00:01
So we're asked to find dx, d, d, y, d, t, d, d, y, d x, is just going to be taking the d, d, y, d x, d equal t minus the natural log of t.
00:08
Y equals t squared minus t to negative 2.
00:15
So dx, t is just going to be taking the derivative of x with respect to t, leaving us with one minus one over t.
00:22
And let's simplify this to be, uh, t minus one over t.
00:28
Then d, y, dt, we just take the derivative of y with respect to t, and we get two t...