We are tasked with differentiating the function $f(x) = \frac{x^3 + 1}{4x^2 + 1}$ using the quotient rule.

The Quotient Rule:

For two functions $u(x)$ and $v(x)$, where $f(x) = \frac{u(x)}{v(x)}$, the quotient rule is given by: $f'(x) = \frac{v(x)u'(x) - u(x)v'(x)}{(v(x))^2}$ In this case:

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

Now, we calculate the derivatives of $u(x)$ and $v(x)$:

1. $u'(x) = \frac{d}{dx}(x^3 + 1) = 3x^2$
2. $v'(x) = \frac{d}{dx}(4x^2 + 1) = 8x$

Apply the Quotient Rule:

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

Now, simplify the numerator: $(4x^2 + 1)(3x^2) = 12x^4 + 3x^2$ $(x^3 + 1)(8x) = 8x^4 + 8x$ Thus, the numerator becomes: $(12x^4 + 3x^2) - (8x^4 + 8x) = 4x^4 + 3x^2 - 8x$

So, the derivative is: $f'(x) = \frac{4x^4 + 3x^2 - 8x}{(4x^2 + 1)^2}$

This is the simplified form of the derivative.

Would you like further clarification or details on any part of this solution?

Related questions:

1. How does the quotient rule differ from the product rule?
2. Can we simplify the quotient rule further in specific cases?
3. What happens when $v(x) = 1$? How does the quotient rule reduce?
4. How does the chain rule apply in more complex functions?
5. Can you differentiate a more complex rational function using the quotient rule?

Tip: Always simplify the terms inside the numerator before attempting to cancel terms when using the quotient rule!