^3+x^4). And now, you'll notice something very

Â exciting. I've got an x^4-x^4 so I can cancel those two terms and I'll be left

Â with a limit of everything else. h^4+4h^3x+6h^2x^2+4hx^3/h.

Â 4:07

But more good news, every single term up in the numerator

Â here, has an h in it. So, I can cancel those h's without

Â affecting the limit. And this limit is the same as the limit

Â of h^3+4h^2x+6hx^2+4x^3. Why? Well, look.

Â h^4/h gives me the h^3. 4h^3x/h gives me the 4h^x/h gives me the

Â 4h^2x, and so forth. Now, we're practically there.

Â I want to evaluate this limit. Most of these terms here have got an h in

Â it, so when I take the limit, these terms are all 0.

Â The only term that survives is this one which as far as h is concerned is a

Â constant. It's the limit of 4x^3 as h approaches 0.

Â That's just 4x^3. And because this whole mess is

Â calculating the derivative of x^4, what I've really done here is shown, from the

Â definition of derivative, that the derivative of x^4 is 4x^3.

Â This limit calculation is perhaps complicated enough to give us a glimpse

Â into the whole story. What's the derivative of x^n? Trying to

Â show the derivative of x^n is nx^n-1. And to do that, we go back to the

Â definition of derivative and try to calculate this limit.

Â The limit is h goes to 0 of (x+h^n)-x^n/h.

Â 5:30

Just like the case when n was 4, the first step is to expand this out.

Â But here, it's a bit trickier, right? To expand out x+h^n, I don't know exactly

Â what n is. n's just some positive whole number so I

Â can't write down exactly what it is. But I can write down enough of it to get

Â a sense of what's going on in the story. h^n+nh^n-1x+, and hidden in this dot,

Â dot, dot is all kinds of other terms that have h's in them,

Â plus nhx^n-1+x^n-x^n/h. Just like before, I've got an x^n and a

Â -x^n, so I can cancel those. And now, I'm left with just these terms,

Â still a bunch of terms with h's in them. And note that every single term in the

Â numerator here has an h, so I can then do the division just like before.

Â The h^n/h becomes h^n-1, and h^n-1x becomes nh^n-2x.

Â 6:31

Everything in the dot, dot, dot here has at least an h^2 in it.

Â So, when I divide it by h, everything that's left over still has at least one h

Â in it. This last term nhx^n-1/h becomes nx^n-1

Â after I divide by h. And now, look.

Â This is a limit. As h approaches 0, this term dies, this

Â term dies, all of these terms with h's in them dies.

Â The only thing that's left is this term here, nx^n-1, and that means that this

Â entire limit is equal to nx^n-1. This limit is calculated in the

Â derivative of x^n. So, what we've really managed to do is

Â show that the derivative of x^n is nx^n-1.

Â [MUSIC]

Â