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.

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.