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To solve this problem, we need to find both \(\Delta y\) and \(dy\) for the given function \(y = x^4 + 4\) at \(x = -3\) with a differential \(\Delta x = dx = 0.01\).

### Step 1: Find \(\Delta y\)
\(\Delta y\) represents the change in the function value when \(x\) changes from \(-3\) to \(-3 + 0.01\). We calculate it as:
\[
\Delta y = f(x + \Delta x) - f(x)
\]
\[
f(x) = (-3)^4 + 4 = 81 + 4 = 85
\]
\[
f(x + \Delta x) = (-3 + 0.01)^4 + 4 = (-2.99)^4 + 4
\]
Now, compute \((-2.99)^4\):
\[
(-2.99)^4 \approx 79.8801
\]
\[
f(-2.99) = 79.8801 + 4 = 83.8801
\]
Thus:
\[
\Delta y = 83.8801 - 85 = -1.1199
\]

### Step 2: Find \(dy\)
\(dy\) represents the differential approximation using the derivative:
\[
dy = f'(x) \cdot dx
\]
First, find \(f'(x)\):
\[
f(x) = x^4 + 4 \implies f'(x) = 4x^3
\]
At \(x = -3\):
\[
f'(-3) = 4(-3)^3 = 4(-27) = -108
\]
Now calculate \(dy\):
\[
dy = -108 \cdot 0.01 = -1.08
\]

### Final Answers:
\[
\Delta y \approx -1.1199
\]
\[
dy = -1.08
\]

Would you like more details or have any questions?

Here are 5 related questions you might explore next:
1. What is the difference between \(\Delta y\) and \(dy\) conceptually?
2. How would \(\Delta y\) change if \(\Delta x\) were smaller, like \(0.001\)?
3. How does the accuracy of \(dy\) as an approximation of \(\Delta y\) vary with different functions?
4. What is the physical interpretation of differentials in calculus?
5. How can Taylor series be used to approximate \(\Delta y\) for higher precision?

**Tip:** Remember, \(dy\) is often used as an approximation of \(\Delta y\) when \(\Delta x\) is very small, but they are not exactly equal!

Use the information to find and compare Δy and dy for the function y = x^4 + 4 at x = -3 with Δx = 0.01.

Explore how to calculate Δy and dy for the function y = x^4 + 4 at x = -3 with a small change Δx = 0.01. Learn the difference between the exact change in the function value and the differential approximation using calculus.

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