Implicit Differentiation: Example 1

by Batool Akmal

My Notes
  • Required.
Save Cancel
    Learning Material 2
    • PDF
      DLM Implicit Differentiation Calculus Akmal.pdf
    • PDF
      Download Lecture Overview
    Report mistake

    00:02 Today, we'll be looking at Implicit Differentiation.

    00:05 Basically, this is just a different way of differentiating.

    00:08 It's a new type of equation that we'll be looking at and learning how to differentiate it.

    00:13 Now, remember that there's a lot of differentiation that you've already done.

    00:17 We've differentiation from first principles which is basically the derivative or the definition of a derivative.

    00:24 We've then moved on to the faster way of differentiating and with that we found the equations of tangents and normals and we also found the gradients of tangents and normals.

    00:35 We then looked at different types of functions.

    00:38 So we looked at what to do or how to differentiate a function within a function.

    00:42 We looked at how to differentiate two functions when they're multiplying together and also two functions when they're dividing one by the other.

    00:51 We then looked at trig differentiation and e of x's and ln of x's.

    00:56 So your mind must be full of differentiation and calculus.

    01:00 But let's now look at another type of equation which we call an implicitly written equation.

    01:06 And we're going to differentiate it implicitly.

    01:09 So here we have our first example. You have x squared plus y squared equals to 25.

    01:16 We're going to talk about how to differentiate it. Let's have a look.

    01:26 First of all, if you didn't know how to differentiate implicitly, you might just rearrange this equation in terms of y and then differentiate it.

    01:34 There's nothing wrong with that. We could do y squared equals to 25 minus x squared.

    01:40 You can get rid of that squared here.

    01:43 So we can say this is the same as the square root of 25 minus x squared and remember that you'd have plus and minus when you square root any value.

    01:51 This time becomes fairly easy to differentiate.

    01:54 We can break it down into its 2 components. So let's do the positive here.

    01:59 So if we have y equals to positive square root of 25 minus x squared and then we'll also differentiate y equals to negative 25 minus x squared.

    02:10 Just makes it easier to separate them out.

    02:12 Remember what we said if it's a square root, you can rewrite this as a power.

    02:18 So this can be the same as to the power of a half and then we differentiate to get the dy/dx equals to chain rule.

    02:27 So you got a big function and a little function so bring the power down.

    02:31 Decrease the power by 1 leaving the inside as it is and then you multiply it with the differential of the inside.

    02:39 So the differential of 25 minus x squared is just going to be minus 2x.

    02:44 The 2's here are cancelled out.

    02:47 So we're left with minus x multiplied by 25 minus x squared to the minus a half, which if you really wanted to, you could rewrite it as minus x over 25 minus x squared to the positive a 1/2.

    03:04 Remember rules of indices or also if you really want to show off your skills we can write it as square root of that. So just back to the form that they gave it in.

    03:14 Remember this is your gradient dy/dx.

    03:16 Similarly, when you differentiate with the positive, not much is going to change apart from the sign. So let's just have a look here.

    03:25 So we rewrite this as dy/dx. You have, if I rewrite this here, it's minus 25 minus x squared to the half.

    03:35 We now have a minus on the outside. You bring the power down so we're applying the chain rule, 25 minus x squared to the minus of half and the differential of the inside is again minus 2x.

    03:48 So you can see the only difference is that you have a minus and a minus there which will make it positive. So you will end up with x over square root of 25 minus x squared and that's the gradients. So you're getting 2 gradients which we'll talk about shortly as to what it means.

    04:04 There is another way of doing this and the other method is called an implicit method which means that you don't have to rearrange.

    04:12 You don't have to rearrange your equation in terms of y, you can just do it implicitly.

    04:19 You can just do it in one go rather than having to rearrange.

    04:22 This equation had an option to rearrange, not all of them do.

    04:26 So, it's important that we learn how to differentiate implicitly because it just makes our lives easier.

    04:32 So let's do the same question and now look at how to differentiate this implicitly.

    04:37 X squared plus y squared equals to 25, and what we're saying here is that we're not going to rearrange this equation, we're just going to differentiate it in one go.

    04:49 So essentially, I want to differentiate d/dx, this term x squared.

