Use the delta method to find the derivative of `f(x)=x^(4)`. Hence find `f^(')(-1/2)" and "f^(')(0)`.

Answer

To find the derivative of the function \( f(x) = x^4 \) using the delta method, we will follow these steps:

### Step 1: Define \( f(x) \) and \( f(x + \Delta x) \)
Let \( f(x) = x^4 \). Then, we can express \( f(x + \Delta x) \):
\[ f(x + \Delta x) = (x + \Delta x)^4 \]

### Step 2: Expand \( f(x + \Delta x) \)
Using the binomial theorem, we expand \( (x + \Delta x)^4 \):
\[ f(x + \Delta x) = x^4 + 4x^3 \Delta x + 6x^2 (\Delta x)^2 + 4x (\Delta x)^3 + (\Delta x)^4 \]

### Step 3: Calculate \( \Delta y \)
Now, we find \( \Delta y \):
\[ \Delta y = f(x + \Delta x) - f(x) = \left( x^4 + 4x^3 \Delta x + 6x^2 (\Delta x)^2 + 4x (\Delta x)^3 + (\Delta x)^4 \right) - x^4 \]
This simplifies to:
\[ \Delta y = 4x^3 \Delta x + 6x^2 (\Delta x)^2 + 4x (\Delta x)^3 + (\Delta x)^4 \]

### Step 4: Calculate \( \frac{\Delta y}{\Delta x} \)
Next, we find \( \frac{\Delta y}{\Delta x} \):
\[ \frac{\Delta y}{\Delta x} = \frac{4x^3 \Delta x + 6x^2 (\Delta x)^2 + 4x (\Delta x)^3 + (\Delta x)^4}{\Delta x} \]
This simplifies to:
\[ \frac{\Delta y}{\Delta x} = 4x^3 + 6x^2 \Delta x + 4x (\Delta x)^2 + (\Delta x)^3 \]

### Step 5: Take the limit as \( \Delta x \to 0 \)
Now, we find the derivative \( f'(x) \):
\[ f'(x) = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} = \lim_{\Delta x \to 0} \left( 4x^3 + 6x^2 \Delta x + 4x (\Delta x)^2 + (\Delta x)^3 \right) \]
As \( \Delta x \to 0 \), the terms involving \( \Delta x \) vanish:
\[ f'(x) = 4x^3 \]

### Step 6: Find \( f'(-\frac{1}{2}) \) and \( f'(0) \)
Now, we will calculate \( f'(-\frac{1}{2}) \) and \( f'(0) \):

1. **For \( f'(-\frac{1}{2}) \)**:
\[ f'(-\frac{1}{2}) = 4 \left(-\frac{1}{2}\right)^3 = 4 \left(-\frac{1}{8}\right) = -\frac{1}{2} \]

2. **For \( f'(0) \)**:
\[ f'(0) = 4(0)^3 = 0 \]

### Final Answers
Thus, the derivative of \( f(x) = x^4 \) is \( f'(x) = 4x^3 \), and the values are:
- \( f'(-\frac{1}{2}) = -\frac{1}{2} \)
- \( f'(0) = 0 \)