[MUSIC] We're going to calculate the 
derivative of x squared with respect to
x. 
maybe a little bit more prosaically, I
want to know how wiggling x affect x 
squared?
There's a ton of different ways to 
approach this.
Let's start by looking at this 
numerically.
So let's start off by just noting that 2 
sqyared is 4,
and I'm going to wiggle the 2 and see how 
the 4 wiggles.
Instead of plugging in 2, let's plug in 
2.01.
2.01 squared is 4.0401. 
And let's just keep on going with some
more examples. 
2.02 squared is 4.0804.
2.003 squared, say, is 4.012009. 
Alright, so those are a few examples.
I've wiggled the inputs, and I've seen 
how the outputs are affected.
And of course the, all the outputs are 
close to 4, alright? But they're not
exactly 4. 
When I wiggled the input from 2 to 2.01,
the output changed by about .04, and a 
little bit more, but that'ts a lot
smaller. 
When I wiggled the input from 2 to 2.02,
the output changed by about .08, not 
exactly .08, but pretty close to .08.
And when I wiggled from 2 to 2.003, the 
output changed by about.
About .012 and a little bit more, but, 
you know, it's close.
Now look at the relationship between the 
input change and the output change.
The input change by .01, the output 
change by 4 times as much, about.
The input change by .02, the output 
change by 4 times as much.
The input change by .003, the output 
change by about 4 times as much.
I'm going to summarize that. 
The output change is the input change
magnified by 4 times. 
Right? The input change by some factor.
And the output change by about 4 times 
that amount.
Let's see this at a different input 
point.
Instead of plugging in 2, let's plug in 3 
and see what happens.
So 3^2 is 9, but what's, say 3.1^2? 
That's 9.61.
Or what's 3.01^2? Well, that's 9.0601. 
maybe wiggle down a little bit.
What's 2.99 squared? That's close to 3 
but wiggling down by .01.
That's 8.9401. 
Let's see how much roughly the output
changed by. 
When I went from 3 to 3.1 the output
changed by Out point 6. 
When I went from 3 to 3.01, the output
changed by about .06, and when I went 
from 3 down to 2.99, the output when down
by about .06 again. 
Little bit less.
Now what's the relationship between the 
input change and the output change? Well
here the input changed by .1, the output 
change by .6, about 6 times as much.
Again, the input change by .01, the 
output changed by About six times as
much. 
And when the input went down by .01 the
output went down by about six times as 
much.
So again, we're seeing some sort of 
magnification of the output change to the
input change, but now it's magnified not 
by four times But by six times.
So the important lesson here is that the 
extent to which wiggling the input
affects the output depends on where 
you're wiggling.
If you're wiggling around 2, the output 
is being changed by about four times as
much. 
If you're wiggling the input around 3,
the output is being change by about six 
times as much.
Instead of doing just a few numerical 
examples, let's generalize this by doing
some algebra. 
So, I'm starting with x^2 and I'm going
to wiggle x and see how x^2 is effected. 
So, instead of plugging in x, I'll plug
in x + something, let's call the change 
in x, h.
Now I want to know, how is this related 
to x ^ 2? Well I can expand out (x+h)^2,
that's x^2+2xh+h^2. 
So when I wiggle the input from x to x+h,
how is the output being affected? Well 
the output, is the old output value plus
this change in output value 2xh+h^2, h^2 
is pretty small.
When h is small, h^2 is really small so 
I'm going to throw that away for now.
And just summarize this by saying that 
the output change is 2xh and the input
change. 
Is h.
Now the derivative is supposed to measure 
the relationship between the output
change and the input change. 
So I'm going to take the ratio of the
output change to the input change, and 
2xh/h=2x, as long as h isn't 0.
This is the ratio of output change to 
input change and that makes sense, right?
Think back to what just happened here a 
minute ago, when we were plugging in some
nearby values and seeing how the outputs 
were affected.
When I was wiggling the input around 2, 
the output was changing by about twice 2.
When I was wiggling the input around 3, 
the output was changing by about twice 3,
alright? 2x is the ratio of output change 
to input change.
If the algebra's not really speaking to 
you, we can also do this geometrically,
like drawing a picture. 
Here's a square of side length x.
The area of this square is not, 
coincidentally, x^2.
