A Level Maths Made Easy - Chain Rule

hey guys welcome back to another one of these AEV maths Made Easy videos today we're looking at the chain roll for differentiation now before we get started again most people that watch these videos aren't subscribed so if you have found any of these videos useful in the past I'd really appreciate it if you could just press subscribe because it really does help me my link to my website and my email address are in the description in case you want tuition or you have any other questions but without further Ado let's jump into the chain Woll now if I gave you something like this and let's just say y equals for now and I asked you to differentiate it in other words to find Dy by DX there isn't really a quick way to do this using your current knowledge You' have to expand the brackets so You' have to use like binomial expansion which yeah that works but it's a bit of a pain in the butt instead something that might be useful is instead if we made this in the form where there is no binomial so instead if I said U is equal x + 3 I would now have y equals U to the^ 5 and differentiating that is super super easy so if I differentiate that we have Dy by du so do you see why it's not DX it's Dy by du because now our variable is U instead of X we just get 5 U to the^ 4 which is really really easy now of course we don't want Dy by du we want Dy by DX so we have Dy by du and what we need to do is find something else that we can Times by in order to to get Dy by DX the way you work this out is actually fairly simple we want Dy by DX so we have the correct numerator we want DX as the denominator now if I times this as it is we're going to get DX as the denominator so instead I need to cancel out the DU and the only way to cancel out the DU is by cross cancelling so that means I need to work out what DX is which as we can see is just one so means D1 by DX it's quite simply Dy by D 5 u^ 4 * 1 and of course we need to sub in the original expression so this would be 5 x + 3 to the^ 4 * 1 but that would be it in terms of your answer this relationship here is what the chain rule is okay and honestly it is extremely useful in many many many many different application so if you're doing anything related to Matts in the future whether it's economics whether it's uh engineering whether it's physics whatever you like you will be using a lot of this because we don't always know exactly what we want instead we need to manipulate the derivatives in order to find what we want and this will also really come into play when we do integration later on now there is a shortcut okay I will show you the shortcut after you do another example so again here we have y = 2x - 1^ 3 so what do we do well again we make that equal to U so we have y = u cubed we then differentiate it which is 3 u^2 now again the question will always be find Dy by DX they will not say differentiate the following they will normally say find Dy by DX because technically I have differentiated it technically they will have to specify now the variables they want you to use and again remember we have Dy by du and I want you to take a second what do you think the other part of this would be remember we want DX on the bottom which means our denominator has to be DX and we want Dy by DX at the moment we have du so we need to cancel that du somehow and we do that by introducing a du at the top now there is a topic that came up last year in your year 12 exams but also it comes up this year which is modeling with differentiation and using like differential equations and you use this a hell of a lot that's why I'm insisting on doing this method for now just so you can get used to the idea of picking your variables and seeing how you can eliminate it so again how did I get that DX is on the bottom because we want DX to be on the bottom judging by the answer we want d on top because we don't have a du in the bottom in the answer so again it's about using the answer to figure out what you need to actually work out so here we would need to have when we have Dy byd already Dy by du is 3 U ^2 but we need D DX so du by DX is just 2 so we're going to times that by 2 so that's 6 U ^2 but again we want it in terms of X because we have d by DX not Dy by du so that would be 6 6 2x - 1^ 2 so what is this shortcut that I keep mentioning well the shortcut is the following if you ever want to differentiate something that has a power so where you would use the chain rule all you do is you bring the power forward in this case I'm going to have a three then you times it by the derivative of what's inside of here so 2x - 1 differentiated would be two and then you times it by the same bracket to the power of nus one and if you notice we get the exact same answer but we get there far quicker now if you notice that I have just done the steps that we've done in this question I've just done them faster so I would recommend practicing this until you're fast enough to do this because you'll get there naturally anyway but if you wanted to write down like a a formal definition if you have y is equal to F ofx that means Dy by DX I should put that to the power of n is just n time the derivative time FX the^ nus 1 that's all the chain rule really is and here we have another example of of a question that we can be given now differentiating this some people might think ah maybe we use the quotient rule which is something we're going to learn later but come on I won't do you that instead you're going to rewrite this using your laws of indices to have minus 3 and again we could use the definition that we wrote above so the answer is going to be n so - 3 * 4 * the derivative with the bracket or you can use as I mentioned before the normal chain rule so let's use the shortcut just to see how it works so if Dy by DX equals we still have that four cuz that's not doing anything the power comes to the front so it's- 3 then you times by the derivative of what's inside so 2x + 3 differentiated is just 2 then you have your normal bracket and you take away one from the power so in other words you have 4 * -3 is -12 * 2 is -4 2x + 3 the -4 that's it again as a bit of a challenge it would be good if you went out now and just quickly tried to do this using the you know the technical in-depth rigorous way just to make sure that you can actually do it uh that would be really really good practice for you now this might seem a bit of a strange way to look at this because this is also technically the chain Rule now I will say that the way that they describe derivatives in the textbook is a little bit wonky so in the textbook they only really tell you that e to the X when you differentiate it is e to the X there's