Let's tackle this tricky Z Planet problem by applying logarithmic differentiation, keeping in mind the modified "power rule" and logarithmic derivative rules from Z Planet.

We are given that:

$f(x) = x^{5x}$

Step 1: Logarithmic Differentiation

We start by taking the natural logarithm (on Z Planet!) of both sides of the equation to simplify the differentiation.

$\ln(f(x)) = \ln(x^{5x})$

Using the properties of logarithms on Earth, $\ln(x^a) = a\ln(x)$. But, since we're on Z Planet, we need to differentiate this expression using their unique laws.

Thus, we rewrite the equation:

$\ln(f(x)) = 5x \cdot \ln(x)$

Step 2: Differentiate both sides

Now, we differentiate both sides with respect to $x$, applying the Z Planet derivative rules.

For the left-hand side, we have:

$\frac{d}{dx} \ln(f(x)) = \frac{f'(x)}{f(x)}$

For the right-hand side, we need to apply the product rule to differentiate $5x \cdot \ln(x)$.

$\frac{d}{dx} \left( 5x \cdot \ln(x) \right) = 5 \left( \frac{d}{dx}(x) \ln(x) + x \frac{d}{dx} \ln(x) \right)$

On Z Planet:

• $\frac{d}{dx}(x) = 1$
• $\frac{d}{dx} \ln(x) = e^x$ (instead of $\frac{1}{x}$ as on Earth)

So, we get:

$\frac{d}{dx} \left( 5x \cdot \ln(x) \right) = 5 \left( 1 \cdot \ln(x) + x \cdot e^x \right)$

$= 5 \left( \ln(x) + x e^x \right)$

Step 3: Solve for $f'(x)$

Now, equating both sides:

$\frac{f'(x)}{f(x)} = 5 \left( \ln(x) + x e^x \right)$

Multiply both sides by $f(x)$ to isolate $f'(x)$:

$f'(x) = f(x) \cdot 5 \left( \ln(x) + x e^x \right)$

Since $f(x) = x^{5x}$, we substitute:

$f'(x) = x^{5x} \cdot 5 \left( \ln(x) + x e^x \right)$

This is the derivative $f'(x)$ in simplified form!

Step 4: Evaluate $f'(1)$

To find $f'(1)$, substitute $x = 1$ into the derivative:

$f(1) = 1^{5 \cdot 1} = 1$

Now, plug $x = 1$ into the derivative expression:

$f'(1) = 1 \cdot 5 \left( \ln(1) + 1 \cdot e^1 \right)$

Since $\ln(1) = 0$ and $e^1 = e$, we get:

$f'(1) = 5 \cdot (0 + e) = 5e$

Thus:

$f'(1) = 5e$

Final Answers:

$f'(x) = x^{5x} \cdot 5 \left( \ln(x) + x e^x \right)$ $f'(1) = 5e$

Would you like more details or further clarification?

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Related Questions:

1. How would the derivative change if the power of $x$ was $3x$ instead of $5x$?
2. Can you explain how the Earth’s power rule would apply to $x^{5x}$?
3. How would the product rule on Z Planet differ from Earth's version?
4. What would be the integral of $e^x$ on Z Planet?
5. If the function were $x^{4x}$, how would the derivative change?

Tip: When using logarithmic differentiation, always remember that the product and chain rules must adapt to local (or Z-Planet) rules for correct application.