The given function is:

$y = \frac{5}{4x^3 + 4}.$

We will use the quotient rule to differentiate this function. Recall that the quotient rule is given as:

$\frac{d}{dx} \left( \frac{f(x)}{g(x)} \right) = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2},$

where $f(x)$ is the numerator and $g(x)$ is the denominator.

Here:

• $f(x) = 5$ (a constant, so $f'(x) = 0$),
• $g(x) = 4x^3 + 4$ (its derivative is $g'(x) = 12x^2$).

Derivative Calculation:

Substitute into the quotient rule:

$\frac{dy}{dx} = \frac{f'(x)g(x) - f(x)g'(x)}{[g(x)]^2}.$

1. Compute $f'(x)g(x)$: Since $f'(x) = 0$, this term becomes $0 \cdot (4x^3 + 4) = 0$.
2. Compute $f(x)g'(x)$: $5 \cdot 12x^2 = 60x^2$.
3. Substitute into the formula:

$\frac{dy}{dx} = \frac{0 - 60x^2}{(4x^3 + 4)^2}.$

This simplifies to:

$\frac{dy}{dx} = \frac{-60x^2}{(4x^3 + 4)^2}.$

Final Answer:

$\boxed{\frac{dy}{dx} = \frac{-60x^2}{(4x^3 + 4)^2}}.$

Let me know if you would like a deeper explanation of any step or further practice problems! 😊

Related Questions:

1. How do you apply the quotient rule to other functions with constants in the numerator?
2. What is the derivative of $y = \frac{1}{4x^3 + 4}$?
3. How can we rewrite $y$ as a product and apply the product rule instead?
4. What happens to the derivative if $f(x)$ or $g(x)$ becomes zero at a point?
5. Can we simplify this expression further by factoring the denominator?

Tip:

Always simplify the numerator and denominator separately before substituting into the quotient rule to avoid mistakes!