To find the gradient of the function $f(x, y, z) = x^2 + y^2 - 2z^2 + z \ln(x)$, we will compute the partial derivatives with respect to $x$, $y$, and $z$, and then evaluate them at the point $(1, 1, 1)$.

Step 1: Compute the partial derivatives

1. Partial derivative with respect to $x$:

$\frac{\partial f}{\partial x} = \frac{\partial}{\partial x} \left( x^2 + y^2 - 2z^2 + z \ln(x) \right)$

The derivatives of the individual terms:

• The derivative of $x^2$ is $2x$.
• The derivative of $y^2$ is 0 (as it doesn't depend on $x$).
• The derivative of $-2z^2$ is 0 (as it doesn't depend on $x$).
• The derivative of $z \ln(x)$ is $z \cdot \frac{1}{x}$, by the chain rule.

Thus, we have:

$\frac{\partial f}{\partial x} = 2x + \frac{z}{x}$

2. Partial derivative with respect to $y$:

$\frac{\partial f}{\partial y} = \frac{\partial}{\partial y} \left( x^2 + y^2 - 2z^2 + z \ln(x) \right)$

• The derivative of $x^2$ is 0.
• The derivative of $y^2$ is $2y$.
• The derivative of $-2z^2$ is 0.
• The derivative of $z \ln(x)$ is 0 (as it doesn't depend on $y$).

Thus, we have:

$\frac{\partial f}{\partial y} = 2y$

3. Partial derivative with respect to $z$:

$\frac{\partial f}{\partial z} = \frac{\partial}{\partial z} \left( x^2 + y^2 - 2z^2 + z \ln(x) \right)$

• The derivative of $x^2$ is 0.
• The derivative of $y^2$ is 0.
• The derivative of $-2z^2$ is $-4z$.
• The derivative of $z \ln(x)$ is $\ln(x)$ (using the product rule).

Thus, we have:

$\frac{\partial f}{\partial z} = -4z + \ln(x)$

Step 2: Evaluate the gradient at the point $(1, 1, 1)$

Now, let's evaluate the partial derivatives at the point $(x, y, z) = (1, 1, 1)$.

1. For $\frac{\partial f}{\partial x}$ at $(1, 1, 1)$:

$\frac{\partial f}{\partial x} = 2(1) + \frac{1}{1} = 2 + 1 = 3$

2. For $\frac{\partial f}{\partial y}$ at $(1, 1, 1)$:

$\frac{\partial f}{\partial y} = 2(1) = 2$

3. For $\frac{\partial f}{\partial z}$ at $(1, 1, 1)$:

$\frac{\partial f}{\partial z} = -4(1) + \ln(1) = -4 + 0 = -4$

Step 3: Gradient at $(1, 1, 1)$

The gradient of $f(x, y, z)$ at the point $(1, 1, 1)$ is the vector of these partial derivatives:

$\nabla f(1, 1, 1) = \left( \frac{\partial f}{\partial x}, \frac{\partial f}{\partial y}, \frac{\partial f}{\partial z} \right) = (3, 2, -4)$

Thus, the gradient of the function at the point $(1, 1, 1)$ is:

$\nabla f(1, 1, 1) = (3, 2, -4)$

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

Here are some related questions you might find interesting:

1. How does the gradient relate to the direction of the steepest ascent of a function?
2. What is the significance of partial derivatives in multivariable calculus?
3. How do you compute the gradient for a vector field?
4. What are some applications of the gradient in physics and engineering?
5. How can the gradient be used to find local maxima or minima of a function?

Tip: The gradient of a function gives you the direction of the steepest increase of the function at any given point. The magnitude of the gradient tells you how steep the slope is in that direction.