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The problem provided asks to calculate the **directional derivative** of the function \( f(x, y) = e^x + \ln(y) + x^2y^3 \) at the point \( (1, 1) \) in the direction of the vector \( \mathbf{v} = \langle -2, 1 \rangle \).

Let's break down the steps to solve this problem:

1. **Gradient of \( f(x, y) \)**:
   The directional derivative requires the gradient of the function, which is the vector of partial derivatives with respect to \( x \) and \( y \).

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

   - The partial derivative with respect to \( x \):
     \[
     \frac{\partial f}{\partial x} = e^x + 2xy^3
     \]
   - The partial derivative with respect to \( y \):
     \[
     \frac{\partial f}{\partial y} = \frac{1}{y} + 3x^2y^2
     \]

2. **Evaluate the gradient at the point \( (1, 1) \)**:
   We now substitute \( x = 1 \) and \( y = 1 \) into the gradient components:

   - \( \frac{\partial f}{\partial x} \Big|_{(1, 1)} = e^1 + 2 \cdot 1 \cdot 1^3 = e + 2 \)
   - \( \frac{\partial f}{\partial y} \Big|_{(1, 1)} = \frac{1}{1} + 3 \cdot 1^2 \cdot 1^2 = 1 + 3 = 4 \)

   Thus, the gradient vector at \( (1, 1) \) is:
   \[
   \nabla f(1, 1) = \langle e + 2, 4 \rangle
   \]

3. **Normalize the direction vector \( \mathbf{v} = \langle -2, 1 \rangle \)**:
   The directional derivative involves the unit vector in the direction of \( \mathbf{v} \). To normalize \( \mathbf{v} \), we find its magnitude:
   \[
   |\mathbf{v}| = \sqrt{(-2)^2 + 1^2} = \sqrt{4 + 1} = \sqrt{5}
   \]
   Therefore, the unit vector \( \mathbf{u} \) is:
   \[
   \mathbf{u} = \frac{1}{\sqrt{5}} \langle -2, 1 \rangle = \langle \frac{-2}{\sqrt{5}}, \frac{1}{\sqrt{5}} \rangle
   \]

4. **Directional Derivative**:
   The directional derivative is the dot product of the gradient and the unit vector in the direction of \( \mathbf{v} \):
   \[
   D_{\mathbf{v}} f(1, 1) = \nabla f(1, 1) \cdot \mathbf{u}
   \]
   Substituting the values:
   \[
   D_{\mathbf{v}} f(1, 1) = \langle e + 2, 4 \rangle \cdot \langle \frac{-2}{\sqrt{5}}, \frac{1}{\sqrt{5}} \rangle
   \]
   Performing the dot product:
   \[
   D_{\mathbf{v}} f(1, 1) = \frac{-2(e + 2)}{\sqrt{5}} + \frac{4}{\sqrt{5}} = \frac{-2e - 4 + 4}{\sqrt{5}} = \frac{-2e}{\sqrt{5}}
   \]

Thus, the directional derivative of \( f(x, y) \) at the point \( (1, 1) \) in the direction of \( \mathbf{v} = \langle -2, 1 \rangle \) is:
\[
D_{\mathbf{v}} f(1, 1) = \frac{-2e}{\sqrt{5}}
\]

Would you like further clarification on any step? 

Here are 5 questions related to this topic:

1. How is the gradient of a multivariable function related to the directional derivative?
2. Why do we need to normalize the direction vector when calculating the directional derivative?
3. How does the magnitude of the gradient affect the value of the directional derivative?
4. Can the directional derivative be negative? What does that indicate geometrically?
5. What is the geometric interpretation of the dot product in the context of directional derivatives?

**Tip**: The gradient vector points in the direction of the steepest ascent of a function, and its magnitude represents the rate of increase in that direction.

Consider the function f(x, y) = e^x + ln(y) + x^2y^3. The directional derivative of f at (1, 1) in the direction of the vector <-2, 1>.

Learn how to find the directional derivative of f(x, y) = e^x + ln(y) + x^2y^3 at the point (1, 1) in the direction of the vector <-2, 1>. This step-by-step solution covers gradient computation, vector normalization, and the directional derivative formula.

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