[John:] ... Have you done much work?
[speaker002:] I've done some work. I've tried
[John:] Good.
[speaker002:] to sort of
[John:] Good.
[speaker002:] what the hell I was supposed to do with.
[John:] [LAUGHTER]... What were you doing?
[speaker002:] There's some I'm totally totally confused on.
[John:] You were doing differentiating
[speaker002:] [cough]
[John:] and then
[speaker002:] Yeah.
[John:] leaving it and then going back to it to see if you could integrate it plus some graphs.
[speaker002:] Some of it. Yeah. No no I don't know what I've done with the graph. I've been concentrating on trying to
[John:] Okay.
[speaker002:] I've gone through these I don't know how many times.
[John:] Right.
[speaker002:] I done them... and then I've took them down and I've gone back. I've g I've got about a half a dozen sheets of paper scattered all over the place, and I've decided to do it in this cos then I can just go to here.
[John:] Right.
[speaker002:] So I've done that. Now I can understand that, understand that.
[John:] Right.
[speaker002:] ... That goes to that, two X, but I'm not sure why.
[John:] Now is that that cosh squared X?
[speaker002:] Yeah.
[John:] Right.
[speaker002:] And that is... goes to two X.
[John:] shine squared X. You mean do you mean?
[speaker002:] No. It's two X in the book.
[John:] Yeah. Okay.
[speaker002:] Now he does make mistakes so that's not a
[John:] Erm...
[speaker002:] problem.
[John:] Mm.
[speaker002:] And then there's That I can understand. But that is actually is not like that. That's X squared but when it's differentiated cos you've gotta go back into the bracket it comes out as that.
[John:] Right.
[speaker002:] And that then goes in
[John:] Okay.
[speaker002:] that. So I understand that bit.
[John:] What you're doing I mean you're coming along very well really.
[speaker002:] But
[John:] Aren't you?
[speaker002:] I'm not doing too bad.
[John:] I think you're doing very well
[speaker002:] But its like that. I don't get Y to twelve. I mean I know th I know that you carry the sine on because
[John:] Mm.
[speaker002:] it goes to sine squared, so you you only differentiate one and according to
[John:] Okay.
[speaker002:] that goes to sine squared.
[John:] Right where your twelve comes from.
[speaker002:] Well it it must be three times four is equal
[John:] Right.
[speaker002:] to twelve.
[John:] Okay.
[speaker002:] But it doesn't always go to the front of them. Like here.
[John:] Yeah. Well
[speaker002:] The use of the the two X. So I'm not conversant with all the rules.
[John:] Mm. Okay. Erm... how did you differentiate these? I mean are you learning these as they are or... I mean
[speaker002:] No.
[John:] do them? How are you doing it?
[speaker002:] No I've I've just All I've done is because... Erm... because that is a function of that
[John:] Right.
[speaker002:] I've differentiated that. Well I didn't differentiate it but I but I did because I couldn't decide.
[John:] As I as I was talking about for a new pattern
[speaker002:] .
[John:] how much you want to write it in there, it's up to you. Just show me how you did that one.
[speaker002:] Well, all I
[John:] That one
[speaker002:] did was that I didn't know whether to do three sine squared... right?
[John:] Mm.
[speaker002:] Then differentiate the four X because that's your X... value. So that'd be three times the four sine squared. Then to differentiate the whole lot because it goes to cos.
[John:] Mm.
[speaker002:] And that is is is not what I get cos what I was doing is doing them meself and then
[John:] Well what Yeah. What
[speaker002:] going and checking the answer.
[John:] Erm
[speaker002:] And then if me answer was wrong trying to
[John:] Mm.
[speaker002:] work out.
[John:] Can you describe your methods. How you how you'd do that. Sort of writing down as many steps as you can. Ah. That's why i couldn't I thought I hadn't got me pen.
[speaker002:] Erm well all
[John:] Draw a nice picture instead.
[speaker002:] All I did is I look at what I've got and if it's just a single term like that
[John:] Right.
[speaker002:] then you
[John:] Yeah.
[speaker002:] know to just differentiate it. When it's two terms... like the the log erm and what like that
[John:] Mm.
[speaker002:] each of the two X multiplied by sine.
[John:] Okay.
[speaker002:] That brings on another rule. Like that
[John:] Okay
[speaker002:] brings another rule.
[John:] Erm... differentiate this for me then. Er... three X squared [cough] plus two. And put some brackets round that and raise that to the power five. Okay Y equals that. Sh show
[speaker002:] .
