The derivative of f (x, y, z) at a point P is greatest in the direction of Ai+jk. In this direction the value of the derivative is 2√3.

We are given the following information:

1. The function is \( f(x, y, z) \).
2. The derivative is greatest in the direction of the vector \( \mathbf{A} = i + j + k \).
3. The value of the derivative in this direction is \( 2\sqrt{3} \).

To analyze this problem, we can use the concept of the gradient. The gradient of a scalar function \( f(x, y, z) \), denoted as \( \nabla f \), points in the direction of the greatest rate of increase of the function and its magnitude is the maximum rate of change.

### Steps to solve:
1. **Gradient relation**: 
   The directional derivative of a function \( f \) at a point in the direction of a unit vector \( \mathbf{u} \) is given by:
   \[
   D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u}
   \]
   where \( \mathbf{u} \) is the unit vector in the direction of \( \mathbf{A} \) (i.e., \( \mathbf{A} = i + j + k \)).

2. **Unit vector**: 
   First, we need to normalize \( \mathbf{A} = i + j + k \) to get the unit vector. The magnitude of \( \mathbf{A} \) is:
   \[
   |\mathbf{A}| = \sqrt{1^2 + 1^2 + 1^2} = \sqrt{3}
   \]
   So, the unit vector \( \mathbf{u} \) in the direction of \( \mathbf{A} \) is:
   \[
   \mathbf{u} = \frac{1}{\sqrt{3}}(i + j + k)
   \]

3. **Directional derivative**:
   The directional derivative in the direction of \( \mathbf{A} \) is given as \( 2\sqrt{3} \). From the formula of the directional derivative:
   \[
   D_{\mathbf{u}} f = \nabla f \cdot \mathbf{u}
   \]
   Substituting \( \mathbf{u} = \frac{1}{\sqrt{3}}(i + j + k) \) and knowing \( D_{\mathbf{u}} f = 2\sqrt{3} \), we get:
   \[
   \nabla f \cdot \frac{1}{\sqrt{3}}(i + j + k) = 2\sqrt{3}
   \]

4. **Dot product**:
   \[
   \frac{1}{\sqrt{3}} (\nabla f \cdot (i + j + k)) = 2\sqrt{3}
   \]
   Let \( \nabla f = (f_x, f_y, f_z) \). Then:
   \[
   \nabla f \cdot (i + j + k) = f_x + f_y + f_z
   \]
   So, the equation becomes:
   \[
   \frac{1}{\sqrt{3}} (f_x + f_y + f_z) = 2\sqrt{3}
   \]
   Multiplying both sides by \( \sqrt{3} \):
   \[
   f_x + f_y + f_z = 6
   \]

Thus, the sum of the partial derivatives of the function \( f(x, y, z) \) with respect to \( x \), \( y \), and \( z \) is \( 6 \).

Would you like further details on any of the steps or have any questions?

Here are some related questions to consider:

1. What is the meaning of the gradient in multivariable calculus?
2. How do you calculate the magnitude of a vector?
3. How is the directional derivative computed?
4. Why is the gradient important for understanding the behavior of functions?
5. How can you apply this knowledge to solve optimization problems?

**Tip**: When working with vectors, always remember to normalize them if the problem requires a unit vector.

The derivative of f (x, y, z) at a point P is greatest in the direction of Ai+jk. In this direction, the value of the derivative is 2√3.

Explore the solution to finding the directional derivative of the function f(x, y, z) in the direction of Ai+jk, where the derivative is 2√3. This problem involves key concepts of multivariable calculus, including gradients and unit vectors, suitable for undergraduate students.

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