We are given the function:

$g(t) = \frac{1}{(2t + 1)^2}$

To find the derivative $g'(t)$, we can use the chain rule combined with the power rule. Here is the step-by-step process:

Step 1: Rewrite the function

We can rewrite the function in a more convenient form for differentiation:

$g(t) = (2t + 1)^{-2}$

Step 2: Apply the chain rule

Using the chain rule, the derivative of $(2t + 1)^{-2}$ with respect to $t$ is:

$g'(t) = -2(2t + 1)^{-3} \cdot \frac{d}{dt}(2t + 1)$

Step 3: Differentiate the inner function

The derivative of $2t + 1$ with respect to $t$ is simply $2$. So the derivative becomes:

$g'(t) = -2(2t + 1)^{-3} \cdot 2$

Step 4: Simplify the result

Simplifying the expression:

$g'(t) = -4(2t + 1)^{-3}$

Thus, the derivative of the given function is:

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

Would you like any further clarification or more details?

Here are 5 related questions to deepen your understanding:

1. How would you compute the second derivative of $g(t)$?
2. How does the chain rule work for more complex functions, like compositions of three functions?
3. Can you apply this same process to differentiate other rational functions?
4. How does the derivative of the function behave as $t \to \infty$?
5. What happens to $g'(t)$ when $t = 0$?

Tip: When differentiating functions involving powers of linear expressions, it's often helpful to rewrite them as exponents to make applying the chain rule easier.