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The problem asks to find the directional derivative \( D_uF(1,2) \) of the function \( F(x,y) = x^3 - 3xy + \ln(y) \), where the unit vector \( \mathbf{u} \) is given by the angle \( \theta = \frac{\pi}{6} \).

### Solution Steps:

#### 1. **Calculate the Gradient of \( F(x, y) \):**

The gradient \( \nabla F(x, y) \) is a vector of partial derivatives:

\[
\nabla F(x, y) = \left( \frac{\partial F}{\partial x}, \frac{\partial F}{\partial y} \right)
\]

- **Partial derivative with respect to \( x \):**
  
  \[
  \frac{\partial F}{\partial x} = 3x^2 - 3y
  \]

- **Partial derivative with respect to \( y \):**

  \[
  \frac{\partial F}{\partial y} = -3x + \frac{1}{y}
  \]

Thus, the gradient is:

\[
\nabla F(x, y) = \left( 3x^2 - 3y, -3x + \frac{1}{y} \right)
\]

#### 2. **Evaluate the Gradient at \( (1, 2) \):**

Substitute \( x = 1 \) and \( y = 2 \) into the gradient:

\[
\nabla F(1, 2) = \left( 3(1)^2 - 3(2), -3(1) + \frac{1}{2} \right)
\]

\[
\nabla F(1, 2) = (3 - 6, -3 + 0.5) = (-3, -2.5)
\]

#### 3. **Unit Vector \( \mathbf{u} \):**

The unit vector \( \mathbf{u} \) is given by the angle \( \theta = \frac{\pi}{6} \). The components of \( \mathbf{u} \) are:

\[
\mathbf{u} = (\cos\theta, \sin\theta) = \left( \cos\frac{\pi}{6}, \sin\frac{\pi}{6} \right) = \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right)
\]

#### 4. **Compute the Directional Derivative:**

The directional derivative is given by:

\[
D_uF(1, 2) = \nabla F(1, 2) \cdot \mathbf{u}
\]

\[
D_uF(1, 2) = (-3, -2.5) \cdot \left( \frac{\sqrt{3}}{2}, \frac{1}{2} \right)
\]

Perform the dot product:

\[
D_uF(1, 2) = (-3) \cdot \frac{\sqrt{3}}{2} + (-2.5) \cdot \frac{1}{2}
\]

\[
D_uF(1, 2) = -\frac{3\sqrt{3}}{2} - \frac{2.5}{2}
\]

\[
D_uF(1, 2) = -\frac{3\sqrt{3}}{2} - 1.25
\]

So, the directional derivative is:

\[
D_uF(1, 2) = -\frac{3\sqrt{3}}{2} - 1.25
\]

This is the value of the directional derivative in the direction of \( \mathbf{u} \) at the point \( (1, 2) \).

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Do you need more details or further clarification?

Here are five related questions:
1. How do you compute a gradient for multivariable functions?
2. What is the physical meaning of a directional derivative?
3. How can you verify if a vector is a unit vector?
4. How does the angle of the unit vector affect the directional derivative?
5. Can the directional derivative be zero? Under what conditions?

**Tip:** The gradient points in the direction of the greatest rate of increase of the function, and its magnitude gives the rate of that increase.

Find the directional derivative DuF(1, 2) of the function F(x, y) = x^3 - 3xy + ln(y), where u is a unit vector given by angle θ = π/6.

Learn how to find the directional derivative of the multivariable function F(x, y) = x^3 - 3xy + ln(y) at the point (1, 2) with respect to a unit vector given by the angle π/6. This solution involves calculating the gradient and performing the dot product with the unit vector.

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