actually a better way to describe this which is y is equal to e to the function of X then it means Dy by DX you just differentiate the function so fime of X and you times it by the original thing so in this case here I could just differentiate and get my answer so that would be two times the derivative of the function so that's differentiate 3 T gives me 3 e to the^ 3 T because we have the original thing again which means it would just be 6 e 3 T but let me show you how to do this with the chain rule just so you're aware if we have Y = 2 e^ 3T the question will say find Dy by DT so if I first because I want this to look like the original definition although again I've just shown you that's not really necessary by the way this works for literally everything so it works for E to X it works for Ln it works for S cosine tan all you need to do is differentiate what's inside the brackets and and then you can get your answer but we'll come to that when we come to that right so I'm going to sub in U is equal to 3T so Y is equal to 2 e to the U which means Dy by du is just 2 e to the U because as we've established it doesn't actually change now I want Dy by DT and I have Dy by du so what do I need to Times by to get Dy by DT so give you a second to a think so we have the Dy good so the numerator doesn't need to change but we have du on the denominator and I don't want that so I'm going to cancel it out with a du in my numerator here and we want a DT in the denominator which we don't have at the moment so I going to have DT there so I need to work out du by DT which is three so that means Dy by DT if we sub that all back in Dy by du is 2 e to the U time d by DT which is three as we got over here which means this is 6 e to the U which if we sub back in for U it is 2 sorry it's 6 e to the^ 3T which as you can see we got there immediately using that result so again using these definitions is in my opinion a lot nicer but again we will get to those when we get to those this also works for integration by the way just so you're aware with the definition of Ln is a little bit odd I'll say it's a little bit different so I would say using the chain R might be nice but I will show you the shortcut for those of you that are interested so once again we only really know that the derivative of Ln X is 1 /x so we need to make it look like Ln X at the moment we have 6t + 1 so the first thing I'm going to do is I'm going to sub in U = 6t + 1 so again the question would be find Dy by DT then I'm going to differentiate I'm going to write what Y is so 3 Ln U which means Dy by du is simply 3 over U then if we want Dy by DT we need to think about what we need to do well we have Dy by du but I want Dy by DT so again take a second to have a think about what your times it by so you want to get rid of the DU over DT so if I differentiate this we get the answer of 6 so means Dy by DT is simply Dy by D which is 3 U * 6 so that's 18 over U and again we want to sub back in for the original uh expression that was in the middle so we get 18 over 6t + 1 now okay what's the shortcut what's the actual definition well if we have y equals Ln of some function whoops that was a bit weird Ln of some function then the derivative is simply the derivative of that function the inside part so whether it's x^2 2x whatever it may be over F ofx and if you have a look at what we did we when we differentiate what's inside we get six so that becomes 6/ 6t + 1 but then it's Times by 3 because there's a three in front of this that's T 18 / 6t + 1 so we actually do get the same thing in this question we would get 3 * 6 over 6t + 1 which is equal to 18 over 6t + 1 and you get there a lot faster but again this result is just because of the chain rule that we just used it's the same thing so let's start get into some more kind of interesting cases so again with this one all you're going to be doing is rewriting it using good oldfashioned laws of indices so you would write the bottom as 2T - 5^ 3 but because it's under a fraction we're going to make it 2T - 5 minus a 3 and again let's use the trick so going go back in the video we're going to use this trick to see if we can do it so we have the power time the derivative time the normal function with the power subtracted by one so we have the six that just stays there the power is minus A3 the derivative where if I differentiate what's inside the bracket I just get two times the original function powerus 4 over 3 CU I have minus one from the the power so let's see actually that wouldn't be y anymore it be Dy by DX or Dy by DT in this case DT Dy by DT and we get so 6 * A3 is - 2 * 2 is -4 2 t - 5 - 4 3 just so you know chain was also used for differentiating parametric equations you do literally the same thing we've done today where you get your two expressions for dyd and DX and you sub it all in and it's super super easy that's why I showed you that way to make sure it's all okay and for the last question this one's a bit more interesting I would say that you absolutely should use the normal definition of the chain rule so the question would be to find Dy by DX now this isn't me saying that you can't use the shortcut because you could write this as two cos x cubed in which case you could still use the shortcut bring the power forward differentiate cos Times by cos squ so you can absolutely use that but again let's use it uh in both ways just just to finish off so I'm going to say that U equal cos x which me Y is equal to 2 U Cub differentiate with respect to U so we want Dy by DX and we have Dy by du so again what do we need to cancel out we need to cancel out the DU and we want DX so if you differentiate cos x to get d by DX you get negative sinx so Dy by DX is just Dy by du in fact I should have put times there whoops 6 U squ screwed time sinx so that's -6 sinx U ^2 sub back in for u - 6 sinx cos 2 x and that will be your answer and again if you did the shortcut method you would get the exact same answer yeah that kind of brings us to the end of differenti using the chain wall again it does come up again again and although it might seem quite easy it is extremely useful in lots of different matths that you're going to be doing over the next few uh years I guess well the next year but also if you do stuff at University like medicine or anything like that you'll use this as well um and it's super super useful so if you found that useful again please do consider subscribing or you know droing me an email or whatever you want to do again you can uh look at the website if you want any kind of tuition CU I do have space in some of the classes but without further Ado I'll leave you to enjoy the rest of of your day and I'll see you in the next one