Now I want to now the derivative of x^2 
with respect to x.
I want to know how changing x would 
affect the area of this square.
Now to see this here is another square. 
This is a slightly larger square of side
length x+h. 
h is a small but positive number.
So how does the area of this new square 
compare to the area of this old square?
Let me put the old square on top of the 
new square, and you can see that when I
change the input from x to x+h, I gained 
a bit of extra area.
The derivative is recording the ratio of 
output change to input change.
So, I want to know what's the ratio of 
this new area as compared to just the
change in the input H. 
So, let me pull off the extra area.
There is extra area, is this L shaped 
region.
How big is this L shaped region? Well, 
this short side here, has side length h.
This side length here, is also h. 
This is the extra length that I added
when I went from x to x+h. 
This inside has length x, and this inside
edge has length x. 
Now I want to know the area of this
region. 
To see that, I'm going to get out my
scissors and cut this region up into 3 
pieces.
Now here's one of those pieces. 
And, here's another one of those pieces.
And, here's the third piece. 
So these are the 2 long thin rectangles,
and they've both got height h, and length 
x.
I'm also left with this little tiny 
corner piece.
And that little tiny corner piece has 
side length h, and the other side is also
length h. 
It's a little tiny square.
Well the limit, this little tiny corner 
piece, is infinitesimal.
I'm going to throw this piece away and 
most of the area is left in these 2 long,
thin rectangles. 
If I rearrange these long, thin
rectangles a bit, can put them end to 
end.
They've both got height h. 
So I can put them next to each other like
this. 
And their base is both length x.
So how much area is in this long thin 
rectangle? Well, it's height h, it's
width is 2x. 
So the area is 2x * h.
Now this is the additional area, 
excepting for that little tiny square,
which we gained when I changed the size 
of the square from x to x + h.
So the change in output is about 2 * x * 
h.
The change in input was h, so the ratio 
of output change to input change Is 2 *
x. 
Maybe what we're doing here seems a
little bit wishy washy, not really 
precise enough.
But we can also calculate the derivative 
of x ^ 2 with respect to x, by just going
back to the definition of derivative in 
terms of limits.
Carefully, f of x is x^2. 
And the derivative of f is by definition
the limit as h approaches 0. 
F of x plus h minus f of x, the change in
output divided by h, the change in input. 
In this case f of x plus h is just x plus
h squared and f of x is just x squared. 
I'm dividing by h.
I can expand this out. 
This is the limit as h approaches 0 of
((x+h)^2-x^2)/h. 
Now I've got an x^2-x^2, so I can cancel
those, and I'm just left with the limit, 
as h approaches 0, of (2xh+h^2)/h.
More good news, in the limit I'm never 
going to be plugging h=0, so I can
replace this With an equivalent function 
that agrees with it when h is close to
but not equal to 0. 
In other words, maybe a little bit more
simply, I'm canceling an h from the 
numerator to the denominator.
So, 2xh over h is just 2x and h^x over h 
is just an h.
Now what's the limit of this sum? Well 
that's the sum of the limits.
It's the limit of 2x as h approaches 0 + 
the limit of h as h approaches 0.
Now as far as wiggling h is concerned, 2x 
is a constant, so the limit of 2x as h
approaches 0 is just 2x. 
And what's the limit of h as h approaches
0? Well, what's h getting close to when h 
is close to 0.
That's just 0. 
So, this limit is equal to 2x and that's
the derivative of x^2. 
What that limit is really calculating is
the slope of a tangent line at the point 
x and we can see that it's working.
This is the graph of y=x^2. 
At -4 the slope of the tangent line is -8
At 2, the slope of the tangent line is 4, 
and, at 6, the slope of the tangent line
is 12. 
There's a ton of different perspectives
here. 
We've been thinking about the derivative
of x ^ 2 with respect to x, numerically, 
algebriaically, geometrically, going back
to the definition of derivative in terms 
of limits, looking at it in terms of
slopes of tangent lines. 
What makes derivatives so much fun is
that there just so many different 
perspectives on this single topic, no
matter how you slice it. 
We've shown that the derivative of x
squared with respect to x is 2 times x. 
Maybe you like algebra, maybe you like
geometry, maybe you just like to play 
with numbers.
But now matter what your interests are, 
derivatives have something to offer you.
[MUSIC]
