00:01
For this problem, we are given the function g of x, y, z equals x plus 2y plus 3z to the power of 3 over 2, the point 112, and the vector, so actually i should label that as v, not u, we have the vector 2j minus k.
00:15
We are asked to find the directional derivative of g in the direction of v at the point 112.
00:22
So our first step here is to find the unit vector, which i'm going to call you, in the direction of v.
00:28
So that will be v divided by its magnitude.
00:31
So it will be v over the square root of 2 squared, so 4 plus 1.
00:36
So it will be v over root 5, or just that would be just 2 over root 5j minus 1 over root 5k.
00:50
Having that, we then want to find the gradient of our function, which is g here, at point x, y, z.
01:00
So we'll have to apply the chain rule here.
01:04
We'll have 3 over 2 times the derivative, well, in this case for our x derivative, at least, we don't need to worry about chain rule.
01:15
Derivative of x with respect to x is just 1.
01:18
So we'll have 3 over 2 times x plus 2y plus 3z to the power of 3 .2 now, it's just to the power of 1⁄2...