00:01
In this question, we're going to compute the directional derivative of the following function at the given point p in the direction of the given vector.
00:07
I'm going to be sure that i use a unit vector for the direction vector.
00:12
So the first thing i do is i find the gradient of my function, f of x, y, which is the ln of 3 plus x squared plus y squared.
00:22
So we're going to have as our first component 1 over 3 plus x squared plus y squared times the derivative of the inside, which is 2x.
00:38
And then for my second component, it's 1 over 3 plus x squared plus y squared times the derivative of the inside, now with respect to y, which is 2y.
00:51
So it's my partial with respect to x in the first component.
00:55
My partial with respect to y in the second component.
00:59
So what i'm getting for my gradient is the vector 2x over 3 plus x squared plus y squared.
01:08
And then 2y over 3 plus x squared plus y squared.
01:16
That's what's going to go in here.
01:20
Now let's evaluate that at the point that i was given, 1, negative 1.
01:27
So let's see.
01:31
Up top, i'm getting 2 times 1 is 2...