Have a look at out first example.
We’ve been given an equation or
a function for drug sensitivity
and we’re calling that R.
M here is the dosage of
the drug in milligrams.
So we’re looking at a drug
sensitivity of a certain drug
and it’s given in terms
of this function.
The question is asking us to find the
rate of change of R with respect to M.
Now, really think about this,
we’ve mentioned this so many
times through this course.
Rate of change with
respect to something.
I’ll take you back to when we looked
at the change in Y over change in X.
How does Y change
with respect to X?
And what does that give you?
It gives you the gradient
and it gives you dy/dx.
So with that thinking,
you can conclude that this
question is actually asking us
to differentiate R
with respect to M.
Let’s try that out on this function
So given function R of M.
So that’s just saying that
R is a function of M.
We have 2M square root of 10
Okay, we’ve discussed that
the question is asking us
for the rate of change
of R with respect to M.
So how does a drug sensitivity
change with respect to the dosage?
Does it increase?
Does it decrease?
And it’s been given to us
in terms of this function.
Let’s have a look at what we need to do.
So if it’s asking us for the rate
of change of R with respect to M,
we need to do dR/dM.
So that’s change in R over change in M.
And that is differentiation.
So we are differentiating
this function here.
Hopefully, lots and lots
of different techniques
are going through
your minds right now.
You’ve got two functions
in terms of M.
You’ve got 2M that’s multiplying
with something else.
You’re starting to think how do you differentiate
two functions that multiply together.
And I hope you’ve concluded that
we need to use the product rule.
Let me remind you of the
product rule here on this side.
dy/dx or a product rule in terms of Y and X
is vdu dx plus udv dx.
So in very simple terms, leave the first
function as it is, differentiate the second,
leave the second function as it
is, differentiate the first.
We’ve done this using this formula
and we’ve done this faster when we
were doing implicit differentiation.
To make this a little bit easier,
let me just write this as 2M
and rather than the square root, I’m
going to write that as a power.
So I have 10 plus 0.5 M
to the power of a half.
And here, you can see two clear functions.
We’ve got this function and
we’ve got this function.
You can call one of them U and one of
them V and we can bring them together.
Or alternatively, we can try
and use the faster method
where we leave one term as it
is, differentiate the second,
leave the second as it is,
differentiate the first.
So remember that we’re now
differentiating with respect to M,
which makes complete sense because
this equation is in terms of M.
I’m going to start with leaving my 2M
as it is, so that’s my first term.
So 2M stays and I now have
to differentiate this.
Look at this closely and decide what
kind of function this is firstly
and what method of
differentiation you’d need.
You can see here that you have
something to the power of a half
and then you have something
inside of that function.
This is a function of a function
and then you have to think back
as to what method you need to use
to differentiate a
function of a function.
We’ll have to use the chain rule method to
differentiate the function of a function.
So the chain rule stated differentiate
the outside function first as a whole
and then multiply it with the
differential of the inside.
So let’s try to –
I’ll put brackets around this because
this seems a bit more complicated.
Bring the power down.
So I’ll bring the half down,
leave the inside
function as it is,
and decrease the power by one.
Don’t forget now to multiply it with
the differential of the inside.
So the inside says that you
have 10 plus a half M,
differential of that is just
going to be a half or 0.5,
however you prefer to write it.
So the differential of the inside,
I’m writing as half because I
already have a half there,
it’s easier for
me to combine it.
That’s the first half of
the product rule then.
The second half, remember that the
first bit I left this term as it is
and I differentiated the second.
I’m now going to differentiate this term
and leave the second term as it is.
So the differential of 2M with
respect to the M is just 2,
and the second term can stay as it is,
we don’t have to fiddle around with it.
That can just stay as it was.
And so we’ve used the
product rule to find dR/dM
or the rate of change
of R with respect to M.
It obviously isn’t very tidy, so we just
need to do some algebra to tidy this up.
I can multiply those two numbers together.
So half and a half.
So I’ve got 2M
and then I’ve got a
quarter on the inside
and then I have 10 plus
0.5M to the minus half.
You can take this
down if you want
or you can leave it for a minute and then
we’ll take it back in the next step.
Here, you can rewrite this as 2
and you can change this to
square root of 10 plus 0.5M.
These two numbers will cancel,
so the 2 and the 4
cancels to give you 2,
giving you M over 2.
And then like I said, you
can bring this down.
So because this is to the
power of minus half,
you can write it as positive half ,
as 10 plus 0.5M
plus 2 square root
of 10, plus 0.5M.
And what you’ve just
found here is dR by dM
or another way of writing this is just
R differentiate it with respect to M.
So if I’m using the similar notation
to what they’ve used in the question,
they gave us R as
a function of M.
If I differentiate it, you
can write it as R’M or dR by dM
and we’ve differentiated it.
So really, you found a function
which represents the gradient
of how drug sensitivity is
changing with respect to dosage.
If we now look at answering the
second part of the question,
so it says find the rate of
change of R with respect to M,
which we found here as an expression.
It then says find R’ of 50.
So it’s saying when you put
50 mg into this equation,
what is the drug sensitivity
just as a number?
So we’ll do that here on the side, it’s just
a simple matter of substituting numbers.
We have R’ of 50.
And all we’re doing is changing
our M, which was here, to 50.
So we’re using exactly
the same equation.
So we’re using this equation and
replacing all our Ms here with 50.
So I’ll end up with 50 over 2,
root of 10 plus a half times 50.
And then I have 2
root of 10 plus a half times 50.
If we continue just simplifying this out,
the 2 and the 50 cancels
out to give you 25.
A half times 50 gives you 25
plus 10, so I have root 35
and then I also have
2 root of 35 here.
You can take 35 as a common denominator.
So you get 25 plus 2 times 35.
So remember when you multiply this –
if I do that on the side –
root 35 multiplied by root 35
is just going to give you 35.
Thirty-five, that should read.
You can try that out
on your calculators
because we said we’re
allowing us calculators here
so that it gives you 35.
But those are just rules
of surds here.
This gives you 25 plus 70 over 35,
which then gives you 95 over 35.
So 95 over 35 gives you an answer of
3 and we’ve got a nice whole number
and so we’re saying, if you
want it to, that R’ of 50
gives you an answer of 3.
And then you can do all types of
mathematical analysis on those results
and you can try and change the dosage
and see how drug sensitivity varies.