[John:] me all the steps.
[speaker002:] Divide by D X two. That goes at the front, right?
[John:] Mm.
[speaker002:] To begin with.
[John:] Right.
[speaker002:] Now you get erm... three X squared plus two and then you differentiate What's in there, which'll give you... six. So if you multiply it by six and it's still raised to the power.
[John:] If you differentiate three X squared what do you get?
[speaker002:] Six.
[John:] Differentiate
[speaker002:] Six X.
[John:] Right. Now.
[speaker002:] .
[John:] Okay. Right. That's, that's the method. Now why're you doing this? What's the theory behind it?
[speaker002:] Dunno.
[John:] [LAUGHTER] Right. Now what you're doing, or what you're, you're attempting to do... erm if I say Y equals P
[speaker002:] Hmm.
[John:] Find D Y by the X. And you can't cos you haven't got a you haven't got an X in sight.
[speaker002:] Mm.
[John:] Now what you've got here is Y equals U to the power five.
[speaker002:] Yeah.
[John:] Right? Now there isn't an X in sight.
[speaker002:] Mm.
[John:] I've only got three X squared plus two there but you can't differentiate whit respect to three X squared plus two and then say you actually done it with respect to X. So... erm you've seen the cha chain rule function and a function.
[speaker002:] Mm.
[John:] This is, this is what you should be doing with it really. Erm we've got that. Y equals that.
[speaker002:] Yeah.
[John:] Right. Let U three X squared plus two. Okay. Now we've got... Y equals U to the power five. So we c can't find D Y by D X
[speaker002:] Mhm.
[John:] But we can find D Y by D U. Right? Which is no problem.
[speaker002:] Mhm.
[John:] Five by D to the four. Yes?
[speaker002:] Mhm.
[John:] Well that's found D Y by D U but what we're looking for is D Y by D X. And D Y by D X... is what? In terms of U and everything else. Remember what that bit? How to work that out?
[speaker002:] Erm isn't it the... U.
[John:] Okay. The easy way to remember it
[speaker002:] Mm.
[John:] is write it like that. D Y by D X equals D Y D Y over something times something over D X with D U in there cancel out. It isn't quite as simple as that but it works.
[speaker002:] Mm.
[John:] Okay. So D Y by D X gives D Y by D U times D U by D X. Well if U is equal to that
[speaker002:] Mhm.
[John:] we can differentiate that with respect to X. We can find D U differentiating this side with respect to X you get D U by D X and
[speaker002:] Mm.
[John:] on that side then you'll get six X.
[speaker002:] Mm.
[John:] Okay. So we want to find D Y by D X and it's D Y by D U then it's D U by D X. Well we've found D U by D X and we've found D Y by D U. So D Y by D U is equal to D Y by D X is equal to D Y by D U, which we've found from here, from Y equals
[speaker002:] Yeah.
[John:] U right? four, times D U by D X which we found up here right? Now we haven't got X in it... yet so we can't give them the answer in that form cos they didn't tell us anything about U. You invented that so where do you put you place U... five times whatever U was... three X squared plus two... all to the power four times six X because normally bring that to the front.
[speaker002:] Three X.
[John:] Right.
[speaker002:] is one.
[John:] And you get thirty X..
[speaker002:] Mm.
[John:] Okay? Now, that's, that's what you should be doing each time with these With that in and and that much in sort of as a minimum. You wil you've obviously sort of seen the method somewhere and you're trying to do it in your head, but cos you're not going through it in a fairly formal way
[speaker002:] Mm.
[John:] you're... most of the time you're getting it right cos this is just making the right sort of guess and you're seeing the sort of When it gets to the awkward ones y you're just sort of Well maybe it's that, maybe its something else. Does that
[speaker002:] Yeah.
[John:] help?
[speaker002:] It does help, I mean that that I have seen before but I didn't sort of understand it and you've actually
[John:] Well the thing is after a while when you've done quite a few it comes it comes fairly easy, and you can do it in your head.
[speaker002:] Mm.
[John:] But when you get to one where you can't do it in your head you've got to go back to this bit.
[speaker002:] Yeah.
[John:] And sometimes I mean there, we sort of let. There might be times when we let you do the whole lot or when you have to have more than one go at it.
[speaker002:] Mm.
[John:] Erm... I mean if you've got something like, for example, erm say we've got the one we've just had. Y equals something to the power.... To the power five right?
