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].