00:01
In this question we are given the function w equal to x cosine y z and x equal to the s square, y equal to t square, z equal to the s minus two t.
00:15
And we have to do this partial derivative by two ways, the first way a by the channel and by the part p by plugin.
00:31
Now let's do the part a first.
00:33
We want to find the dw over the ds.
00:37
Applaying the formula for the general.
00:39
First it will equal to the dw over the x.
00:42
So we have here will be coside y, z, and then times the x over the ds, which is 2s, and then plus dw over the dy.
00:54
So we have a minus here.
00:57
And then we have the x, z, side, y z times dy over the ds so have the 0 and then plus d w over the dz so we have here will be minus and then we have the x y times cosine under y z and times dz over the s so will be 1 looking the value of the x y z we should have 2 s coside y times z so we should have the s t square minus two t about three and then minus x times y we should have the s square t square and then we have coposite of the s t square minus two t about three so that will be the first partial derivative now the second partial derivative to equal to 2.
01:57
Now again we should have the cosine y z times d x over the d t will be 0 minus x z side of the y z times d y z times 2 t and minus x y cosy x x x over d t will be times y z times d z 2 in front...