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.