Chain, Product and Quotient Rule: Examples

00:01 First example, we have y=3x-1 to the 5.

00:06 The first thing that we need to do now because we know so many different ways of differentiating and because we are dealing with so many different types of mathematical functions, the first thing we have a moment to just look at this and to recognize what kind of problem this is and what rule we should be using.

00:30 So when you look at this closely you see that you have a large function something to the power of 5 and then you have an inside function which is just 3x-1.

00:41 We can use the chain rule but we can use the faster version of it, so we’ll just say dy/dx.

00:47 You differentiate the outer function first so 5(3x-1) stays as it is and decrease the power by 1 so this is just bringing the power down, decrease the power by 1 general differentiation and then you multiply it with the differential of the inside which in our case is just going to be 3.

01:06 Your final answer for the derivative is 15 (3x-1) to the 4.

01:12 And it doesn’t take us more than a minute to just use the faster version of the chain rule on questions like this.

01:18 Have a look at this second example.

01:21 Look at it for a moment and just think what kind of a function are you dealing with now? So we have y equals to x squared, (2x+1) to the power of 6.

01:33 I hope you recognized that there are in fact two functions here.

01:38 So you have this function and you have this function which are multiplying together.

01:43 You now have to think back and try and remember what you have to do in order to differentiate two functions that are multiplying together.

01:52 So have a little moment and think about what you'd have to do in order to differentiate a function like this? I hope you've recalled that in order to differentiate two functions that are multiplying together you will have to use the product rule because these are two product functions.

02:11 So you then ask yourself what is the product rule, you can make a little note of it for yourself so you can say that dy/dx is just vdu dx plus udv dx or however it is that you've learned it, and we then apply that straight to this question.

02:27 So work ahead of me if you'd like. Whilst I just do this example for you.

02:32 So we’ll say, u equals to x squared and v equals to (2x+1) to the power of 6.

02:40 Remember that we now need to differentiate both of them separately and then we're going to put them into our rule.

02:47 So u dash equals to 2x, v dash have a look at this, this is the chain rule so in order to apply the chain rule you bring the power down, (2x+1) to the 5, and you multiply this with 2 which is the differential of the inside.

03:08 We now tidy this up to get 12(2x+1) to the power of 5.

03:13 It’s all done now, all we have to do is put it into our product rule, so we can say that dy/dx is vdu dx.

03:22 You can draw little arrows if that helps so you don’t get confused with the terms so you multiply v first, well, in the product rule it doesn’t matter what you do first but just to stay consistent with the quotient rule, you'll find me always starting with the v.

03:37 So you multiply that with the differential of u, so we have (2x+1) to the power of 6 multiply it by 2x and then a plus in the middle and you're now multiplying u with vdu dx.

03:51 So we have 12(2x+1) to the 5, multiply it by x squared.

03:57 We can tidy this up so we've got (2x+1) to the 6, I should put the 2x at the front just to stay consistent so 2x here, and then I have the 12 and the x squared can add together, it’s not an equals, so plus 12 x squared, (2x+1) to the 5.

04:20 There's more you can do with these obviously. You can simplify this.

04:24 We look at an example where you had common factors and you can see that we've got a common factor of (2x+1) and (2x+1) to the power of 5 which is the common factor so you can take out (2x+1) to the power of 5 out of both terms, this now leaves you with the 2x(2x+1), and you also got 12x squared.

04:50 Tidy this up a little bit more so I can see that when I multiply this through I will get 4x squared plus 2x, plus 12x squared and the only reason I multiplied it through whereas usually I prefer to just leave it factor wise is because I can add those two terms together, so we got (2x+1) to the 5, we have 16x squared plus 2x.

05:19 And by using the product rule we have found the derivative of two products x squared multiplied by 2x+1 to the power of 6.

05:29 Let’s look at our last example.

05:33 Observe for a second what's happening in this function.

05:36 We have y = 5x-1 divided by 3x-2.

05:41 There should be lots of rules and lots of different functions going through your minds right now and what rule would be appropriate to differentiating this.

05:50 I’ll write it down while you have a little think of what kind of function this is, so we have y = 5x-1 over 3x-2.

06:00 You can see, I hope, that this is a function that's being divided by a function so you have a top function and you have a bottom function.

06:08 You also know a method to differentiate this.

06:12 So think through all the rules that we've done and see if you can remember how to differentiate two functions when they're dividing each other.

06:20 I hope you recalled that in order to differentiate two functions that are dividing you will have to use the quotient rule which states vdu dx minus udv dx over v squared.

06:35 With practice I hope you will actually manage to learn this rules but you must continue to practice in order to get good and to get faster at applying these rules.

06:44 Okay, so looking at the rule that does mean we need a u and a v and the differential of them.

06:50 Remember what I said to always take the top function as your u and your bottom function as your v and we're going to differentiate both of them individually.

06:59 U=5x-1, and v=3x-2, like we do in the product rule we differentiate each one of them separately, so du/dx, u dashed in this case just 5, nice and easy and v dash is going to be 3.

07:17 Luckily, there was no chain rule this time which doesn’t happen very often, so let’s be happy with that.

07:23 Right, let’s apply dy/dx. Remember what the rule says, vdu dx, so we're going to multiply that with that and then we're going to multiply u with dy/dx.

07:37 Remembering this time that there is a minus in between which is important and then we're also going to divide it by v squared.

07:43 So if I write the 5 first and then I have (3x-2) subtract 3(5x-1), and then all over v squared which is (3x-2) squared.

07:58 I’ll leave the (3x-2) at the bottom factorized, so rather than expanding out just cuz it’s tidier.

08:04 At the top, I should be able to simplify some because they're all to the power of 1 The x’s and we've got some numbers multiply that 2, that gives me 15x-10, and that gives me another 15x+3, when I multiply that through, so remember that you're multiplying it through with a negative 3 so -3 multiplied by a -1 will give you a +3.

08:29 We have 14x positive and -15x which cancels out leaving us with just -10 +3 which gives me -7/(3x-2) all squared.

08:43 And if I could see that things would simplify further I could have expanded the denominator but because nothing else is going to cancel, we can just think one step ahead of ourselves and leave it as it is.

08:54 So we have used the quotient rule to derive or to come to the derivative of y=5x-1 divided by 3x-2, giving us an answer of -7/(3x-2) all squared.

09:10 Have a look at the exercise calculations now. There's quite a few different questions to try out.

09:18 All of them will be using a variety of different rules. Make sure you look at each question first.

09:24 Decide what kind of function it is and then go for the rule that would suit your function best.

09:31 Remember for each different type of function we have a different rule so make sure you apply it correctly and good luck.