Given `f(x)=ax^(2)`, where 'a' is a constant, find `f^(')(x)` by the delta method. Hence find `f^(')(2)`.

Answer

To find the derivative of the function \( f(x) = ax^2 \) using the delta method, we will follow these steps:

### Step 1: Define the function and its increment
Let \( f(x) = ax^2 \). We will consider a small increment \( \Delta x \) in \( x \), which gives us a new point \( x + \Delta x \). The corresponding function value at this new point is:
\[ f(x + \Delta x) = a(x + \Delta x)^2 \]

### Step 2: Expand the function at the new point
Now, we expand \( f(x + \Delta x) \):
\[ f(x + \Delta x) = a((x + \Delta x)^2) = a(x^2 + 2x\Delta x + (\Delta x)^2) = ax^2 + 2ax\Delta x + a(\Delta x)^2 \]

### Step 3: Calculate the change in the function
The change in the function value, denoted as \( \Delta y \), is:
\[ \Delta y = f(x + \Delta x) - f(x) = (ax^2 + 2ax\Delta x + a(\Delta x)^2) - ax^2 \]
This simplifies to:
\[ \Delta y = 2ax\Delta x + a(\Delta x)^2 \]

### Step 4: Formulate the derivative
Now, we can express the derivative \( f'(x) \) using the definition of the derivative:
\[ f'(x) = \lim_{\Delta x \to 0} \frac{\Delta y}{\Delta x} \]
Substituting \( \Delta y \):
\[ f'(x) = \lim_{\Delta x \to 0} \frac{2ax\Delta x + a(\Delta x)^2}{\Delta x} \]
We can factor out \( \Delta x \) from the numerator:
\[ f'(x) = \lim_{\Delta x \to 0} \left( 2ax + a\Delta x \right) \]

### Step 5: Evaluate the limit
As \( \Delta x \) approaches 0, the term \( a\Delta x \) approaches 0:
\[ f'(x) = 2ax \]

### Step 6: Find \( f'(2) \)
Now, we need to find \( f'(2) \):
\[ f'(2) = 2a(2) = 4a \]

### Final Result
Thus, the derivative \( f'(x) \) is \( 2ax \) and \( f'(2) \) is \( 4a \).

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