To solve the derivative problem on the spaceship using Z-Planet rules, we begin with the function $f(x) = x^{x^3}$.

Step 1: Logarithmic Differentiation

Take the natural logarithm of both sides: $\ln f(x) = \ln (x^{x^3}) = x^3 \ln x$ Now differentiate both sides with respect to $x$, using the Z-Planet rules for differentiation.

Step 2: Differentiate Using Z-Planet Rules

On Z-Planet, the derivative rules are:

1. $\frac{d}{dx} x^n = n^2 x^{n-1}$
2. $\frac{d}{dx} \ln(x) = e^x$

Using these, we differentiate $\ln f(x)$: $\frac{d}{dx} \ln f(x) = \frac{d}{dx} (x^3 \ln x)$

Apply the product rule (unchanged from Earth's rules): $\frac{d}{dx} (x^3 \ln x) = (x^3) \frac{d}{dx} (\ln x) + (\ln x) \frac{d}{dx} (x^3)$

Compute each derivative:

1. $\frac{d}{dx} (\ln x) = e^x$
2. $\frac{d}{dx} (x^3) = 3^2 x^{3-1} = 9x^2$

Thus, $\frac{d}{dx} (x^3 \ln x) = x^3 e^x + (\ln x)(9x^2)$

Step 3: Relate to $f'(x)$

Recall that: $\frac{1}{f(x)} f'(x) = x^3 e^x + 9x^2 \ln x$ Therefore, solving for $f'(x)$, we get: $f'(x) = f(x) (x^3 e^x + 9x^2 \ln x)$ Substitute $f(x) = x^{x^3}$: $f'(x) = x^{x^3} (x^3 e^x + 9x^2 \ln x)$

Step 4: Evaluate $f'(1)$

To find $f'(1)$, we evaluate $f(1)$, $1^3 e^1$, and $9 \ln 1$:

• $f(1) = 1^{1^3} = 1$
• $1^3 e^1 = e$
• $9 \ln 1 = 0$

Hence, $f'(1) = 1 \cdot (e + 0) = e$

Final Answers

$f'(x) = x^{x^3} (x^3 e^x + 9x^2 \ln x)$ $f'(1) = e$

Do you want details or have any questions?

Here are some related questions:

1. How would you solve $f(x) = x^{x^2}$ using Z-Planet rules?
2. What happens if we change the derivative rule for logarithmic differentiation again?
3. Can you explain the differences in using Z-Planet rules versus Earth's rules?
4. How would you evaluate higher derivatives on Z-Planet?
5. What if $f(x)$ was a combination of powers and logarithms?

Tip: When working on problems with unusual derivative rules, carefully rewrite each rule and proceed step by step to avoid errors.