Here Is The Story that happened to you earlier...

but today you're on board the spaceship #1625325007, 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 1010710107 , 2296222962)


Now, let f(x)=x3xf(x)=x3x. 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, ddxxn=nxn−1ddxxn=nxn-1 , on Z planet is "twisted" as the following:  

 ddxxn=n2xn−1ddxxn=n2xn-1  (ddxconst=0ddxconst=0) , and

the derivative of log function, ddxln(x)=1xddxln(x)=1x , turns into "ZP" oddness:  ddxln(x)=exddxln(x)=ex .

all the other rules have not been changed.

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.

Here Is The Story that happened to you earlier... but today you're on board the spaceship #1625325007, 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 1010710107 , 2296222962) Now, let f(x)=x^3x. 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 = n x^(n−1), on Z planet is "twisted" as the following: d/dx x^n = n^2 x^(n−1) (d/dx const = 0) , and the derivative of log function, d/dx ln(x) = 1/x, turns into "ZP" oddness: d/dx ln(x) = e^x. All the other rules have not been changed.

Explore the solution to a unique derivative problem on the Z-Planet, where derivative rules differ from Earth's. Learn how to apply logarithmic differentiation to find the derivative of f(x) = x^{x^3} using Z-Planet's special rules for power and logarithmic functions. Suitable for undergraduate-level math students.

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