So as we move further, we need to consider
all mighty chain rule, in order to calculate
derivatives here. Firstly, we need to introduce
ourselves to the concept of composite functions or
composition of two functions. Well, assumes that,
as you do remember, functions are always considered as a relation between two sets:
arguments and [inaudible]. So imagine that we
have three sets now x,y and z and our function. Let's start with, surprise, the function g
actually maps x to y, and our function f marks y to z. So as a result of which, if we take some argument
from the function x, we're having a conversation
here, conveyor belt. Firstly, we have moved
from x to y by function g, and then the result of function g is moved from
y to z by function f, and that's what's the
composition of two functions is its consequent application of two functions 2, 1
initial argument. The idea basically is
that we need to assess how much the value
intercepts of that changes if we slightly change the value of the argument in the set of x. The idea here is that we do
not know the structure of the function from x to z, all over, which is
our composition here. We do know how it changes between our blue arrows and
not our red one. So we need to rewrite our derivatives in
terms of derivative of function f and dutiful function g. So let's
proceed with it. Firstly, here is the rule. The rule basically states, it's quite something. Let's see. Assume that we have
for example two pipes, two pipes with filters. I'll try to draw
some basic thing. It's one thing and that
the other thing is called g and seconds called
f. Inflow goes to g, and it goes out from
f. So how g works? g works as a filter. So assume that, for example, if the water goes to the
g pipe, that will take, which is somewhere near here, is actually multiplied by 0.9, 10 percent of all inflow
are lost, is filtered out, and the outtake from
the f function is, for example, 0.8 of the
intake of the f function. So we are losing
20 percent of what actually count towards the
beginning of the f pipe. So as a result, what we are going to lose? Well, it's quite a progress. We need to multiply our outtake, our
filtration coefficient, if you may say, pipe g ends of the pipe f, which is the same
thing is actually displayed by our rule of
the duty of composition. In order to understand
how much function changes the composite function changes by the change of its
initial arguments. We need to understand how
much changes inside function, and then how much changes our outer function to avoid the change of
our inside function. Basically, the idea here
is that you can't put derivative inside your brakes. The only thing that you
can do is understand how one function changes all over the argument in each backends. For function F, the
argument here is called function g. So let us come to the
some basic example, and [inaudible] will be much
more simple and decent. So first of all, let's start with painfully famous and
painfully important function, exponent of minus x squared. You will use it for
all of your life as a Gaussian distribution and it's must f for every
decent human being. So it's basically
a composition of two functions: exponent
of x and minus x squared. So let us consider our rule. Our rule tells us the following. First of all, we need to
learn how to draw here. Second of all, we need to decide what is mean function and
what is outer function here. The rule, the intuitive
rule here is simple one. The outer function is the one
which you calculate last. So assume that they are trying to find the value of this
function at some point, for example, one, one
is the closing point. So in order to do it, you first start with x squared. You are considering x
squared, that's 0.1. If that's your forced function, then you are moving forward
to minus x squared, and then you will substitutes your resulting value and
put it inside the exponent. So the last function that we've
talked about is exponent, so it's outer function, and the inner function
is minus x squared. So let us write it. Let us start with outer function. The outer function, as
we said, is exponent, a derivative of
exponent as we know it's the same function. So it's exponent of
minus x squared. Remember, we're
finding a derivative, two of which are inner functions. Our argument is called inner
function minus x squared. It's our artificial x. x, let's write it like this. Then we proceed with the
derivative of minus x squared, which is formula is minus two x. So the answer here is minus two exponent of minus
x squared multiplied by x. That's tricky. You need to get in touch with it. You need practice, and actually, all the examples and tests that we are going to give you in this week is the thing
which is going to help you. So as a last point of our search for
calculating derivatives, we're going to start with some basic trick which is
logarithmic derivative, which is the following we draw. It's not necessary,
but it's nice.