So now we know how to find derivatives by definition.
Well it's actually not the case,
nobody finds derivatives only by definition.
There is a set of arithmetic rules here and you're not
destined to do this limit calculations
for all alternative for the rest of your life.
So let us proceed with some basic,
the most obvious rules here which is
inherited from the properties of limits.
Let us start with the most basic ones,
it's the derivative of sum and derivative of difference,
it's actually the simplest one
and the most intuitive one.
The derivative of sum is the sum of derivatives,
same applies for the difference case.
Let us just assume some basic example here.
For example, let us take
the function x squared plus cosine x,
and assume we're trying to find the derivative here.
So by our rule,
we need to write the sum
of two derivatives of x squared and cosine function,
and then it is one we've actually seen in
our magic both radar,
it's 2x plus minus sine of x,
and that's actually [inaudible] and well congrats,
you're now have
a very decent tool to calculate derivatives.
Let's move forward, and further we are trying
to think about some [inaudible] pretty obvious stuff,
it's multiplication by number.
Multiplication by number it's
a general case for derivative of sum.
Let's just assume the following,
imagine you have sum function f and you
need to find the derivative of 2f,
one can think of 2f as
derivative of a sum f plus f which is an easy transition.
By our summation rule it
is derivative of f plus derivative of f
which is easily by arithmetic from
kindergarten to derivatives of f, that's it.
More general case assume that we have some thing
like three Pi x power three,
and we need to find derivative of this.
Well, you don't need to be frightened by three Pi,
three Pi is just a number or
whatever it means, it's just a number.
So we need to move it from
the derivative and multiply it by the derivative
of x power three which is
actually results in nine Pi x squared,
that's it, doing great.
We now know, that's the next rule.
So let's move further and
let's consider honest multiplication,
multiplication function by function.
It's a bit complex at the very moment,
but let's try it on some example.
For example, let us consider function
x multiply it by natural logarithm of x,
so it's a product of two functions x and algorithm x,
so we need to use our final,
final here implies that we are
getting a sum of two products here.
Firstly, we need to multiply derivative of the first one,
multiply it by the second function.
The second term is derivative of the second one
multiplied by the first one.
So all derivatives here we've actually
seen in the previous videos so we just write them down,
derivative of x equals 1 and derivative
of natural logarithm is 1 divided by x,
so we get in the result natural logarithm of x plus 1.
It's more tricky but it's still rather easier.
So let's move to the last arithmetical rule here,
and it's the rule of division.
So it's far more complex than others,
but we are going to live with it quite happily.
In order to understand it,
we are going with some most basic example of all.
You probably remember that in the table of derivatives,
we have function such as 1 divided by x,
or in other words it's x powered minus 1.
Since it's a power function,
there is a common rule for it and
yet you need to multiply
the power by x powered with the same power minus 1,
simply writing it is minus 1 divided by x squared.
That's how it's done by our table,
but how it's done by
our division rule it's
actually division of two functions,
one and x, so let us consider this case also,
of course results are expected to coincide,
but who knows maybe I'm lying.
Well let's start actually with denominator,
it's quite easy you need to
square the denominator of the original fraction,
and for the nominator,
we are going to go with this rather complex rule.
First of all, we need to write
the derivative of the nominator multiplied by
the denominator minus the derivative
of the denominator multiplied by nominator,
and derivative of one is actually zero,
so the first term is
absent and the derivative of the x equals 1,
so we are getting
our familiar minus 1 divided by x squared.
Well, they coincided so
no problem here and our rule is actually working.
Now we're actually in capacity of quite a decent amount
of basic rules to go with derivative calculations,
but in order to do any derivative,
we need to study a bit more which
is our next [inaudible].