The question asks to show that the first derivative $f'(z)$ and the second derivative $f''(z)$ exist everywhere, and then to find both $f'(z)$ and $f''(z)$ when $f(z) = z^3$.

Step 1: Show that $f'(z)$ exists and find it

We are given the function $f(z) = z^3$. To find $f'(z)$, we take the derivative of $f(z)$ with respect to $z$.

$f'(z) = \frac{d}{dz}(z^3) = 3z^2$

Thus, $f'(z) = 3z^2$, and since $3z^2$ is a polynomial function, the derivative exists everywhere on the complex plane or the real number line.

Step 2: Show that $f''(z)$ exists and find it

Next, we find $f''(z)$ by differentiating $f'(z)$:

$f''(z) = \frac{d}{dz}(3z^2) = 6z$

So, $f''(z) = 6z$, and since $6z$ is also a polynomial, the second derivative exists everywhere as well.

Conclusion:

• $f'(z) = 3z^2$
• $f''(z) = 6z$

Both the first and second derivatives exist everywhere since they are polynomials.

Would you like more details on any step?

Here are five related questions you might find useful:

1. How do you calculate higher-order derivatives for a given function?
2. What is the geometric interpretation of the first and second derivatives?
3. Can we determine points of inflection using the second derivative for this function?
4. How does the behavior of the function $f(z) = z^3$ change based on the values of $z$?
5. What role does differentiability play in analyzing polynomial functions?

Tip: When working with polynomials, the derivatives are always well-defined and exist everywhere, as polynomials are infinitely differentiable functions.