[MUSIC]
Hi, welcome to the second part of our week.
In this part, we're going to speak more about linear approximations and tangent
lines, actually using our knowledge about the derivatives and what it stands for.
So, let us start with a remembering basic things,
which we've learned in the previous video in this week.
First of all, we've learned the definition of differentiable functions.
It's stands for functions which can be easily approximated and
easily fitted sufficiently with a linear functions and
the change of the functions are close to its differential.
So let us extrapolate and rewrite this statement further.
First of all, we need to rewrite by its definition,
functions change and differential so we get this nice trick.
And also let's move our -f(a) term to the right part and so
f(a) comes as part of the right part.
It's kind of nice.
So what do we have here, we have here in equation for
all our values our functions as arbitrary point are which is close to
our given point a, and it's kind of nice.
Let's take a look because that's our values, that's what we are aiming at.
And here, we have some constant thing,
then we have constant value multiplied linear function and
then we have some kind of arrow.
So the first part, first two jumps,
I was the right side of this equation is actually stands for linear function.
What linear function?
Well, you can easily guess, what linear function is
connected with derivative and differentiability?
Well, it's tangent line.
So, we now have the equation of a tangent line, which is kind of awesome here.
And well, so it's quite easy, but you need to
understand both basic rules to compute this line.
Since you need to understand the value of derivative and given point and
the value function itself.
And if you substitute x with our given point a,
You just go at the second term to disappear, it's equal to 0.
And you just left with the first term which is f(a) which stands for
our tangent line intersects our curve,
our graph of our function f at point a which is actually what we're aiming for.
Well, that's how I figure for this and
that's our green line, which we are talking about.
So, let us consider some example.
Well, you are going to go with our awesome example x powered x, and so
let's just find the derivative in the point 1 divided by exponent foundation, e.
So, what are we going to do here?
We are going to compute two things that were
need to write tangent line equation here.
First of all, we need to find a derivative in the given point.
So we need to find f, the derivative
of f at the point e powered minus 1.
So, let us remember that it stands for
x powered x multiplied by logarithm of x plus 1.
And I'm going to use the following notation,
you're unfamiliar with it and we are going to use it in all our lecture.
So, we are going to draw this trade to line and
write here that you need to
substitute x with 1 divided by e.
So, let's just write this thing down.
It's going to be,
e powered -1 and the power of e powered -1.
Well, we are going to simplify obviously, but
let's just proceed with.
We need to write a logarithm of e powered -1 plus 1.
And well, you can easily understand why I'm not going to simplify
the first drama of the project because the second will actually equals to
zero because natural logarithm of e powered -1 is actually -1 for definition.
So, the second term is 0 and all of it equals to 0.
So we've computed the derivative in our points and
we've just need to write runs here.
Let us write it in some as a corner.
So, y equals to actually 0
multiplied by x minus e powered -1,
plus the value of the function in
this point, which is z term.
We are going to multiply our powers here and write it like this.
So, another words y equals to e powered -1 divided by e,
which is kind of tricky, but it's actually means
that we have a horizontal tangent line at this point,
which kind of means something.
And we're going to learn about it more in the last week and
in the last part of this week.
[MUSIC]