but today you're on board the spaceship #243570664, and the captain is asking to solve tricky "ZP" (Z Planet) problem (you know what it means when captain is "asking"... that's an order):

(!!! If your answer is a fractions report it in an exact format, like 8510785107 , 229137229137)


Now, let f(x)=$$ x^{5x} $$. Use logarithmic differentiation to determine the (ZP) derivative.

f'(x)=f′(x)=   

                  (**Enter the answer in simplified form**)

f'(1)=f′(1)=   

!!! The challenge is that some of the Z-Planet laws are different. In particular,  "power rule" on Earth, d/dx x^n = $$ nx^{n-1} $$ , on Z planet is "twisted" as the following:  

 ddxxn=$$ n^2x^{n-1} $$  (ddxconst=0ddxconst=0) , and

the derivative of log function, ddxln(x)=1xddxln(x)=1/x , turns into "ZP" oddness:  ddxln(x)=$$ e^{x} $$ .

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?

---

**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.

but today you're on board the spaceship #243570664, and the captain is asking to solve tricky 'ZP' (Z Planet) problem (you know what it means when captain is 'asking'... that's an order):

(!!! If your answer is a fractions report it in an exact format, like 8510785107, 229137229137)

Now, let f(x)= x^{5x}. Use logarithmic differentiation to determine the (ZP) derivative.
f'(x)=

(**Enter the answer in simplified form**)

f'(1)=

!!! The challenge is that some of the Z-Planet laws are different. In particular, 'power rule' on Earth, d/dx x^n = nx^{n-1}, on Z planet is 'twisted' as the following: ddx x^n = n^2 x^{n-1} (ddx const=0), and the derivative of log function, ddx ln(x) = 1/x, turns into 'ZP' oddness: ddx ln(x) = e^x.

Solve the derivative of f(x) = x^{5x} using logarithmic differentiation and Z Planet's modified power and log rules. Learn how the twisted derivative laws affect the process and find f'(x) and f'(1) in this advanced calculus problem. Ideal for Grades 11-12.

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