The problem asks us to find the first derivative of the function: $f(x) = (x^2 + 1)(4x + 1).$

To solve this, we can use the product rule for derivatives, which states: $(f \cdot g)' = f' \cdot g + f \cdot g',$ where $f(x) = x^2 + 1$ and $g(x) = 4x + 1$.

Step-by-Step Solution

1. Define the two parts:

   • $f(x) = x^2 + 1$
   • $g(x) = 4x + 1$

2. Compute $f'(x)$: $f'(x) = \frac{d}{dx}(x^2 + 1) = 2x.$

3. Compute $g'(x)$: $g'(x) = \frac{d}{dx}(4x + 1) = 4.$

4. Apply the Product Rule: $f'(x) = f'(x) \cdot g(x) + f(x) \cdot g'(x).$ Substituting the values, we get: $f'(x) = (2x)(4x + 1) + (x^2 + 1)(4).$

5. Expand and Simplify: $f'(x) = 2x \cdot 4x + 2x \cdot 1 + 4 \cdot x^2 + 4 \cdot 1.$ This simplifies to: $f'(x) = 8x^2 + 2x + 4x^2 + 4.$

6. Combine Like Terms: $f'(x) = 12x^2 + 2x + 4.$

Final Answer

The first derivative of $f(x)$ is: $f'(x) = 12x^2 + 2x + 4.$

Would you like more details on any part of this solution, or do you have any questions?

Here are some related questions for further practice:

1. How would you find the second derivative of $f(x)$?
2. What is the derivative of a function of the form $(ax + b)(cx + d)$?
3. How can the product rule be applied to three functions multiplied together?
4. What if $f(x) = (x^2 + 1)(4x + 1)^2$? How would the differentiation change?
5. How can we apply the product and chain rules together?

Tip: Always double-check each step in differentiation to ensure no terms are missed, especially when applying product and chain rules.