    04:55 So I'm going to try and differentiate each term individually.

    04:59 I want to differentiate d/dx, y squared and then I also want to differentiate d/dx of 25.

    05:07 You don't have to write this out each time.

    05:09 You'll see the more practice we do, the less we'll have to write and we can just go straight into differentiating.

    05:15 As the first time, as we explain what's going on, I'd like to write as a whole.

    05:19 Okay, let's just read out what we've written. We're saying differentiate x squared with respect to x.

    05:26 So that makes perfect sense.

    05:27 We're differentiating x with respect to x and we know that when we differentiate x squared, we'll get 2x. I'm gonna leave this term out for a second.

    05:37 Let's just come to this term here. We're saying differentiate 25 with respect to x.

    05:43 Remember when you differentiate any constant, it just goes to zero.

    05:46 So that part is fixed as well. Let's read this statement here.

    05:51 We are now saying, differentiate y squared with respect to x.

    05:56 Well, that doesn't make any sense. We're saying differentiate y with respect to x, they're two different things. So we now need to find another way of how we can do this.

    06:07 Basically, here's what you could do. You can differentiate d/dy.

    06:13 This is something that I'm adding myself y squared but also remember that the d/dx or this term here will still remain.

    06:23 So, this is just dy/dx. See what we've done here. We've tweaked it a little bit.

    06:31 We're using the chain rule in a sense because we are differentiating the outside and we're differentiating the inside. We are leaving the y squared as a differential of x.

    06:45 So we leave that term here but we're also adding a little bit so that we're actually able to differentiate y. Now, I can leave this as 2x.

    06:54 This part of the equation I can do. I am now saying differentiate y squared with respect to y.

    07:00 That I know is 2y but I also add in an extra term dy/dx and that needs to say equals to zero.

    07:10 If someone asks you to find the gradient of a function, that means dy/dx.

    07:15 My objective eventually is to say dy/dx equals to something because that is the gradient.

    07:20 All I need to do from this equation is rearrange it.

    07:24 I want to keep dy/dx as the subject of the equation so I can leave 2y, dy/dx, on this side of the equation. And I can take the 2x over.

    07:35 I want to make dy/dx the subject of my equation so I can say dy/dx equals to minus 2x divided by 2y, just taking that over to the other side, dividing it 'cause it's multiplying on this side.

    07:51 You can cancel the 2's, so our gradient is minus x over y and this is the gradient implicitly.

    07:58 So we're saying dy/dx equals to minus x over y.

    08:04 Now this looks a little bit different but numerically, it will be the same value as the previous gradient.

    08:09 I'll explain what this actually is because it's nice to understand what this graph is actually doing and then we can substitute the numbers in just to check that it works.

    08:19 So if you recall, x squared plus y squared equals to 25 is actually the equation of the unit circle. So this will give us the radius.

    08:33 So the radius is the square root 25 to give us 5, plus or minus 5, the radius is a positive length.

    08:39 So essentially, we have a circle which is placed at the origin zero, zero because of this x squared, y squared.

    08:49 There's no values or numbers being added to it so it's going to get the center and we know that the radius is going to be 5.

    08:55 So if I attempt to draw a circle that has radius 5.

    09:01 Okay, we found 2 gradients. We said implicitly dy/dx equals to minus x over y and we also found that the gradient earlier which we said that dy/dx equals to minus x over square root of 25 minus x squared.

    09:25 We had 2 answers. We obviously had plus and minus.

    09:29 So the positive gradient or the positive value of this graph would be up here and the positive quadrant and the negative one would be here.

    09:39 So we're getting 2 different types of gradients.

    09:42 Now, if we for example just take a number.

    09:45 So let's just say, let's substitute or let's find dy/dx, so not substitute, let's find dy/dx at 3, minus 4. Let's just check if they're numerically correct.

    10:01 So in this case if I use the negative gradient because 3, minus 4 will be somewhere around here.

    10:07 We're looking at this point here, 3 minus 4.

    10:11 So if I use the gradient dy/dx, not negative, positive gradient because that would give you a positive gradient line.

    10:18 X over 25 minus x squared, so I'm just taking the positive version of this gradient because I can see that the line is positive here.