[speaker002:] Mm.
[John:] But it wasn't three X squared plus two.
[speaker002:] Mm.
[John:] It was erm sine sine squared six X... plus cos three X
[speaker002:] Mm.
[John:] Right. Now doing that in your head would be a bit awkward.
[speaker002:] Mm....
[John:] But there you'd do it in three goes I mean, you could probably do it in two goes, but to be safe you could split it up into three goes so...
[speaker002:] What rule does that bring out because that.
[John:] Well
[speaker002:] Is that still the same, you just
[John:] There's an there's an add in there.
[speaker002:] Mm.
[John:] Right. So lets... erm.
[speaker002:] Yeah....
[John:] L equal seven X... okay and... M equals D X. Right we don't we can probably do these these bits in your head quite easily.
[speaker002:] Mm.
[John:] Okay. So we've got here, differentiate that we get D L by D X.... D M by D X is true.
[speaker002:] Mm.
[John:] Now what were trying to find inside. We haven't done anything about this. It's all to the power five yeah?
[speaker002:] Mm.
[John:] So we could maybe do what we did last time. Let U equal this lot in the brackets. Sine squared seven X. three X... Now we're trying to find D U by D X.
[speaker002:] Mm.
[John:] At some stage. Well you could probably do that one in your head.
[speaker002:] Mm.
[John:] And what would that come to?
[speaker002:] Erm that'd be fourteen cos squared seven X.
[John:] So if you differentiate sine squared what do you get?
[speaker002:] That. Cos squared.
[John:] Cos squared
[speaker002:] Well that's what I wanted to ask you. What happens to them? Because I know I asked one of my friends
[John:] Mm.
[speaker002:] because I wasn't I said how come it ends up being that and they said you just carry. When you're using the trig function and you're differentiating
[John:] Mm.
[speaker002:] it, that you do use because it goes to the front but it stays the same. It carries on right the way through. You don't lose the value of it. [cough]
[John:] Good. Right. Okay.
[speaker002:] But the squares I wasn't sure what happens to them and I confused him by asking him.
[John:] Okay. Let's look at this. Erm... L is equal to seven X
[speaker002:] Mm.
[John:] So
[speaker002:] [cough]
[John:] ... So it'll be about five stages in this. working it out any of them them all and then you can miss out the ones we do you feel you don't need.... Just look at that bit. [cough] You're trying differentiate what's inside there with respect to something. Well if you've got something like erm... that one.... Tan plus seven X. So we've got Y equals... sine L What're we having? seven X.
[speaker002:] Mm.
[John:] Squared.
[speaker002:] Squared.
[John:] We write it like that because it's sine squared, it is more obvious what the meaning is.
[speaker002:] [cough] Mm.
[John:] Now we can't differentiate that... with respect to L.
[speaker002:] Mm.
[John:] Because it's not L, it's sine L. Okay. If that was i if that was L squared Y equals X squared, you could differentiate that with respect to X. If it's Y equals L squared you could differentiate it
[speaker002:] Mm.
[John:] to L. It's not. It's sine. So... we'll differentiate... gonna have another substitution on this one. So it's all just goes on and on and on until you can get something that's erm a straight forward one letter to the power so you can differentiate it. So let H equal sine L.
[speaker002:] Mhm.
[John:] now we've got Y equals H squared. Well that's no problem. D Y by D H... cos now two H.
[speaker002:] Mm.
[John:] but now we've got H equals sine L. Does that define D H by D L?...
[speaker002:] Mm.
[John:] And if you differentiate sine sine L you get?
[speaker002:] Five sine L erm cos L.
[John:] Erm... Well we've got cos. Now we're trying to find from this one. Let's work out what we're going to do. We're trying to find D Y by D X... but we can't find that,
[speaker002:] Mm.
[John:] so we're going for D L... D L by D X
[speaker002:] Mm.
[John:] times D Y by D L.
[speaker002:] Mhm.
[John:] Okay?
[speaker002:] Why isn't that er cos?
[John:] If you differentiate sine X what do you get?
[speaker002:] Cos X.
[John:] We don't get cos X or
[speaker002:] So because you took the number of X value away.
[John:] Erm
[speaker002:] If you substituted L. You've taken that, you've squared X so you've got You've made that into another value. Right? So
[John:] Mm.
[speaker002:] you've made that Y equal to H squared. So that becomes H so the value of that
[John:] I started off let H equal sine L.
