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Let's solve each derivative step by step:

### 1. \( z(t) = \ln\left((e^t + 2)^2\right) \)

To find the derivative, we will use the chain rule:
\[
\frac{d}{dt} \ln(u) = \frac{1}{u} \frac{du}{dt}
\]
where \( u = (e^t + 2)^2 \). We also apply the chain rule again to differentiate \( u \).

#### Step 1: Differentiate the outer function
\[
z'(t) = \frac{1}{(e^t + 2)^2} \cdot \frac{d}{dt} \left((e^t + 2)^2\right)
\]

#### Step 2: Differentiate the inner function
Using the chain rule for \( (e^t + 2)^2 \), we get:
\[
\frac{d}{dt} (e^t + 2)^2 = 2(e^t + 2) \cdot \frac{d}{dt} (e^t + 2)
\]
and
\[
\frac{d}{dt} (e^t + 2) = e^t
\]

Thus,
\[
\frac{d}{dt} (e^t + 2)^2 = 2(e^t + 2) \cdot e^t
\]

#### Step 3: Combine the results
\[
z'(t) = \frac{2(e^t + 2) \cdot e^t}{(e^t + 2)^2}
\]

Simplifying,
\[
z'(t) = \frac{2e^t}{e^t + 2}
\]

### 2. \( z(t) = \frac{1}{\left(e^{t^3 + t} + 3t\right)^3} \)

Rewrite \( z(t) \) as:
\[
z(t) = \left(e^{t^3 + t} + 3t\right)^{-3}
\]

Now, use the chain rule and power rule to differentiate:
\[
z'(t) = -3\left(e^{t^3 + t} + 3t\right)^{-4} \cdot \frac{d}{dt} \left(e^{t^3 + t} + 3t\right)
\]

#### Step 1: Differentiate the inner function
We need to differentiate \( e^{t^3 + t} + 3t \).

The derivative of \( e^{t^3 + t} \) is:
\[
\frac{d}{dt} e^{t^3 + t} = e^{t^3 + t} \cdot (3t^2 + 1)
\]

And the derivative of \( 3t \) is:
\[
\frac{d}{dt} 3t = 3
\]

Thus,
\[
\frac{d}{dt} \left(e^{t^3 + t} + 3t\right) = e^{t^3 + t} \cdot (3t^2 + 1) + 3
\]

#### Step 2: Combine the results
\[
z'(t) = -3\left(e^{t^3 + t} + 3t\right)^{-4} \left(e^{t^3 + t} \cdot (3t^2 + 1) + 3\right)
\]

Would you like further details on any of these steps or have any questions?

### Related Questions:
1. How can we apply the product rule to derivatives involving multiple functions?
2. What are the general steps for finding the derivative of a logarithmic function?
3. How does the chain rule apply when differentiating composite functions?
4. What are the specific rules for differentiating exponential functions?
5. How would the process change if we were working with partial derivatives?

### Tip:
Always simplify expressions before differentiating if possible; it can make the process easier and reduce the chance of errors.

Compute the derivatives of the following functions: z(t) = ln((e^t + 2)^2) and z(t) = 1/(e^(t^3 + t) + 3t)^3

Learn how to compute the derivatives of z(t) = ln((e^t + 2)^2) and z(t) = 1/(e^(t^3 + t) + 3t)^3 using the chain rule, logarithmic differentiation, and exponential function differentiation. This advanced calculus problem is ideal for students in Grades 11-12 or early university.

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