    10:26 That's a negative quadrant positive under the gradients here should make that a little bit clear, it's positive. So now we've put the numbers in.

    10:35 So we've put x in as 3 here and we've replaced our x here as 3.

    10:43 If we just work that out, that gives us 3 over root 25 minus 9, which gives you 3 over root 16 and root 16 is 4 so that gives us a gradient of So remember what I've done? There are 2 types of gradients that I got from this equation here.

    11:01 It's telling me that you could get a positive and a negative gradient just by filling out what the graph looks like, I know that I have in the negative quadrant, let's get rid of this so it's not so confusing.

    11:13 In the negative quadrant, I get a positive gradient and in the positive quadrant I get negative gradient.

    11:19 So we're just dealing with this 0.3 minus 4 here, which I've kind of thought, had it in myself, I knew that this would give more a positive gradient. So, only use the positive part of the gradient.

    11:28 Similarly, when I did it implicitly, I got this answer which looks different but all I'm trying to do here is prove that they are the same values.

    11:37 So if we do this at 3, minus 4 as well dy/dx, when you put your values in, will give you minus 3 over minus 4, the minuses will cancel out to give you as well.

    11:52 So, if you're ever worried that implicit differentiation doesn't look like the same answer, they will be the same.

    11:59 It's the same gradient that we've calculated.

    12:02 It's just done a different way and an easier way.

    12:05 So what we recommend, is when you see equations like these, which involves x's and y's and a mixture of them and it's difficult to rearrange or even now because you have the skill, you don't even have to rearrange, so it's not a matter of your thinking whether it's easy or not.

    12:22 You can just go straight away and differentiate it implicitly but remember one thing, when you differentiate y, every time you differentiate y, you must multiply with a factor of dy/dx.

    12:36 So if we just make a little note here, so if we say when differentiating y with respect to x, always multiply with dy/dx.

    12:52 So remember this and that's the only important little thing that you need to remember for implicit differentiation and then we rearrange dy/dx to find our gradient.

    13:02 We'll do a couple more examples to make this idea and this concept a bit clearer.

    About the Lecture

    The lecture Implicit Differentiation: Example 1 by Batool Akmal is from the course Implicit Differentiation.

    Included Quiz Questions

    1. y = ±√(-x² + 25x)
    2. x² + y² = 25x
    3. y² = -x² + 25x
    4. y² = x² + 25x
    5. y = ±√(x² + 25x)
    1. 4x + 3y²y' = 5
    2. 4x + 3y² = 5
    3. 4x + 3y²y' + 5 = 0
    4. 4x + y²y' = 5
    5. 4x + 3y' = 5
    1. dy/dx = [-1 / 2y].[3x² + 5]
    2. dy/dx = [1 / 2y].[3x² + 5]
    3. dy/dx = [1 / y].[3x² + 5]
    4. dy/dx = [-1 / y].[3x² + 5]
    5. dy/dx = [-1 / 2].[3x² + 5]
    1. 2xy³ + 3x²y²y'+ 2yy' = 0
    2. 2xy³ + 3x²y²+ 2y = 0
    3. 2xy³ + 3x²y²y'+ 2y' = 0
    4. 2xy³ + x²y²y'+ 2yy' = 0
    5. 2xy³ + 3x²y²y'+ 2yy' = 100
    1. dy/dx = (14x) / (3y²)
    2. dy/dx = (14x) / (y²)
    3. dy/dx = x / (3y²)
    4. dy/dx = -14x / (3y²)
    5. dy/dx = (14x + 23) / (3y²)
    1. dy/dx = [234x² + 13]/3y²
    2. dy/dx = [1 / y²].[234x² + 13]
    3. dy/dx = [1 / 3y].[234x² - 13]
    4. dy/dx = [-1 / 3y²].[234x² + 13]
    5. dy/dx = [1 / 3y²].[234x + 13]
    1. dy/dx = 2/3
    2. dy/dx = 1/3
    3. dy/dx = -2/3
    4. dy/dx = 8/3
    5. dy/dx = 1/2

    Author of lecture Implicit Differentiation: Example 1

     Batool Akmal

    Batool Akmal

    Customer reviews

    5,0 of 5 stars
    5 Stars
    4 Stars
    3 Stars
    2 Stars
    1  Star