Find the derivatives of the following at any point of their domains :
`f(x)=(ax^(2)+b)^(2)`

Answer

To find the derivative of the function \( f(x) = (ax^2 + b)^2 \) at any point in its domain, we will use the chain rule of differentiation. Here are the steps:

### Step 1: Identify the outer and inner functions
In the function \( f(x) = (ax^2 + b)^2 \), we can identify:
- The outer function \( u^2 \) where \( u = ax^2 + b \)
- The inner function \( u = ax^2 + b \)

### Step 2: Apply the chain rule
According to the chain rule, the derivative of \( f(x) \) can be expressed as:
\[ f'(x) = \frac{d}{dx}(u^2) \cdot \frac{du}{dx} \]
where \( \frac{d}{dx}(u^2) = 2u \) and \( \frac{du}{dx} = \frac{d}{dx}(ax^2 + b) \).

### Step 3: Differentiate the inner function
Now, we differentiate the inner function:
\[ \frac{du}{dx} = \frac{d}{dx}(ax^2 + b) = 2ax + 0 = 2ax \]

### Step 4: Substitute back into the chain rule
Now substituting back into the chain rule:
\[ f'(x) = 2u \cdot \frac{du}{dx} = 2(ax^2 + b) \cdot (2ax) \]

### Step 5: Simplify the expression
Now we simplify the expression:
\[ f'(x) = 2(ax^2 + b)(2ax) = 4ax(ax^2 + b) \]

### Final Answer
Thus, the derivative of the function \( f(x) = (ax^2 + b)^2 \) is:
\[ f'(x) = 4ax(ax^2 + b) \]

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