[speaker002:] Yeah.
[John:] Right. Now... If if all let's say all we've got to differentiate was Y equals sine squared X.
[speaker002:] Mm.
[John:] Now I could find D Y by D sine X.
[speaker002:] Mm.
[John:] differentiate that, with respect to sine X but not with respect to X.
[speaker002:] No.
[John:] Right. And if I if I found D Y by D sine X that would show me, not the gradient of this curve,
[speaker002:] Mm.
[John:] the gradient of the curve that I would get if I plotted Y against sine X.
[speaker002:] Mm.
[John:] But we're plotting Y against X which'll obviously be a very different curve with
[speaker002:] Mm.
[John:] lots of different gradients... Thank you very much sir.
[speaker002:] Thank you.
[speaker003:] Right I'll leave you to it.
[John:] So for that there's there's no way of differentiating that with respect to X. [door closing] There is eventually because you learn it. You know Oh it comes out as cos X.
[speaker002:] Mm.
[John:] Or comes out as whatever it comes out as. Right? You can differentiate sine X on its own like that. Y plus cos X, but we can't differentiate that straight off without changing and doing the substitution. So we'll let er what letter haven't we used yet? Q. So if we let Q equal sine X we can find D Q by D X... we can differentiate this side with respect to X and that side Q by D X when we differentiate.
[speaker002:] Mm.
[John:] Okay? Now we've let Q equal sine X so we've got Y equals Q squared. We'll differentiate that with respect to Q. No problem. Right? differentiate both sides with respect to Q
[speaker002:] To Q.
[John:] Right. Okay? But we're looking for here, D Y by D X so we have D Y by D X is equal to and we just write, D Y D X. D Y by D Q there D Y by D Q
[speaker002:] Two Q.
[John:] Is two Q... times Q by D X.... Okay?
[speaker002:] Mhm.
[John:] Now we've found D Y by D X all we need to do is put back Q to what it was.
[speaker002:] Mm.
[John:] Right which is sine squared X. So s two... sine squared X
[speaker002:] Mhm.
[John:] Okay. So you're sort of right in a way that it's... two sine squared X but this wrong. Are you are you happy with that??
[speaker002:] Yeah. Mhm.
[John:] Okay get Y equals X squared no problem. Two X.
[speaker002:] Mm.
[John:] Y equals... log X squared you can't just picture log X.
[speaker002:] No.
[John:] very very different. We're finding the gradient... the gradient on a specific graph when we plot the X against the Y, the Y against the X. Right?
[speaker002:] Mm.
[John:] If we say well that looks a bit awkward so we're going to plot the Y against sine X or the y against log X
[speaker002:] Mm.
[John:] Erm or the the Y against I mean in this one if you You could plot that against X which is the normal way. If you tried to plot that against Think of how you'd do it if you had a graph and you plotted Y against three X squared plus two.
[speaker002:] Mm.
[John:] Well you'd just get Y equals A graph that would look exactly the same as Y equals X to the.
[speaker002:] Yeah.
[John:] Which is why the gradient of it would be five times that expression to the power four.
[speaker002:] Yeah.
[John:] Okay. But that would not be the gradient of plotting Y against X. A very different graph.
[speaker002:] Mm.
[John:] So. Try and think of what you do when you find D Y by D X. It becomes oh well you you do some little tricks on numbers and you shuffle them about and that's the answer they want.
[speaker002:] Mhm.
[John:] If you can get it back that's why I say do the graph if you can get it back to erm... but it's not just something to the fifth, it's really It's something to the tenth. Right Okay one way differentiating that
[speaker002:] Mm.
[John:] is multiply it out. You've got five lots of brackets there. Three X squared plus two every one.
[speaker002:] Mm.
[John:] In the half an hour multiply them all out, right? And you get terms in X to the tenth.
[speaker002:] Yeah.
[John:] Everything else all the way down. And you could just differentiate each term quite simply then. once you've done the multiplying, the differentiation would be very simple. there'd just these straight powers of X, and you could do that and you'd get to the same answer as we get to it's just that this is a quicker way of doing it.
[speaker002:] Mm.
[John:] But that, that is a Y equals something X to the tenth, so it's it's not going to have a gradient that requires Y equals something to the tenth.
[speaker002:] Mm.
[John:] Which is why you must... do that sort of That that's probably all you need... for that bit. To know what you're doing right. Erm... I think it would be useful for you do one the same way. Erm... we'll do a fairly simple one and then use the same one but make it a little bit more complicated.
