We are tasked with finding the derivative of the function:

$p(t) = \frac{(2t + 3)^3}{4t^2 - 1}$

This is a quotient of two functions, so we will use the quotient rule. The quotient rule states:

$\left(\frac{f(t)}{g(t)}\right)' = \frac{f'(t)g(t) - f(t)g'(t)}{g(t)^2}$

Here, $f(t) = (2t + 3)^3$ and $g(t) = 4t^2 - 1$.

Step 1: Differentiate $f(t) = (2t + 3)^3$

We apply the chain rule to differentiate $f(t)$.

First, differentiate the outer function:

$f'(t) = 3(2t + 3)^2 \cdot \frac{d}{dt}(2t + 3)$

Now, differentiate the inner function $2t + 3$:

$\frac{d}{dt}(2t + 3) = 2$

So, we have:

$f'(t) = 3(2t + 3)^2 \cdot 2 = 6(2t + 3)^2$

Step 2: Differentiate $g(t) = 4t^2 - 1$

Now, let's differentiate $g(t)$:

$g'(t) = \frac{d}{dt}(4t^2 - 1) = 8t$

Step 3: Apply the Quotient Rule

Using the quotient rule formula, we substitute:

$p'(t) = \frac{f'(t)g(t) - f(t)g'(t)}{g(t)^2}$

Substitute the values of $f(t)$, $f'(t)$, $g(t)$, and $g'(t)$:

$p'(t) = \frac{6(2t + 3)^2(4t^2 - 1) - (2t + 3)^3(8t)}{(4t^2 - 1)^2}$

Step 4: Simplify the Expression

The derivative $p'(t)$ is:

$p'(t) = \frac{6(2t + 3)^2(4t^2 - 1) - 8t(2t + 3)^3}{(4t^2 - 1)^2}$

This is the derivative of the given function, and it can be further simplified by expanding the terms in the numerator if needed.

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Would you like further simplification or details on any step?

Related Questions:

1. What is the chain rule and how is it applied in calculus?
2. Can you provide more examples of using the quotient rule in derivatives?
3. How does one simplify complex rational expressions after applying the quotient rule?
4. Can you show an example of a product rule applied to a function similar to this one?
5. What is the geometric interpretation of a derivative for rational functions?

Tip:

Always verify whether the expression can be simplified after applying the derivative rules to make further calculations easier.