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To find the derivative of the given function \( f(z) = \left( 6z + e^{-z^2} \right)^3 \), we will use the **chain rule** and the **power rule** for differentiation.

### Step 1: Differentiate the Outer Function
The function can be written as:
\[
f(z) = \left( u(z) \right)^3, \quad \text{where } u(z) = 6z + e^{-z^2}.
\]
Using the chain rule:
\[
f'(z) = 3 \left( u(z) \right)^2 \cdot u'(z).
\]

### Step 2: Differentiate the Inner Function \( u(z) \)
Now, we need to find \( u'(z) \):
\[
u(z) = 6z + e^{-z^2}.
\]
The derivative of \( u(z) \) is:
\[
u'(z) = 6 + \frac{d}{dz} \left( e^{-z^2} \right).
\]
Using the chain rule for \( e^{-z^2} \):
\[
\frac{d}{dz} \left( e^{-z^2} \right) = e^{-z^2} \cdot (-2z) = -2z e^{-z^2}.
\]
Therefore:
\[
u'(z) = 6 - 2z e^{-z^2}.
\]

### Step 3: Substitute Back into the Derivative of \( f(z) \)
Now, substitute \( u(z) \) and \( u'(z) \) back into \( f'(z) \):
\[
f'(z) = 3 \left( 6z + e^{-z^2} \right)^2 \cdot \left( 6 - 2z e^{-z^2} \right).
\]

### Final Answer
\[
f'(z) = 3 \left( 6z + e^{-z^2} \right)^2 \cdot \left( 6 - 2z e^{-z^2} \right).
\]

Let me know if you'd like further details on any part of the solution.

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Here are some related questions to practice further:

1. How would you apply the chain rule to differentiate \( g(x) = \left( 5x^2 + \sin(x) \right)^4 \)?
2. How do you differentiate \( h(x) = x^2 e^{-x} \)?
3. What is the derivative of \( k(t) = \ln(2t + e^t) \)?
4. How can you simplify derivatives that involve exponentials and products?
5. What steps would you take to find the second derivative of \( f(z) \)?

**Tip:** Remember that the chain rule is essential whenever you have a function raised to a power. First differentiate the outer function, then multiply by the derivative of the inner function.

Find the derivative of the following function: f(z) = (6z + e^{-z^2})^3.

Learn how to find the derivative of the function f(z) = (6z + e^{-z^2})^3 using the chain rule and power rule. This solution provides a step-by-step approach to differentiating complex functions, suitable for undergraduate or advanced high school calculus students.

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