To differentiate $f(x) = \frac{x^3 + 1}{4x^2 + 1}$ using the quotient rule, we apply the formula:

$\frac{d}{dx} \left( \frac{u(x)}{v(x)} \right) = \frac{v(x) \cdot u'(x) - u(x) \cdot v'(x)}{(v(x))^2}$

Where:

• $u(x) = x^3 + 1$
• $v(x) = 4x^2 + 1$

Step 1: Differentiate the numerator $u(x)$

$u'(x) = \frac{d}{dx} (x^3 + 1) = 3x^2$

Step 2: Differentiate the denominator $v(x)$

$v'(x) = \frac{d}{dx} (4x^2 + 1) = 8x$

Step 3: Apply the quotient rule

Now plug everything into the quotient rule formula:

$f'(x) = \frac{(4x^2 + 1)(3x^2) - (x^3 + 1)(8x)}{(4x^2 + 1)^2}$

Step 4: Simplify the numerator

First, expand both terms in the numerator:

$(4x^2 + 1)(3x^2) = 12x^4 + 3x^2$ $(x^3 + 1)(8x) = 8x^4 + 8x$

So the numerator becomes: $(12x^4 + 3x^2) - (8x^4 + 8x) = 12x^4 + 3x^2 - 8x^4 - 8x = 4x^4 + 3x^2 - 8x$

Step 5: Write the final result

The derivative of $f(x)$ is:

$f'(x) = \frac{4x^4 + 3x^2 - 8x}{(4x^2 + 1)^2}$

Let me know if you'd like further details or clarification!

Related questions:

1. How would the quotient rule change if $f(x)$ had more complicated polynomials?
2. Can you show how to use the product rule in combination with the quotient rule?
3. What happens if the denominator has a higher degree than the numerator?
4. How would you compute the second derivative of $f(x)$?
5. What is the physical interpretation of the derivative of this function?

Tip: Always simplify the numerator carefully after applying the quotient rule, as this reduces the complexity of the derivative expression.