[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]