00:01
So we're going to be doing implicit differentiation for this problem.
00:04
Implicit differentiation is really useful when we have both y's and x's playing an important role in an equation that we need to differentiate.
00:13
So the first thing we're going to do is we're going to plug in one for all of the x values.
00:23
So when we do that, we see that we get y cubed plus 1 times y.
00:31
Minus y equals 8 times 1 to the 4th.
00:37
And that ultimately gives us y cubed equals 8, which means y equals 2.
00:44
So now we know that the point on the curve is 1 -2 that we're focusing on.
00:50
And now what we need to do is calculate d, y, d, x.
00:53
So this is going to be the inputs differentiation.
00:56
We take the derivative on both sides.
00:58
When we do that, we get 3y squared times y prime plus y plus xy prime minus y prime.
01:19
And that's going to be equal to 32x cubed.
01:27
Then we can factor out the y prime since it's in all of these different terms...