Mathematics in Medicine: Exercise 1

00:01 Have a look at out first example.

00:03 We’ve been given an equation or a function for drug sensitivity and we’re calling that R.

00:10 M here is the dosage of the drug in milligrams.

00:13 So we’re looking at a drug sensitivity of a certain drug and it’s given in terms of this function.

00:19 The question is asking us to find the rate of change of R with respect to M.

00:24 Now, really think about this, we’ve mentioned this so many times through this course.

00:29 Rate of change with respect to something.

00:32 I’ll take you back to when we looked at the change in Y over change in X.

00:36 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.

00:44 So with that thinking, you can conclude that this question is actually asking us to differentiate R with respect to M.

00:54 Let’s try that out on this function So given function R of M.

00:59 So that’s just saying that R is a function of M.

01:02 We have 2M square root of 10 plus 0.5M.

01:10 Okay, we’ve discussed that the question is asking us for the rate of change of R with respect to M.

01:16 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.

01:27 Let’s have a look at what we need to do.

01:29 So if it’s asking us for the rate of change of R with respect to M, we need to do dR/dM.

01:35 So that’s change in R over change in M.

01:39 And that is differentiation.

01:41 So we are differentiating this function here.

01:46 Hopefully, lots and lots of different techniques are going through your minds right now.

01:50 You’ve got two functions in terms of M.

01:53 You’ve got 2M that’s multiplying with something else.

01:57 You’re starting to think how do you differentiate two functions that multiply together.

02:03 And I hope you’ve concluded that we need to use the product rule.

02:06 Let me remind you of the product rule here on this side.

02:10 dy/dx or a product rule in terms of Y and X is vdu/dx plus udv/dx.

02:18 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.

02:26 We’ve done this using this formula and we’ve done this faster when we were doing implicit differentiation.

02:32 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.

02:39 So I have 10 plus 0.5 M to the power of a half.

02:44 And here, you can see two clear functions.

02:47 We’ve got this function and we’ve got this function.

02:50 You can call one of them U and one of them V and we can bring them together.

02:54 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.

03:04 So remember that we’re now differentiating with respect to M, which makes complete sense because this equation is in terms of M.

03:13 I’m going to start with leaving my 2M as it is, so that’s my first term.

03:16 So 2M stays and I now have to differentiate this.

03:21 Look at this closely and decide what kind of function this is firstly and what method of differentiation you’d need.

03:29 You can see here that you have something to the power of a half and then you have something inside of that function.

03:35 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.

03:44 We’ll have to use the chain rule method to differentiate the function of a function.

03:49 So the chain rule stated differentiate the outside function first as a whole and then multiply it with the differential of the inside.

03:58 So let’s try to - I’ll put brackets around this because this seems a bit more complicated.

04:02 Bring the power down.

04:03 So I’ll bring the half down, leave the inside function as it is, and decrease the power by one.

04:12 Don’t forget now to multiply it with the differential of the inside.

04:17 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.

04:26 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.

04:33 That’s the first half of the product rule then.

04:35 The second half, remember that the first bit I left this term as it is and I differentiated the second.

04:42 I’m now going to differentiate this term and leave the second term as it is.

04:46 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.

04:54 That can just stay as it was.

04:57 And so we’ve used the product rule to find dR/dM or the rate of change of R with respect to M.

05:05 It obviously isn’t very tidy, so we just need to do some algebra to tidy this up.

05:10 I can multiply those two numbers together.

05:12 So half and a half.

05:13 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.

05:22 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.

05:28 Here, you can rewrite this as 2 and you can change this to square root of 10 plus 0.5M.

05:36 These two numbers will cancel, so the 2 and the 4 cancels to give you 2, giving you M over 2.

05:42 And then like I said, you can bring this down.

05:45 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.

05:57 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.

06:07 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.

06:13 If I differentiate it, you can write it as R’M or dR by dM and we’ve differentiated it.

06:19 So really, you found a function which represents the gradient of how drug sensitivity is changing with respect to dosage.

06:31 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.

06:40 It then says find R’ of 50.

06:43 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.

06:56 We have R’ of 50.

06:58 And all we’re doing is changing our M, which was here, to 50.

07:03 So we’re using exactly the same equation.

07:05 So we’re using this equation and replacing all our Ms here with 50.

07:10 So I’ll end up with 50 over 2, root of 10 plus a half times 50.

07:20 And then I have 2 root of 10 plus a half times 50.

07:28 If we continue just simplifying this out, the 2 and the 50 cancels out to give you 25.

07:33 A half times 50 gives you 25 plus 10, so I have root 35 and then I also have 2 root of 35 here.

07:42 You can take root 35 as a common denominator, so remember when you multiple root 35 by root 35 in the enumerator, it's such going to give you 35, so this gives us 25 + 2 x 35 all over root 35, which when you add give you 95 over root 35.

08:00 As this doesn't simply you can leave this as it is or work it out as a decimal on your calculator? Therefore the value R dash at 50 is 95 over root 35.

08:13 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.