00:01
So in this problem, we are given a function of two variables, f of x comma y is equal to x times y times e to the power of x minus y.
00:12
We're also given a point, let's call the point p, which is 5 comma 5.
00:16
Next, we're going to start by finding the gradient of the function.
00:19
The gradient of the function is the vector f of x comma f of y.
00:23
So in our case, f of x, by using the product rule, we're going to have y times e to the power of x minus y, then plus the root of the second times the first, so that's going to be xy times e to the power of x minus y comma, and then the derivative of f, with respect to y.
00:42
That's going to be the derivative first, which is x times the second e to the power of x minus y, and plus the second times the first as negative xy times e to the power of x minus y.
00:53
We're going to evaluate this gradient of f at the point p, which is going to equal to replace y and x, both of them with 5, so we're going to have 5 times e to the 0, which is 1, plus 25 times e to the 0 comma 5 minus 25...