A Level Maths Made Easy - Product Rule

hey guys welcome back to another a Lev maths Made Easy video this time we're going to be looking at differentiating products so product meaning we have two functions of X that are multiplying each other so in this first case we have X which is the function of X and e x which is a different function of X how we can differentiate those and essentially do anything we need differentiation for um so if we start as we've established there are two different functions of X here x and e to the x uh so what do we need to do well essentially when you differentiate all you really need to do is you take the first thing so I like to call these u and v the two different functions this is how it's written in your textbook as well and we're going to differentiate both of them so DX is one dvdx is just e to the X because of course it stays exactly the same and then all we really do is we multiply and add the opposite terms so 1 * e to X is e to X Plus x e to the X and that's what our differentiation is going to be so we could call that Dy by DX and that as y now if you notice I've given you a bit of a formula so to speak but there is a easier way to think of this all you do is you're going to differentiate both of these so for example the first one is x * the differential of the second one plus this times the differential of the first one so essentially all you need to do is it's almost like you're expanding brackets we have to do it to both of the two different functions and times them together so if I show you something like this for example this is of course combined with the chain rule so again we have U is equal to x v isal to x + 1^ 5 and essentially we want to differentiate U times it by V differentiate V times it by U so hopefully you can see what I mean by doing it to both of them so DDX is of course one dvdx if we use the shortcut we discussed in the last video you bring the power forward five you times it by this differentiated which is of course one then you times it by the same thing to the power of 4 so it's just 5 x + 1 to the power of four and again all we need to do is make sure we times each derivative by the opposite function so x + 1^ 5 plus X so X x * 5 I'm just going to write that as 5x x + 1^ 4 and that's what you're kind of doing so the differentiation doesn't really change all that much and let's do one more example of the basics again slightly I'm not even going to say It's Tricky really as long as you know how to differentiate the basic functions you're basically doing the exact same thing again but with two functions instead of one so again here it would be 8x cubed and here would be again bring the power forward differentiate that you just get one so I'm not going to write it and again times the opposite values now of course because we're adding technically speaking it doesn't matter which way round we add them so it doesn't matter which way round we actually add them so if you did the first one at the second it's the same thing because adding is commutative so it's the same either way around so that would be 2 * 3 6 WR d by DX 6 x 4 5 + x^2 Plus 8 x cubed just like that so go with this one I would like you to try it for yourself but you may need to kind of rewrite the functions as something a bit easier to to use so X2 and I'm going to write this in terms of an Indy to make it a bit easier but I have a go at trying to do it in terms of the product rule and I'm going to start going through it now so again bring the power forward differentiate the inside that's still just one the negative a half and then yeah just times the opposite things so we have 2X X in for Dy DX x + 6^ half plus half x^2 x + 6^ negative a half so when do you use the product rule you use product rule whenever you have two functions of X multiplied by each other so as you can see in all of these examples we have two different functions of X that are multiplying by each other that's the really key thing to remember it's not too tricky after that I think that this is quite nice it's not too bad um and yeah let's go on to an exam style question we have to apply this in a slightly different context so here is a past exam question so we have a curve with equation blah blah blah passes through the point a which is 1 one find an equation for the tangent to the curve at a so if you imagine that we have some kind of curve and let's say that's a a tangent will have the same gradient as the curve at that point a so we need to find the gradient of the curve at a and that will be equivalent to our gradient of the line so remember whenever you want an equation of a line so a tangent or a normal or whatever you need two things need a coordinate to find plus C and you need a gradient as well so we need to differentiate this expression here x cubed again this would be a good time to pause the video and actually have a go at differentiating this for yourself see if you can apply what you've learned but I will be going through it just just so we can save some time so that be 2X and DV by DX would be -3 because don't forget when we differentiate this we do get a minus one 2 - x^ 2 and then times the cross terms equals so X2 * this so - 3x^2 2 - x all 2 plus 2x 2 - x or Cub cuz remember I'm just doing the cross terms and let's rewrite that that's a little bit ugly not going to lie and then to find the gradient all we need to do is sub in the x coordinate of the point we're looking at so subbing in xal 1 so x = 1 we get Dy by DX equal something so I'm going to use a calculator because again while this is quite easy to do without a calculator it's a good it's good to use a calculator just to make sure you don't make any mistakes so 1^ 2 2 - 1^ 2 + 2 1 2 - 1 cub and we get ne 1 nice and easy so that means our gradient of this line is -1 Sox + C and if you remember the coordinates are 1 1 so we're going to sub in y = 1 x = 1 so then we get 2 is equal to C so our final answer is is just x + 2 and that's our final answer and that'll be four or so marks maybe five marks if we get lucky but part B will also give us another two marks so show the normal to the curve passes through the origin so the normal will pass through a still but the gradient is the negative reciprocal so the gradient we're going to do minus1 over the gradient of this line which is min-1 which gives us 1 so we have y is = x + C and again we're going to sub in x = 1 y = 1 So Cal 0 so the Y intercept is zero so it's just the equation of yal X and that's it so in terms of the product rule I don't think this is too crazy the next thing we have to go through is the quotient Rule now I'm actually going to show you that the quotient rule you don't really need it instead you can use the product rule and the chain Rule and that is usually easier because with the with the quotient rule the actual order that you do things in does matter so it is a bit tricky it's easier to make m Stakes so I'm going to show you a method and way you can use the chain Rule and the product rule instead to get the quotient rule which in my opinion is a lot easier because it doesn't matter on the order then but hopefully you found that somewhat useful again there are loads of questions that you can use to practice this I don't think this is too challenging personally but when we get into some of the laser stuff like modeling and integration that's when it gets a bit more fun so yeah if you did enjoy the video or you found it useful please do consider subscribing but I'll see you in the next one