00:01
For this problem, we are given the function w equals x over y plus 2z, and we are asked to find d cubed w by dz, d, y, dx, and d cubed w by dx squared dy, so to begin, we can actually do that second partial derivative, or the second requested derivative, pretty easily, because when we take the derivative of w with respect to x, we get one over, it's one over y plus 2z, which means that the second derivative with respect to x is going to be 0.
00:34
So that second requested derivative, wxxy, is going to be 0.
00:41
Then we can look at the other requested derivative.
00:46
So we want to do wx as our first derivative is going to be 1 over y plus 2z.
00:55
Then the second derivative that we take is with respect to y.
00:59
So that we'll essentially be looking at this most easily as y plus 2 z to the power of negative 1...