[speaker002:] Mm. What what rule is that? The chain rule?
[John:] a function of a function
[speaker002:] Mm?
[John:] the chain rule. You can't do it in one go, so you split it down into bits.
[speaker002:] Because it is a function of a function of X?
[John:] Yeah. Right. And then they want the gradient of Y against X. Not Y against the function, a sub-function of X.
[speaker002:] Mm.
[John:] So if we get something like...
[speaker002:] ...
[John:] Right. Have a go at that one. Yeah.
[speaker002:] So we put the or sine squared... think it's better leaving it like that because it's sine brackets
[John:] Well what do you think?
[speaker002:] Well I think
[John:] [cough]
[speaker002:] Erm... think it's probably better if.
[John:] Okay. Go on. Have a have a have a go, See what happens with that.
[speaker002:] Alright. Well I know now that that's not right. [LAUGHTER]
[John:] Good. Okay. Well okay. Fine. If you know it's not right, scrub that. Think of something more useful to let U be equal to Cos this is the the main part of it. Working out what you're going to put U equal to.
[speaker002:] Well if I put U equal because brackets the bracket there.
[John:] Right.
[speaker002:] . And.
[John:] Okay.
[speaker002:] and let that equal sine squared.
[John:] Well we've had that already, but sine squared doesn't mean anything.
[speaker002:] No.
[John:] Does it? I'd like you to get your calcu calculator out, look up the sine of some some number but I'm not going to tell you what it is and then square it.
[speaker002:] Mm. But how do you separate that then? That's it. That's that's it until I know or I know what I'm actually looking for in it I can run around chasing me tail all day which
[John:] Right.
[speaker002:] is basically what I do know.
[John:] Right. Okay. Okay what've we got here? [cough] Erm... if I said Y equals sine squared X,
[speaker002:] Mm.
[John:] you couldn't differentiate that.
[speaker002:] Mm.
[John:] You'd have to do what we did on the other page. You'd have to do this.
[speaker002:] Mm.
[John:] So that was sort of on this one. What do we do there? What do we let U equal to?
[speaker002:] Squared...
[John:] Let Y equal Q squared.
[speaker002:] Mm.
[John:] Yeah, so we let the whole thing there that was raised to the power be equal to something. So... if we let V not U Right, we let U be that,
[speaker002:] Mm.
[John:] and let V If if U is zero,
[speaker002:] Mhm.
[John:] then V will be You can write it as sine squared but I'm just writing this way cos it it means more.
[speaker002:] Sine V brackets raised to the power.
[John:] Three equals sine U squared. Okay? Right, can you take it from there do you think?
[speaker002:] Well, I'll have a go.
[John:] Try to sort of keep in mind what you're aiming for but don't rush getting there. See the steps along the way....
[speaker002:] [cough].... Is that right?...
[John:] Right, can you see what you're doing?
[speaker002:] Yeah.
[John:] This isn't a simple two stage one like the other one. You've got to try and work out what this middle step's going to be.
[speaker002:] Mm....
[John:] I'll give you I'll give you a clue and see if you can work out how you're going to get towards it. What we're going to do The way we can work it is something like D Y by D X is equal to, now we've got that,
[speaker002:] Mm.
[John:] so we can differentiate both sides there with respect to U.
[speaker002:] Mm.
[John:] And find D V by D X. So we can D V by D U. D V by D U... times What do you want. Then come across D Y by And at the end, D X. So what've we got in terms of X? We've got D U.
[speaker002:] Yeah.
[John:] So we can find D U by D X. So we can find something by D U.
[speaker002:] Mm.
[John:] What've you got in terms of U? We've got D V in terms of U so we can find D V by D U. So we get D Y by D Y by D V
[speaker002:] Mm.
[John:] Right? Times D V by D U.
[speaker002:] Mm.
[John:] Times D U by D X. Which is simply those and those cancelling out. And
[speaker002:] Mm.
[John:] it comes to?
[speaker002:] D Y by D U
[John:] That'll come to D Y by D X. So that that'll go out with that that'll go out with X an you get D Y by D X. So you've got you've got the V in terms of U.
[speaker002:] Yeah.
[John:] So you can find D V by D U. Okay? We've got U in terms of X so you can find D U by D X. And you've got V, now.... Erm you've got Y. Where's Y gone? You haven't got Y in this one. Okay. Then what's Y equal to? [recording ends]