Question

Find the directional derivative of the function if f(x, y, z) = 3e^x cos(yz) at point (0, 0, 0) in the direction of the vector (V = 2i + j - 2k).

Answer 1

The directional derivative of the given function at point (0, 0, 0) in the given direction is 6.

Step-by-step explanation:

To find the directional derivative of the function f(x, y, z) = 3eˣ cos(yz) at point (0, 0, 0) in the direction of the vector V = 2i + j - 2k, we can use the formula:

Directional Derivative = ∇f · V

Where ∇f is the gradient of the function f, and · represents the dot product.

In this case, the gradient of f(x, y, z) = 3eˣ cos(yz) is:

∇f = (∂f/∂x)î + (∂f/∂y)ĵ + (∂f/∂z)k

Calculating the partial derivatives and substituting the values at (0, 0, 0), we get:

∇f = (3 cos(0))î + (-3eˣ z sin(yz) + 0)ĵ + (-3eˣ y sin(yz) + 0)k

Now, we can take the dot product with the direction vector V = 2i + j - 2k:

∇f · V = (3 cos(0))(2) + 0 + 0 = 6

So, the directional derivative of the function f(x, y, z) = 3eˣ cos(yz) at point (0, 0, 0) in the direction of the vector V = 2i + j - 2k is 6.