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# Calculus: Derivatives 2

Video transcript

In the last presentation, I
hopefully gave you a little bit of an intuition of
what a derivative is. It's really just a way to
find the slope at a given point along the curve. Now we'll actually apply
it to some functions. So let's say I had
the function f of x. f of x is equal to x squared. And I want to know what is
the slope of this curve. What is the slope at x is equal
to-- let's say at x equals 3. What is the slope of x? Let's draw out what I'm asking. Coordinate axis. x-coordinate, that's
the y-coordinate. And then if I were to draw--
let me pick a different color. So we want to say what is the
slope when x is equal to 3. This is x equals 3. And of course when x equals
3, f of x is equal to 9. We know that, right? So what we do is we take a
point, maybe a little bit further along the curve. Let's say this point
right here is 3 plus h. And I keep it abstract as h
because as you know we're going to take the limit
as h approaches 0. And at this point
right here is what? It's 3 plus h squared, right? Because the function is f of
x is equal to x squared. So this point right here is
3 plus h, 3 plus h squared. Because we just take the 3 plus
h and put it into x squared and we get 3 plus h squared. And this point here
is of course 3, 9. What we want to do is we
want to find the slope between these two point. I really have to
find a better tool. This one keeps freezing, I
think it's too CPU intensive. But anyway. So we want to find the slope
between these two points. So what's the slope? so it's a
change in y, so it's 3 plus h squared minus this y minus
9 over the change in x. Well that's 3 plus h minus 3. So if we simplify this top
part or we multiply it out, what's 3 plus h squared? That's 9 plus 6h plus h
squared, and then get the minus 9, and all of that is over--
well this 3 and this minus 3 cancel out, so all
you're left is with h. And even if we simplify
this, this 9 minus 9, they cancel out. So let me go up here. We're left with-- this pen
keeps freezing-- it's 6h plus h squared over h. And now we would simplify this,
right, because we can divide the top and the bottom, that
numerator and the denominator by h. And you get 6 plus h squared. So that's the slope
between these two points. It's 6 plus h squared. So if we want to find the
instantaneous slope at the point x equals 3, f of x is
equal to 9, or the point 3,9, we just have to find the limit
as h approaches 0 here. So we'll just take the
limit as h approaches 0. Well this is an easy
limit problem, right? What's the limit of 6 plus h
squared as h approaches 0? Well it equals 6. So we now know that the
slope of this curve at the point x equals 3 is 6. So if you actually did a
traditional rise over run, the slope, this change in
y over change in x is 6. So we have the instantaneous
slope at exactly the point x is equal to 3. So that's useful. You know if this was a graph of
someone's position, we would then know kind of the
instantaneous velocity, which is-- well I won't go into that. I'll do a separate
module on physics. But this was useful, but
let's see if we can do more generalized version where we
don't have to know ahead of time what point we want
to find the slope at. If we can get a generalized
formula for the slope at any point along the graph f of
x is equal to x squared. So let me clear this. So we're going to stick with f
of x is equal to x squared. And we know that the slope at
any point of this is just going to be the limit as h approaches
0 of f of x plus h minus f of access. All of that over h. This part right here, this is
just the slope formula that you learned years ago. It's just change in
y over change in x. And all we're doing is we're
seeing what happens as the change in x gets smaller and
smaller and smaller as it actually approaches 0. And that's why we can get
the instantaneous change at that point in the curve. So let's apply this
definition of a derivative to this function. And actually if you want to
know the notation, I think this is the notation
Lagrange came up with. This is equal to f prime of x. Don't take my word
on it on Lagrange. You might want to look
it up on Wikipedia. But this [UNINTELLIGIBLE] derivative of f of
x is f prime of x. Let's apply it to x squared. So we're going to say the
limit as h approaches 0 of f of x plus h. Well, f of x plus h is
just-- this pen driving me crazy-- x plus h squared. I just took the x plus h
and put it into f of x. Minus f of x--well that's
just x squared-- over h. And this is equal to the
limit as h approaches 0. Just multiply this out of. x squared plus 2xh plus h
squared minus x squared-- running out of space--
all of that over h. Let's simplify this. This x squared cancels out
with this minus x squared. And then we can divide the
numerator and the denominator by h, and we're left with the
limit as h approaches 0. Numerator and denominator
by h of 2x plus h. Well this is easy. This goes to 0, this
is just equal to 2x. So there we have it. The limit as h approaches
0 is equal to 2x. And this is equal to f prime of
x, so the derivative of f of x, which is the denoted by f
prime of x is equal to 2x. Well what does it tell us? What have we done
for ourselves? Well now I can give you any
point along the curve. Let's say we want to know the
slope at the point 16, right. When at the point 16,256. That's a point along f
of x equals x squared. It's just 16 and
then 16 squared. What's the slope at that point? Well we now know the
slope is 2 times 16. So the slope is equal to 32. Whatever the x value is you
just put into this f prime of x function or the derivative
function, and you'll get the slope at that point. I think that's pretty neat and
I'll show you how in future presentations how we can
apply this to physics and optimization problems and a
whole other set of things. And I'm also going to show you
how to find the derivatives for a whole set of other functions. I'll see in the
next presentation.