00:01
In the first problem we are given that w equals to xy plus x z plus y z where x equals to t minus 1, y equals to t squared minus 1 and z equals to t.
00:20
We are asked to find out dw by dt using chain rule in subpart a.
00:28
So by chain rule we can write d w by d t to be equal to do w over do x times d x over d t plus do w over do y times d y over d t plus do do d d t.
00:44
Plus do do d d t.
00:49
Now let us evaluate each of these derivatives.
00:52
The partial derivative of w with respect to x gives y plus z and the partial derivative of x with respect to t is 1 plus the partial derivative of w with respect to y is x plus z and the partial derivative of y with respect to t is 2 t plus the partial derivative of w with respect to z is x plus y times the partial derivative of z with respect to t is 1 so now let us simplify this and substitute the values of x y and z we have have t squared minus 1 plus t plus t minus 1 plus t times 2 t plus t minus 1 plus t squared minus 1.
01:45
So now simplifying this and canceling out the terms we get 6 t squared minus 3.
01:52
So therefore this is the required answer for subpart a.
01:57
In subpart b of this problem, we are asked to substitute the values of x, y and z in w and then differentiate it with respect to t.
02:09
So let us substitute the values we have w to be equal to t minus 1 times t squared minus 1 plus t minus 1 times t plus t squared minus 1 times t.
02:23
Simplifying this and canceling out terms we obtain t2.
02:28
2 t cubed minus 3 t plus 1.
02:31
Now let us differentiate this with respect to t.
02:34
We get 2 times 3 t squared minus 3 times 1 plus 0 which simplifies to 6 t squared minus 3.
02:44
So therefore this is the required answer for subpart b.
02:50
In the second problem we are given that w equals to x times y times x where x equals to s plus t y equals to s minus t and z equals to s t squared we are asked to find out do w over do s and do w over do t by making use of the chain rule in subpart a so first let us find out do w over do s this can be found out by making use of the chain rule which is do w over do x times do x over do s plus, sorry, this is x plus do w over do y times do y over du s plus do w over do z times do z over do t.
03:43
Now let us evaluate each of these partial derivatives.
03:48
We get the partial derivative of w with respect to x is y z and the partial derivative of x with respect to s is 1.
03:57
Partial derivative of w with respect to y is x z times the partial derivative of y with respect to s is 1 plus the partial derivative of w with respect to z is x y and the partial derivative of z with respect to s sorry this is s is t squared so now let us substitute the value of x y and z we have s minus t times s 2.
04:25
T squared plus s plus t times s t squared plus s plus t times s minus t times t squared...