00:01
Okay, so for this problem, we are given z equals 3x squared, y to the third.
00:07
We're given that x equals t to the fourth and y equals t squared.
00:13
So now we're asked to find, i'm going to kind of write this here at the bottom, dz over dt, using the chain rule, since t is not in the equation of z, we're going to have dz over dx, dx over dt, plus dz.
00:32
Dz over d .y and then d .y over dt.
00:38
So i want to look at dz and integrate, or i'm sorry, derive it in terms of x.
00:46
So if we look at this, we're going to have y to the third is a constant.
00:50
So that is going to stay since we x has a, since x is present.
00:55
And then we're going to have 6x.
01:00
And then if we look at this in terms of y, we have 3x squared times.
01:08
Times 3y squared.
01:11
So this is going to be 9x squared, y squared.
01:17
And then now we're going to do dx over d t, which is going to be 4 t to the third, and then d y over d t is 2t.
01:32
So now what we're going to do is we're going to plug these in.
01:36
So i have 6xy to the third, and then 4 t to the third, plus 9x squared y squared times 2t...