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We are tasked with finding the directional derivative of the function \( f(x, y) = 4x^2 + y^2 \) at the point \( (-1, 2) \) in the direction pointing to the origin. Let's go through the steps.

### Step 1: Gradient of the function
The directional derivative requires the gradient \( \nabla f(x, y) \), which is the vector of partial derivatives:

\[
f(x, y) = 4x^2 + y^2
\]
- \( f_x(x, y) = \frac{\partial}{\partial x}(4x^2 + y^2) = 8x \)
- \( f_y(x, y) = \frac{\partial}{\partial y}(4x^2 + y^2) = 2y \)

Thus, the gradient is:
\[
\nabla f(x, y) = (8x, 2y)
\]

### Step 2: Compute the gradient at the point \((-1, 2)\)
Substitute \( x = -1 \) and \( y = 2 \) into the gradient:
\[
\nabla f(-1, 2) = (8(-1), 2(2)) = (-8, 4)
\]

### Step 3: Direction vector
We are given that the direction points towards the origin from the point \((-1, 2)\). The vector pointing to the origin from this point is:
\[
\mathbf{v} = (0, 0) - (-1, 2) = (1, -2)
\]

### Step 4: Normalize the direction vector
To use this direction, we need to normalize it. The magnitude of \( \mathbf{v} = (1, -2) \) is:
\[
|\mathbf{v}| = \sqrt{1^2 + (-2)^2} = \sqrt{1 + 4} = \sqrt{5}
\]
So, the unit vector in the direction of \( \mathbf{v} \) is:
\[
\hat{\mathbf{v}} = \left( \frac{1}{\sqrt{5}}, \frac{-2}{\sqrt{5}} \right)
\]

### Step 5: Compute the directional derivative
The directional derivative \( D_{\hat{\mathbf{v}}} f(-1, 2) \) is the dot product of the gradient and the unit vector:
\[
D_{\hat{\mathbf{v}}} f(-1, 2) = \nabla f(-1, 2) \cdot \hat{\mathbf{v}} = (-8, 4) \cdot \left( \frac{1}{\sqrt{5}}, \frac{-2}{\sqrt{5}} \right)
\]
This gives:
\[
D_{\hat{\mathbf{v}}} f(-1, 2) = -8 \left( \frac{1}{\sqrt{5}} \right) + 4 \left( \frac{-2}{\sqrt{5}} \right)
\]
\[
= \frac{-8}{\sqrt{5}} + \frac{-8}{\sqrt{5}} = \frac{-16}{\sqrt{5}}
\]

### Step 6: Conclusion
Thus, the directional derivative is \( \frac{-16}{\sqrt{5}} \). This matches option **B**.

Would you like further clarification or details? Here are some related questions:

1. What is the meaning of a directional derivative geometrically?
2. How would the directional derivative change if the point was further from the origin?
3. How can you interpret the gradient vector in the context of this function?
4. Could we have solved this without normalizing the vector?
5. How would the calculation change if the function had a \(xy\) term?

**Tip**: The directional derivative gives the rate of change of the function in the direction of a given vector, so normalizing that vector is essential to get the correct rate.

Find the directional derivative of the function f(x, y) = 4x^2 + y^2 at the point (-1, 2) in the direction pointing to the origin.

Learn how to calculate the directional derivative of f(x, y) = 4x^2 + y^2 at the point (-1, 2) in the direction towards the origin. This step-by-step guide includes the gradient calculation and normalizing vectors, essential for college-level calculus students.

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