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[MUSIC] We're going to calculate the 
derivative of x squared with respect to 

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x. 
maybe a little bit more prosaically, I 

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want to know how wiggling x affect x 
squared? 

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There's a ton of different ways to 
approach this. 

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Let's start by looking at this 
numerically. 

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So let's start off by just noting that 2 
sqyared is 4, 

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and I'm going to wiggle the 2 and see how 
the 4 wiggles. 

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Instead of plugging in 2, let's plug in 
2.01. 

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2.01 squared is 4.0401. 
And let's just keep on going with some 

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more examples. 
2.02 squared is 4.0804. 

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2.003 squared, say, is 4.012009. 
Alright, so those are a few examples. 

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I've wiggled the inputs, and I've seen 
how the outputs are affected. 

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And of course the, all the outputs are 
close to 4, alright? But they're not 

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exactly 4. 
When I wiggled the input from 2 to 2.01, 

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the output changed by about .04, and a 
little bit more, but that'ts a lot 

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smaller. 
When I wiggled the input from 2 to 2.02, 

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the output changed by about .08, not 
exactly .08, but pretty close to .08. 

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And when I wiggled from 2 to 2.003, the 
output changed by about. 

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About .012 and a little bit more, but, 
you know, it's close. 

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Now look at the relationship between the 
input change and the output change. 

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The input change by .01, the output 
change by 4 times as much, about. 

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The input change by .02, the output 
change by 4 times as much. 

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The input change by .003, the output 
change by about 4 times as much. 

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I'm going to summarize that. 
The output change is the input change 

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magnified by 4 times. 
Right? The input change by some factor. 

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And the output change by about 4 times 
that amount. 

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Let's see this at a different input 
point. 

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Instead of plugging in 2, let's plug in 3 
and see what happens. 

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So 3^2 is 9, but what's, say 3.1^2? 
That's 9.61. 

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Or what's 3.01^2? Well, that's 9.0601. 
maybe wiggle down a little bit. 

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What's 2.99 squared? That's close to 3 
but wiggling down by .01. 

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That's 8.9401. 
Let's see how much roughly the output 

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changed by. 
When I went from 3 to 3.1 the output 

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changed by Out point 6. 
When I went from 3 to 3.01, the output 

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changed by about .06, and when I went 
from 3 down to 2.99, the output when down 

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by about .06 again. 
Little bit less. 

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Now what's the relationship between the 
input change and the output change? Well 

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here the input changed by .1, the output 
change by .6, about 6 times as much. 

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Again, the input change by .01, the 
output changed by About six times as 

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much. 
And when the input went down by .01 the 

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output went down by about six times as 
much. 

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So again, we're seeing some sort of 
magnification of the output change to the 

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input change, but now it's magnified not 
by four times But by six times. 

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So the important lesson here is that the 
extent to which wiggling the input 

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affects the output depends on where 
you're wiggling. 

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If you're wiggling around 2, the output 
is being changed by about four times as 

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much. 
If you're wiggling the input around 3, 

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the output is being change by about six 
times as much. 

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Instead of doing just a few numerical 
examples, let's generalize this by doing 

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some algebra. 
So, I'm starting with x^2 and I'm going 

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to wiggle x and see how x^2 is effected. 
So, instead of plugging in x, I'll plug 

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in x + something, let's call the change 
in x, h. 

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Now I want to know, how is this related 
to x ^ 2? Well I can expand out (x+h)^2, 

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that's x^2+2xh+h^2. 
So when I wiggle the input from x to x+h, 

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how is the output being affected? Well 
the output, is the old output value plus 

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this change in output value 2xh+h^2, h^2 
is pretty small. 

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When h is small, h^2 is really small so 
I'm going to throw that away for now. 

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And just summarize this by saying that 
the output change is 2xh and the input 

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change. 
Is h. 

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Now the derivative is supposed to measure 
the relationship between the output 

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change and the input change. 
So I'm going to take the ratio of the 

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output change to the input change, and 
2xh/h=2x, as long as h isn't 0. 

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This is the ratio of output change to 
input change and that makes sense, right? 

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Think back to what just happened here a 
minute ago, when we were plugging in some 

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nearby values and seeing how the outputs 
were affected. 

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When I was wiggling the input around 2, 
the output was changing by about twice 2. 

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When I was wiggling the input around 3, 
the output was changing by about twice 3, 

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alright? 2x is the ratio of output change 
to input change. 

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If the algebra's not really speaking to 
you, we can also do this geometrically, 

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like drawing a picture. 
Here's a square of side length x. 

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The area of this square is not, 
coincidentally, x^2. 

