Implicit Differentiation: Exercise 2

by Batool Akmal

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

    00:01 Example 2: We now have xy plus 15x equals to 2y plus 5.

    00:12 A very obvious-looking implicit equation because it has a mixture of x?s and y?s.

    00:17 We are going to differentiate this implicitly. So we?ve looked at it.

    00:21 We?ve seen that there is a mixture of x?s and y?s.

    00:24 I really don?t want to go into trying to rearrange it. We have an easier method.

    00:28 Let?s just do this implicitly.

    00:31 Okay, thinking about each term I can see here that this x times y so you must think product rule there. I can see that we have a y term here, so we?re thinking dy/dx and then we?ve got some straightforward terms in between.

    00:46 So I?m gonna make some space for my product rule here.

    00:49 Remember what we said. Leave 1 term as it is, so let?s leave x as it is.

    00:56 Differentiate y. When you differentiate y that goes to 1 and then remember to multiply this with the factor dy/dx.

    01:04 Second term, we now differentiate this, which gives us 1 and we leave the y as it is.

    01:13 Easy. So we?ve just used the product rule and managed to differentiate x times y.

    01:19 Our next term, 15x just differentiate to 15. 2y, we can see another y term.

    01:27 If you were differentiating 2y with respect to y, that would just give you 2 but remember to multiply this with dy/dx.

    01:35 And the 5 at the end, it?s just a constant, so that goes to zero.

    01:40 We?re gonna tidy this equation up. So I have x times dy/dx plus y and take it out of the brackets and multiply it with appositives, so that?s fine.

    01:52 I have plus 15 equals to 2dy/dx, so not just 2y.

    02:02 Okay, we?re now going to move this often, the option is to keep things positive.

    02:10 So, we?re going to leave the 2dy/dx on this side. And then you can move the other 2dy/dx term here.

    02:18 So my whole objective is to keep many terms positive as I can.

    02:24 We've got y plus 15 on this side. You can take dy/dx as a common factor leaving you with 2 minus x and then on this, remember you still have y plus 15 and lastly, divide by 2 minus x because it?s multiplying on this side so you end up with y plus 15 over 2 minus x equals to dy/dx, which is your gradient.

    About the Lecture

    The lecture Implicit Differentiation: Exercise 2 by Batool Akmal is from the course Implicit Differentiation.

    Author of lecture Implicit Differentiation: Exercise 2

     Batool Akmal

    Batool Akmal

    Customer reviews

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