Example 2: We now have xy plus 15x equals to 2y plus 5.
A very obvious-looking implicit equation because it has a mixture of x?s and y?s.
We are going to differentiate this implicitly. So we?ve looked at it.
We?ve seen that there is a mixture of x?s and y?s.
I really don?t want to go into trying to rearrange it. We have an easier method.
Let?s just do this implicitly.
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.
So I?m gonna make some space for my product rule here.
Remember what we said. Leave 1 term as it is, so let?s leave x as it is.
Differentiate y. When you differentiate y that goes to 1
and then remember to multiply this with the factor dy/dx.
Second term, we now differentiate this, which gives us 1 and we leave the y as it is.
Easy. So we?ve just used the product rule and managed to differentiate x times y.
Our next term, 15x just differentiate to 15. 2y, we can see another y term.
If you were differentiating 2y with respect to y,
that would just give you 2 but remember to multiply this with dy/dx.
And the 5 at the end, it?s just a constant, so that goes to zero.
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.
I have plus 15 equals to 2dy/dx, so not just 2y.
Okay, we?re now going to move this often, the option is to keep things positive.
So, we?re going to leave the 2dy/dx on this side. And then you can move the other 2dy/dx term here.
So my whole objective is to keep many terms positive as I can.
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.