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Now I want to now the derivative of x^2 
with respect to x. 

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I want to know how changing x would 
affect the area of this square. 

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Now to see this here is another square. 
This is a slightly larger square of side 

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length x+h. 
h is a small but positive number. 

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So how does the area of this new square 
compare to the area of this old square? 

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Let me put the old square on top of the 
new square, and you can see that when I 

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change the input from x to x+h, I gained 
a bit of extra area. 

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The derivative is recording the ratio of 
output change to input change. 

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So, I want to know what's the ratio of 
this new area as compared to just the 

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change in the input H. 
So, let me pull off the extra area. 

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There is extra area, is this L shaped 
region. 

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How big is this L shaped region? Well, 
this short side here, has side length h. 

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This side length here, is also h. 
This is the extra length that I added 

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when I went from x to x+h. 
This inside has length x, and this inside 

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edge has length x. 
Now I want to know the area of this 

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region. 
To see that, I'm going to get out my 

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scissors and cut this region up into 3 
pieces. 

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Now here's one of those pieces. 
And, here's another one of those pieces. 

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And, here's the third piece. 
So these are the 2 long thin rectangles, 

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and they've both got height h, and length 
x. 

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I'm also left with this little tiny 
corner piece. 

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And that little tiny corner piece has 
side length h, and the other side is also 

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length h. 
It's a little tiny square. 

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Well the limit, this little tiny corner 
piece, is infinitesimal. 

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I'm going to throw this piece away and 
most of the area is left in these 2 long, 

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thin rectangles. 
If I rearrange these long, thin 

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rectangles a bit, can put them end to 
end. 

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They've both got height h. 
So I can put them next to each other like 

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this. 
And their base is both length x. 

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So how much area is in this long thin 
rectangle? Well, it's height h, it's 

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width is 2x. 
So the area is 2x * h. 

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Now this is the additional area, 
excepting for that little tiny square, 

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which we gained when I changed the size 
of the square from x to x + h. 

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So the change in output is about 2 * x * 
h. 

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The change in input was h, so the ratio 
of output change to input change Is 2 * 

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x. 
Maybe what we're doing here seems a 

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little bit wishy washy, not really 
precise enough. 

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But we can also calculate the derivative 
of x ^ 2 with respect to x, by just going 

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back to the definition of derivative in 
terms of limits. 

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Carefully, f of x is x^2. 
And the derivative of f is by definition 

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the limit as h approaches 0. 
F of x plus h minus f of x, the change in 

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output divided by h, the change in input. 
In this case f of x plus h is just x plus 

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h squared and f of x is just x squared. 
I'm dividing by h. 

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I can expand this out. 
This is the limit as h approaches 0 of 

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((x+h)^2-x^2)/h. 
Now I've got an x^2-x^2, so I can cancel 

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those, and I'm just left with the limit, 
as h approaches 0, of (2xh+h^2)/h. 

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More good news, in the limit I'm never 
going to be plugging h=0, so I can 

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replace this With an equivalent function 
that agrees with it when h is close to 

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but not equal to 0. 
In other words, maybe a little bit more 

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simply, I'm canceling an h from the 
numerator to the denominator. 

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So, 2xh over h is just 2x and h^x over h 
is just an h. 

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Now what's the limit of this sum? Well 
that's the sum of the limits. 

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It's the limit of 2x as h approaches 0 + 
the limit of h as h approaches 0. 

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Now as far as wiggling h is concerned, 2x 
is a constant, so the limit of 2x as h 

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approaches 0 is just 2x. 
And what's the limit of h as h approaches 

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0? Well, what's h getting close to when h 
is close to 0. 

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That's just 0. 
So, this limit is equal to 2x and that's 

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the derivative of x^2. 
What that limit is really calculating is 

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the slope of a tangent line at the point 
x and we can see that it's working. 

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This is the graph of y=x^2. 
At -4 the slope of the tangent line is -8 

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At 2, the slope of the tangent line is 4, 
and, at 6, the slope of the tangent line 

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is 12. 
There's a ton of different perspectives 

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here. 
We've been thinking about the derivative 

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of x ^ 2 with respect to x, numerically, 
algebriaically, geometrically, going back 

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to the definition of derivative in terms 
of limits, looking at it in terms of 

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slopes of tangent lines. 
What makes derivatives so much fun is 

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that there just so many different 
perspectives on this single topic, no 

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matter how you slice it. 
We've shown that the derivative of x 

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squared with respect to x is 2 times x. 
Maybe you like algebra, maybe you like 

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geometry, maybe you just like to play 
with numbers. 

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But now matter what your interests are, 
derivatives have something to offer you. 

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[MUSIC]