hi guys Levi here from
BRIGHT YOUNG BRAINS
so this is the second portion right of
us going through grade 12
differentiation
um so in the previous tutorial right on
differentiation
we were using y and y prime
so we are going to start ...
... making use of
f(x) and f prime right because in
grade12 of those are the symbols you're
going to use
more often so we already said that this
is the main
way to differentiate what we need to
keep in mind is that
you will not always have a single term
that you need to differentiate
there are cases where you have multiple
terms in an expression
and differentiating that expression it's
a matter of just differentiating each
term individually
an example of that
is if we had g(x) and they want us to
differentiate this expression
we are just going to end up with 15x
squared
minus 4x and all we did was 3
multiplied by 5 gave us 15 and then 3
minus 1
gave us a 2. and then here we just did
the exact same thing again we just used
to this law
2 multiplied by negative 2 this gives us
a negative 4
and then 2 minus 1 this gives us a 1.
that's why we didn't write anything
there
because the value there is just a one
so what we're going to do is we're going
to work on how
to differentiate expressions
where x has an exponent of one
and where x has an exponent of zero so
i'm
deliberately saying where x has an
exponent of zero
because you are probably going to be
asking what?
was that uh and all we're trying to
refer to is a constant
right a constant is literally just a
term where
x has an exponent of zero so you want to
just
keep in mind stuff like that so to give
meaning to what we've just mentioned now
if g of x was a bit more chunky right
let's say it had
more terms than just two let's say it
had
also negative three x and then a minus
one at the end let's say it was like a
full-blown
cubic function if you had to
differentiate this expression
nothing changes really you just still
differentiate each term individually
so the first two
we already know that that's just going
to be 15x squared
minus 4x now when you differentiate
these last two first key thing you want
to remember is
anytime you differentiate anything with
one
as an exponent for example what we have
here right
if you're going to differentiate
negative 3x...
negative 3x differentiated all that
you're going to say is
negative 1 or rather sorry 1 multiplied
by negative 3
that's going to give you a negative 3
right and then here you're going to say
where one is you're going to obviously
just finish
the process right one minus one where
one is and this will leave you with
x to the power of zero and we know that
anything where
it's a variable to the power of zero
this
is just a one so here we will just have
negative three on its own
so this makes it look like uh it makes
it look like a complicated process but
it's actually quite simple
this is all a way of us trying to say
that
if you ever have to differentiate
anything where the exponent is a one
this expression differentiated will
always just be the coefficient on its
own
right because of our differentiation law
our main one in group 12. we know that
nothing funny happened we still followed
the same steps right
we know that 1 multiplied by a negative
3
gave us a negative 3 and then where one
is we know we have to still
minus a one there so this ended up being
one minus one which is just a zero
and we know that anything to the power
of zero is just a one so we know that
this is basically just saying negative
three times one
which is just a negative 3 there that's
why
we just have a negative 3 for this term
there now
our other law or not a law per say
now our other method of differentiating
a constant
here we have a constant and we said uh
firstly this is an entire cubic function
and they wanted to differentiate that
cubic function
and we said all you do is differentiate
each term individually one by one
this we differentiated this we
differentiated
that we differentiated now this negative
one we also need to
differentiate it right so
law is or the law whenever you need to
differentiate a constant the law is that
the constant just falls away
literally we have a constant if you're
going to differentiate this constant
it just falls away i'm almost tempted to
put a plus
zero there right however i'll just leave
it open
right what i will put down here is
the constant disappears and i hope our
spell disappears correctly there
i'm not very good at spelling but am
good at math though...
anyways so yeah the negative one just
disappears there
right and that's it
this function differentiated will just
end up as
15x squared minus
4x minus three and then we said that
a constant differentiated it just
disappears
if i have edited that out it just means
i spelled it wrong
if i left it then i spelt it right so
let's just see what happens when i edit
this video
hi guys Levi here from
BRIGHT YOUNG BRAINS
so we're working on differentiation
right specifically for grade
12s and today we're mainly going to
showcase three main scenarios
that we want to keep in mind whenever we
differentiate
so there are cases where
you will not be bluntly given
expressions
nicely packed like this so what we mean
is there are cases where you will be
given expressions
yes asking you to differentiate however
you will have negative exponents
there and cases where you have surds
there
and cases where your expressions are in
fraction format
so anytime you have to differentiate
and you are not given something ...
something that's
this blunt you will always need to
firstly
put it into a scenario or put it in a
format that is
this blunt before
you can execute this step right so from
the top when you're given
an expression that you need to
differentiate and it's in one of these
three formats
you cannot differentiate it in grade 12
from the get go you will have to
simplify it first
and then differentiate
so that's what we're going to show
so what we have here right is an
expression
and with an expression like this
if they ask us to differentiate we won't
be able to differentiate immediately
main reason is this is something that is
not
in our normal differentiable
format right this expression here
is not in this format we cannot clearly
see a coefficient
and a value for a and an exponent
that's why we are not able to
differentiate this immediately
especially in grade 12
and grade 12 if you're going to
differentiate expressions like this
you will need to put them in this format
first so what we're going to work on is
how you go from this format
to a format that is differentiable
if i can put it that way so if we're given
f(x) in this format
you will first need to simplify f(x)
and simplifying f(x)
this is what it's going to look like now
all that has happened is anytime you
have
a fraction with
your variable to the power a value
under or where the denomination where
the
denominator position is anytime you have
anything in this format, it's a fraction
and our variable is on the denominator
to the power of an exponent
and we need to differentiate the whole
thing if you're going to need to
differentiate
we need to keep in mind that all that
happens is you take
your value for x with its exponent up
you take it to the numerator position
and then once you take it to the
numerator position
you need to make sure that you change
your
sign for your exponent into
a NEGATIVE/POSITIVE value or rather you just
need to INTERCHANGE your sign so meaning
if you were given a negative
value down here just rewrite it with a
positive value up there
if you're given a positive value down
here just rewrite it with a
negative value at the top so once you
have this
in this format we can now differentiate
this
because all these terms right this term
and this term
are in this format that we can
differentiate so anything in this format
we already mentioned that if we have to
differentiate
it's all a matter of multiplying the
coefficient in front
and then subtracting one from the
exponent
so that's what we're going to do here
for these two
there we go now what you want to take
note of
is how here when we were differentiating
the negative x to the power of negative
three
keep in mind your signs as always here
we had a negative three multiplied by a
negative one
this is going to give us a positive
three and then another key thing is you
see where the exponent
is this is going to read negative 3
minus 1 because of our law
so negative 3 minus 1 that's going to
give you a negative 4.
so this here you mainly
usually don't want to just leave it like
that the reason is anytime you're
working with differentiation right
you are normally asked to leave your
answers
with positive exponents what that means
is
you are given a question like this, yes
you have differentiated
however if you have a negative exponent
there you will usually...
you will usually lose at least about a
mark or so if you don't
put it in a format where the exponents
are all positives
so what we mean is if you have something
like this with a negative exponent
all you want to do is you've already
differentiated right
all you'll need to do is whatever
variable has
that negative exponent you just take it
under the
coefficient right just like what we've
done here
the x to the power of negative four the
whole thing went
under the three and then it gets a
positive value/exponent there
so in the beginning before you
differentiate you have to take it up
it gets a negative exponent when you're
done
because you will usually be left with
the negative exponent you just reverse
the process you just take it back down
and then it is now finished and key
thing is you see how here
when we have differentiated we have made
sure that we write
the appropriate symbol and then here
even though we are now
just simplifying our already
differentiated
equation you always want to keep in mind
just write the
f prime in front as well so that you
don't get
penalized unnecessarily
