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today we'll be looking at implicit
differentiation basically this is just a
different way of differentiating it's a
new type of equation that we will be
looking at and learning how to
differentiate it now remember that
there's a lot of differentiation that
you've already done we've done
differentiation from first principles
which is basically the derivative or the
definition of a derivative we then moved
on to the faster way of differentiating
and with that we found the equations of
tangents and normals and we also found
the gradients of tangents and normals we
then looked at different types of
functions so we looked at what to do or
how to differentiate a function within a
function we looked at how to
differentiate two functions when they're
multiplying together and also two
functions when they're dividing one by
the other we then looked at trig
differentiation and ease of access and
Ln of X's so your mind must be full of
differentiation in calculus but let's
now look at another type of equation
which we call an implicit implicitly
written equation and we are going to
differentiate it implicitly so here we
have our first example you have x
squared plus y squared equals 225 we're
going to talk about how to differentiate
it let's have a look first of all if you
didn't know how to differentiate
implicitly you might just rearrange this
equation in terms of Y and then
differentiate it there's nothing wrong
with that we could do y squared equals
225 minus x squared you can get rid of
that squared here so we can say this is
the same as the square root of 25 minus
x squared and remember that you'd have
plus and minus when you square root any
value this then becomes fairly easy to
differentiate we can break it down into
its two components so let's do the
positive here so if we have y equals to
a positive square root of 25 minus x
squared and then we'll also
differentiate y equals to negative 25
minus x squared just make
easier to separate them out remember
what we said if it's a square root you
can rewrite this as a power so this can
be the same as to the power of 1/2 and
then we differentiate to get dy by DX
equals 2 chain rule so you got a big
function and a little function so bring
the power down decrease the power by 1
leaving the inside as it is and then you
multiply it with the differential of the
inside so the differential of 25 minus x
squared is just going to be minus 2x the
twos here cancel out so we left we are
left with minus X multiplied by 25 minus
x squared to the minus 1/2 which if you
really wanted to you could rewrite it as
minus x over 25 minus x squared to the
positive 1/2 remember rules of indices
or also if you really want to show off
your skills we can write it as square
root of that so just back to the form
that they gave it and remember this is
your gradient dy by DX similarly when
you differentiate with the positive not
much is going to change apart from the
sign so let's just have a look here so
we rewrite this as dy by DX you have if
I rewrite this here it's minus 25 minus
x squared to the 1/2 we now have a minus
on the outside you bring the power down
so we're applying the chain rule 25
minus x squared to the minus 1/2 and the
differential of the inside is again
minus 2x so you can see the only
difference is that you have a minus and
a minus there which will make it
positive so you will end up with x over
square root of 25 minus x squared and
that's the gradient so you're getting
two gradients which we'll talk about
shortly as to what it means there is
another way of doing this and the other
method is called an implicit method
which means that you don't have to
rearrange you don't have to go in terms
you don't have to rearrange your
equation in terms of Y you can just do
it implicitly you can just do it in one
go rather than having to rearrange this
equation had an all
to rearrange not all of them do so it's
important that we learn how to
differentiate implicitly because it just
makes our lives easier so let's do the
same question and now look at how to
differentiate this implicitly x squared
plus y squared equals 225 and what we're
saying here is that we're not going to
rearrange this equation we're just going
to differentiate it in one go so
essentially I want to differentiate D by
DX this term x squared so I'm going to
try and differentiate each term
individually I want to differentiate D
by DX y squared and then I also want to
differentiate D by DX of 25 you don't
have to write this out each time you'll
see the more practice we do the less
we'll have to write and we can just go
straight into differentiating where as
the first time as we explain what's
going on I'd like to write it as a whole
okay let's just read out what we've
written we're saying differentiate x
squared with respect to X so that makes
perfect sense we're differentiating X
with respect to X and we know that when
we differentiate x squared we'll get 2x
I'm going to leave this time out for a
second let's just come to this term here
we're saying differentiate 25 with
respect to X remember when you
differentiate any constant it just goes
to 0 so that part is fixed as well let's
read the statement here we are now
saying differentiate Y squared with
respect to X now that doesn't make any
sense we're saying differentiate Y with
respect to X there are two different
things
so we now need to find out a way of how
we can do this basically here's what you
could do you can differentiate D by dy
this is something that I'm adding myself
