# Chain and Product Rule – Implicit Differentiation

by Batool Akmal
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00:01 So after those 2 examples, we can now summarize what we?ve just learned.

00:06 Remember what we?re saying, if you have a function of x in terms of y and you?re trying to differentiate y with respect to x, you can do that but you'll have to differentiate y with respect to y because that?s the only way we can differentiate y.

00:23 So we can write that as d/dy of y and you must remember to multiply it with the factor dy/dx to make up for the extra factor that you have just multiplied in.

00:35 You might find doing this numerically a lot easier than this actual definition so we?ll keep practicing.

00:41 Let?s move on to a more difficult example.

00:45 Let?s look at our next example.

00:49 You can see already that this has a mixture of x?s and y?s, so hopefully you are recognizing that this is an implicit equation.

00:57 Let?s write this out and talk about each term separately.

01:01 We have 2x cubed plus xy equals to y squared plus 10.

01:11 Start to look at each term and thinking about what we?re going to do.

01:16 We?ve already decided that this is an implicit differentiation question so we?re not going to attempt to rearrange for y, we?re just going to differentiate it all in one go.

01:26 Now, keep your objective clear. We are trying to find dy/dx, that is the gradient.

01:32 So every time you are doing these questions, remember that you want to find dy/dx not dx/dy or no other variations of it.

01:41 You have to get dy/dx by itself and then get your answer.

01:46 So let?s start to look at each individual term.

01:49 This we can differentiate with respect to x, this we can differentiate with respect to y but we?re to multiply this with dy/dx.

01:58 Here is a constant which will just disappear so no matter what you would differentiate it with respect to x or y, you are going to get a value of zero and this little part here is 2 functions that are multiplying together.

02:13 Now, this is going to need a little bit more of thinking but let?s just do the easy parts first and then we?ll come to x times y.

02:20 So 2x cubed will differentiate 2, bring the power down, decrease the power by 1 so that gives you 6x squared.

02:31 So I?ve multiplied the 2 numbers together and I?ve decreased the x cubed by 1.

02:36 Here the xy, I?m going to leave some space for and we?ll come back to that.

02:42 The y squared, we?re going to differentiate with respect to y but remember that when you do that, you?ll have to multiply it with the factor of dy/dx.

02:52 So y squared simply differentiates to 2y and then don?t forget to multiply this with dy/dx and any constants just disappear or go to zero.

03:03 Back to x times y, here you can see 2 functions that are timesing together.

03:10 Think of all the methods you know that help you differentiate a function like this.

03:15 Is this a chain rule? Is this a function inside of a function? Is this a product rule question? Are there 2 functions timesing together or is it a quotient question, are they 2 functions dividing? Hopefully, we have established that this is 2 functions timesing together.

03:33 So this will need the product rule.

03:35 I?m going to remind you what the product rule is here on the side, dy/dx is udvdx plus vdudx or the other way around. It really doesn?t matter.

03:46 So what we?re trying to do is leave 1 term as it is and multiply it with a differential of the other and leave the other term as it is and multiply it with the differential of the first.

03:58 You can think about it as the first term, second term.

04:01 So leave first term as it is, differentiate second.

04:04 Leave second term as it is, differentiate the first and multiply.

04:08 Now, you could do this for xy here you can do this on the side and apply the entire rule to it but it?s a lot easier just to use logic and knowing that rule and just applying it straight away in your question here.

04:22 So let?s attempt to do that because it?s faster, it?s more efficient and it saves time.

04:26 Let?s apply the product rule to x times y.

04:29 The first thing we?ll leave, while this our first term, we?ll live it as it is.

04:34 So leave x as it is. We?re now going to differentiate second term.

04:39 So we?re going to differentiate y, remember what we do when we differentiate y.

04:43 We differentiate it with respect to y and multiply it with dy/dx.

04:48 So y will just differentiate to 1 and you must multiply it with a factor dy/dx.

04:54 The second part of the product rule, we?ll now have to differentiate the first term.

05:02 So the first term x, just differentiates to 1 and we?ll leave the second term as it is.

05:08 So the y term will just stay as y. Once again, we?ve applied the product rule.

05:15 Remember what we?ve done. We?ve left the first term as it is which is x, differentiated the y, so we did differential of that and then for the second half, we?ve differentiated the first term which is this term here and then we left the y as it was.

05:32 And you can see that we have the entire product rule here in these brackets and we didn?t really have to do it on this side or apply uv and with more practice, you?ll see that you will just be able to do the product rule a lot faster.

05:44 We?ll tidy this up and then rearrange it.

05:46 Remember what we said, its dy/dx that you?re looking for.

05:49 You can see it here already that you have 2 dy/dx terms, here and here.

05:54 So we?ll gonna have to take it all on 1 side and rearrange but let?s tidy up firstly.

05:59 So I have 6x squared plus x, dy/dx, plus y equals to 2y, dy/dx.

06:10 Now, you can leave all the terms that don?t have dy/dx on one side of the equation then take everything else to the other. So we have 2y, dy/dx minus x, dy/dx.

06:27 So we?re just taking like terms to one side.

06:32 Here you have a common factor of dy/dx in both terms.

06:36 So you can take it out as a common factor.

06:38 So we have dy/dx and then you have 2y minus x, on the other side, there?s not much else we can do. So it just stays as 6x squared plus y and lastly because the 2y minus x is multiplying on this side of the equation and you want to get dy/dx by itself, you can divide it on the other side.

06:58 So we can rewrite this as dy/dx equals 6x squared plus y over 2y minus x and that is your gradient for this fairly complicated looking implicit equation and you can find the gradient at particular points just by substituting values of x and y into this.

07:20 So like I mentioned before, practice is key in this topic.

07:25 So it?s really important that you try these exercise questions.

07:29 Take each step slowly, break it down.

07:31 Remember to always look at your question and to think about it.

07:34 Think about what kind of equation it is and then you have a lot of choice of differentiation method to choose from.

07:41 You choose the correct method and then you differentiate.

07:44 Watch out for x?s and y?s or functions that are multiplying and remember that you?d have to use the product rule for that.

07:50 Good luck and I?ll see you in the exercises lecture.

The lecture Chain and Product Rule – Implicit Differentiation by Batool Akmal is from the course Implicit Differentiation.

### Included Quiz Questions

1. f(x) g'(x) + g(x) f'(x)
2. f(x) g(x) + f(x) g(x)
3. f(x) + g'(x)
4. g'(x) f'(x) + f(x) g(x)
1. dy/dx=[2xy³+3x²y²] / [1-3x²y²-2x³y]
2. dy/dx=[2xy³+3x²y] / [1+3x²y²-2x³y]
3. dy/dx=[2xy³+3x²y] / [1-3x²y²-2x³y]
4. dy/dx=[2xy³+3x²y²] / [1-3xy²+x³y]
5. dy/dx=[2xy³-3x²y²] / [1-3xy²-2x³y]
1. y'= -x² / y²
2. y'=x² / y²
3. y'= -x/y
4. y'=x/y
5. 1
1. dy/dx=(2y-2x+1) / (2y-2x-1)
2. dy/dx=(2x-2y+1) / (2y-2x-1)
3. dy/dx=(2x-2y+1) / (2y+2x-1)
4. dy/dx=(2y-2x+1) / (2y+2x-1)
5. dy/dx=(y-x+1) / (2y-2x+1)

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