00:01
In this problem, we need to find out the directional derivative.
00:03
We are given a function f of x, y, z which is equivalent to x, y, z and this could be written as x, y, z raised to the power of positive 1 divided by 2.
00:18
Now let us differentiate the function with respect to x and we obtain half of y, z times of x, y, z raised to the power of minus half.
00:30
Differential of the function with respect to y is equivalent to half times of x, z of x, y, z raised to the power of minus half and lastly we have to differentiate the function with respect to z and we obtain it as half times of x, y of x, y, z raised to the power of minus.
01:01
Next let us find out the gradient of f.
01:05
So, gradient of f is given as f of x, f of y, f of z.
01:14
So, we could basically write this as half of y, z times x, y, z raised to the power of minus half then we have half of x, z times x, y, z raised to the power of minus half and we have half of x, y times x, y, z raised to the power of minus half.
01:41
We need to find the directional derivative at the point p which is given to us as 3, 1, 3.
01:49
So, we can say that x is equivalent to 3, y equivalent to 1 and z is equivalent to 2.
01:58
So, let us substitute the value of x and y and z...