Y squared but also remember that the d
by DX or this term here will still
remain so this is just dy by DX so see
what we've done here we've tweaked it a
little bit we're using the chain rule in
a sense because we are differentiating
the outside and we're differentiating
the inside
we are leaving the y-squared as a
differential of X so we leave that term
here but we're also adding a little bit
so that we're actually able to
differentiate Y now I can leave this as
2x this part of the equation I can do I
am now saying differentiate Y squared
with respect to Y that I know is 2y but
I also add in an extra term dy by DX and
that needs to stay equals to zero if
someone asks me to find the gradient of
a function that means dy by DX my
objective eventually is to say dy by DX
equals to something because that is the
gradient so all I need to do from this
equation is rearrange it I want to keep
dy by DX as the subject of the equation
so I can leave 2y dy by DX on this side
of the equation and I can take the 2x
over I want to make dy by DX the subject
of my equation so I can say dy by DX
equals to minus 2x divided by 2y just
taking that over to the other side
dividing it because it's multiplying on
this side you can cancel the tooth so
our gradient is minus x over y and this
is the gradient implicitly so we're
saying dy by DX equals 2 minus x over y
now this looks a little bit different
but numerically it will be the same
value as the previous gradient I'll
explain what this actually is because
it's nice to understand what this graph
is actually doing and then we can
substitute those numbers in just to
check that it works so if you recall x
squared plus y squared equals 225 is
actually the equation of a unit circle
so this will give us the radius so the
radius is the square root of 25 to give
us 5 plus or minus 5 where radius is a
positive length so essentially we have a
circle which is placed at the origin 0 0
because of this x squared y squared
there's no values or number
being added to it so it's going to be at
the center and we know that the radius
is going to be five so if I attempt to
draw a circle that has radius five okay
we found two gradients so we said
implicitly that dy by DX equals two
minus x over y and we also found it the
gradient earlier which we said that dy
by DX equals two minus x over square
root of 25 minus x squared we had two
answers we obviously had plus and minus
so the positive gradient or the positive
value of this graph would be up here in
the positive quadrant and the negative
one would be here so we're getting two
different types of gradients now if we
for example just take a number so let's
just say let's substitute or let's find
d-y-by-dx
so not substitute let's find dy by DX at
3 minus 4 let's just check if they are
numerically correct so in this case if I
use the negative gradient because 3
minus 4 will be somewhere around here
so we're looking at this point here 3
minus 4 so if I use the gradient dy by
DX not negative positive gradient
because that will give you a positive
gradient line x over 25 minus x squared
so I'm just taking the positive version
of this gradient because I can see that
the line is positive here that's a
negative coordinate positive and the
gradients here should make that a little
bit clearer is positive so now we've put
the numbers in so we've put X in as 3
here and we've replaced our X here as 3
if we just work that out that gives us 3
over root 25 minus 9 which gives you 3
over root 16 and root 16 is 4 so that
gives Anna gives us a gradient of 3 over
4 so remember what I've done there are
two types of gradients that I got from
this equation here it's telling me that
you could get a positive and
negative gradient just by signaling out
what the graph looks like I know that I
have in the negative coordinate let's
get rid of this it's not so confusing in
the negative quadrant I get a positive
gradient and in the positive quadrant I
get negative gradient so we're just
dealing with this point 3 minus 4 here
which I've kind of thought ahead of
myself and knew that this will give me a
positive gradient so only used the
positive part of the gradient similarly
when I did it implicitly I got this
answer which looks different but all I'm
trying to do here is prove that they are
the same values so if we do this at 3
minus 4 as well dy by DX when you put
your values in will give you minus 3
over minus 4 the minuses will cancel out
to give you 3 over 4 as well so if
you're ever worried that implicit
differentiation doesn't look like the
same answer they will be the same it's
the same gradient that we've calculated
it's just done a different way and an
easier way so what we recommend is when
you see equations like these which
involve X's and Y's and a mixture of
them and it's difficult to rearrange or
even now because you have the skill you
don't even have to rearrange so it's not
a matter of you thinking whether it's
easy or not you can just go straight
away and differentiate it implicitly but
remember one thing when you
differentiate Y every time you
differentiate Y you must multiply with a
factor of d-y-by-dx so if we just make a
little note here so if we say when
differentiating Y with respect to X
always multiply just by that with dy by
DX so remember this and that's the only
important little thing that you need to
remember for implicit differentiation
and then we rearranged dy by DX to find
our gradient we'll do a couple more
examples to make this idea and this
concept a bit clearer
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