Welcome back.
In the last presentation I showed you that if I had the
function f of x is equal to x squared, that the derivative of
this function, which is denoted by f-- look at that, my pen
is already malfunctioning.
The derivative of that function, f prime of
x, is equal to 2x.
And I used the limit definition of a derivative.
I used, let me write it down here.
This pen is horrible.
I need to really figure out some other tool to use.
The limit as h approaches 0 -- sometimes you'll see delta x
instead of h, but it's the same thing-- of f of x plus
h minus f of x over h.
And I used this definition of a derivative, which is really
just the slope at any given point along the curve,
to figure this out.
That if f of x is equal to x squared, that
the derivative is 2x.
And you could actually use this to do others.
And I won't do it now, maybe I'll do it in a
future presentation.
But it turns out that if you have f of x is equal to x to
the third, that the derivative is f prime of x is
equal to 3x squared.
If f of x is equal to x to the fourth, well then the
derivative is equal to 4x to the third.
I think you're starting to see a pattern here.
If I actually wrote up here that if f of x -- let me see
if I have space to write it neatly.
If I wrote f of x -- I hope you can see this -- f
of x is equal to x.
Well you know this.
I mean, y equals x, what's the slope of y equals x?
That's just 1, right?
y equals x, that's a slope of 1.
You didn't need to know calculus to know that.
f prime of x is just equal to 1.
And then you can probably guess what the next one is.
If f of x is equal to x to the fifth, then the derivative is--
I think you could guess-- 5 x to the fourth.
So in general, for any expression within a polynomial,
or any degree x to whatever power-- let's say f of x is
equal to-- this pen drives me nuts.
f of x is equal to x to the n, right?
Where n could be any exponent.
Then f prime of x is equal to nx to the n minus 1.
And you see this is what the case was in all
these situations.
That 1 didn't show up.
n minus 1.
So if n was 25, x to the 25th power, the derivative
would be 25 x to the 24th.
So I'm going to use this rule and then I'm going to show
you a couple of other ones.
And then now we can figure out the derivative of pretty much
any polynomial function.
So just another couple of rules.
This might be a little intuitive for you, and if you
use that limit definition of a derivative, you could
actually prove it.
But if I want to figure out the derivative of, let's say, the
derivative of-- So another way of-- this is kind of, what is
the change with respect to x?
This is another notation.
I think this is what Leibniz uses to figure out the
derivative operator.
So if I wanted to find the derivative of A f of x, where A
is just some constant number.
It could be 5 times f of x.
This is the same thing as saying A times the
derivative of f of x.
And what does that tell us?
Well, this tells us that, let's say I had f of x.
f of x is equal to-- and this only works with the constants--
f of x is equal to 5x squared.
Right?
Well this is the same thing as 5 times x squared.
I know I'm stating the obvious.
So we can just say that the derivative of this is just 5
times the derivative of x squared.
So f prime of x is equal to 5 times, and what's the
derivative of x squared?
Right, it's 2x.
So it equals 10x.
Right?
Similarly, let's say I had g of x, just using
a different letter.
g of x is equal to-- and my pen keeps malfunctioning.
g of x is equal to, let's say, 3x to the 12th.
Then g prime of x, or the derivative of g, is equal
to 3 times the derivative of x to the 12th.
Well we know what that is.
It's 12 x to the 11th.
Which you would have seen.
12x to the 11th.
This equals 36x to the 11th.
Pretty straightforward, right?
You just multiply the constant times whatever the
derivative would have been.
I think you get that.
Now one other thing.
If I wanted to apply the derivative operator-- let me
change colors just to mix things up a little bit.
Let's say if I wanted to apply the derivative of operator-- I
think this is called the addition rule.
It might be a little bit obvious.
f of x plus g of x.
This is the same thing as the derivative of f of x plus
the derivative3 of g of x.
That might seem a little complicated to you, but all
it's saying is that you can find the derivative of each of
the parts when you're adding up, and then that's the
derivative of the whole thing.
I'll do a couple of examples.
So what does this tell us?
This is also the same thing, of course.
This is, I believe, Leibniz's notation.
And then Lagrange's notation is-- of course these were the
founding fathers of calculus.
That's the same thing as f prime of x plus g prime of x.
And let me apply this, because whenever you apply it, I think
it starts to seem a lot more obvious.
So let's say f of x is equal to 3x squared plus 5x plus 3.
Well, if we just want to figure out the derivative, we say f
prime of x, we just find the derivative of each
of these terms.
Well, this is 3 times the derivative of x squared.
The derivative of x squared, we already figured
out, is 2x, right?
So this becomes 6x.
Really you just take the 2, multiply it by the 3, and
then decrement the 2 by 1.
So it's really 6x to the first, which is the same thing as 6x.
Plus the derivative of 5x is 5.
And you know that because if I just had a line that's y equals
5x, the slope is 5, right?
Plus, what's the derivative of a constant function?
What's the derivative of 3?
Well, I'll give you a hint.
Graph y equals 3 and tell me what the slope is.
Right, the derivative of a constant is 0.
I'll show other times why that might be more intuitive.
Plus 0.
You can just ignore that.
f prime of x is equal to 6x plus 5.
Let's do some more.
I think the more examples we do, the better.
And I want to keep switching notations, so you don't get
daunted whenever you see it in a different way.
Let's say y equals 10x to the fifth minus 7x to the
third plus 4x plus 1.
So here we're going to apply the derivative operator.
So we say dy-- this is I think Leibniz's
notation-- dy over dx.
And that's just the change in y over the change in x,
over very small changes.
That's kind of how I view this d, like a very small delta.
Is equal to 5 times 10 is 50 x to the fourth minus 21 --
right, 3 times 7-- x squared plus 4.
And then the 1, the derivative of 1 is just 0.
So there it is.
We figured out the derivative of this very
complicated function.
And it was pretty straightforward.
I think you'll find that derivatives of polynomials are
actually more straightforward than a lot of concepts that you
learned a lot earlier in mathematics.
That's all the time I have now for this presentation.
In the next couple I'll just do a bunch of more examples, and
I'll show you some more rules for solving even more
complicated derivatives.
See you in the next presentation.