To differentiate the function $f(x) = \frac{x^3 + 1}{4x^2 + 1}$ using the quotient rule, we apply the formula for the derivative of a quotient of two functions:

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

Here:

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

Step 1: Differentiate $u(x)$ and $v(x)$

Now we find the derivatives of $u(x)$ and $v(x)$.

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

Step 2: Apply the quotient rule formula

Now substitute these values into the quotient rule formula:

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

Step 3: Simplify the numerator

Expand both terms in the numerator:

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

Now, subtract the second term from the first:

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

Step 4: Final derivative

Thus, the derivative of the function is:

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

This is the simplified expression for the derivative.

Would you like more details or have any questions on this topic?

Here are 5 related questions:

1. How does the quotient rule differ from the product rule?
2. Can the quotient rule be applied to functions with more complex numerators and denominators?
3. What happens when the denominator of a quotient function is a constant?
4. How would you apply the quotient rule to trigonometric functions?
5. How does the chain rule integrate with the quotient rule for nested functions?

Tip: Always simplify the expression for the numerator after applying the quotient rule to avoid errors in the final derivative expression.