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 a positive, 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 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.