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

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

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

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

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

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The input and the output are exactly the 
same. 

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

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

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the derivative of the identity function 
is 1. 

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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pieces, these blue pieces and this red 
corner piece. 

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

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

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

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The change in the side length of cube is 
h. 

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

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

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

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

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

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

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

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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. 

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

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

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

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

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

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

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

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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 

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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definition of derivative and try to 
calculate this limit. 

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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. 

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

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

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

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

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

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

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

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

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

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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. 

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

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

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after I divide by h. 
And now, look. 

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

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

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

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

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

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

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[MUSIC]