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The problem involves calculating both the change in \( y \) (\( \Delta y \)) and the differential of \( y \) (\( dy \)) for a given function \( y = 4\sqrt{x} \).

### Part 1: Find the change in \( y \), \( \Delta y \)

We are given \( y = 4\sqrt{x} \), \( x = 3 \), and \( \Delta x = 0.4 \). The change in \( y \), \( \Delta y \), is defined as:

\[
\Delta y = y(x + \Delta x) - y(x)
\]

1. First, compute \( y(3) \):
\[
y(3) = 4\sqrt{3} \approx 4(1.732) = 6.928
\]

2. Then compute \( y(3 + 0.4) = y(3.4) \):
\[
y(3.4) = 4\sqrt{3.4} \approx 4(1.844) = 7.376
\]

3. Now, calculate \( \Delta y \):
\[
\Delta y = y(3.4) - y(3) \approx 7.376 - 6.928 = 0.448
\]

So, \( \Delta y \approx 0.448 \).

### Part 2: Find the differential \( dy \)

The differential \( dy \) is defined as:
\[
dy = \frac{dy}{dx} \cdot dx
\]

1. First, find the derivative of \( y = 4\sqrt{x} \):
\[
\frac{dy}{dx} = \frac{4}{2\sqrt{x}} = \frac{2}{\sqrt{x}}
\]

2. Evaluate the derivative at \( x = 3 \):
\[
\frac{dy}{dx}\bigg|_{x=3} = \frac{2}{\sqrt{3}} \approx \frac{2}{1.732} = 1.154
\]

3. Now, compute \( dy \) using \( dx = 0.4 \):
\[
dy = 1.154 \cdot 0.4 = 0.462
\]

So, \( dy \approx 0.462 \).

### Summary:
- \( \Delta y \approx 0.448 \)
- \( dy \approx 0.462 \)

Would you like more details on how these results were obtained? Here are 5 follow-up questions you might find helpful:
1. What is the difference between the actual change \( \Delta y \) and the differential \( dy \)?
2. How does the approximation \( dy \) compare to \( \Delta y \) when \( \Delta x \) is small?
3. How is the concept of differentials used in real-world applications?
4. Can you find the second derivative of the function \( y = 4\sqrt{x} \)?
5. How would the results change if \( \Delta x \) were smaller, say \( 0.1 \)?

**Tip**: Remember, \( dy \) approximates \( \Delta y \) for small \( \Delta x \). This is because \( dy \) represents the linear approximation of the function change at a specific point.

Let y = 4√x. Find the change in y, Δy when x = 3 and Δx = 0.4. Find the differential dy when x = 3 and dx = 0.4.

Learn how to compute the change in y (Δy) and the differential dy for the function y = 4√x at x = 3 with a step-by-step explanation. This problem explores key calculus concepts such as differentiation and approximation techniques. Suitable for high school calculus students.

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