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 a positive, 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 side, 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.
The lecture Implicit Differentiation: Exercise 1 by Batool Akmal is from the course Implicit Differentiation.
What is the expression for dy/dx given the equation x⁷ + y⁷ = 7 ?
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