Math Problem Statement

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, 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)=1/x , turns into "ZP" oddness:  ddxln(x)=exddxln(x)=ex .

all the other rules have not been changed.

Solution

We are tasked with finding the derivative of f(x)=x3xf(x) = x^{3x} under the unique Z-Planet differentiation laws. Let's break this problem down step-by-step, incorporating logarithmic differentiation and applying Z-Planet-specific rules.

Step 1: Express f(x)=x3xf(x) = x^{3x}

This expression is a power function where both the base and the exponent depend on xx. To differentiate it, logarithmic differentiation is helpful.

Step 2: Take the natural logarithm of both sides

Taking the natural logarithm of both sides:

ln(f(x))=ln(x3x)\ln(f(x)) = \ln(x^{3x})

Using the logarithm property ln(ab)=bln(a)\ln(a^b) = b \ln(a), we can rewrite this as:

ln(f(x))=3xln(x)\ln(f(x)) = 3x \ln(x)

Step 3: Differentiate both sides using the Z-Planet rules

Now, differentiate both sides with respect to xx. On the left-hand side, apply the chain rule:

ddxln(f(x))=1f(x)f(x)\frac{d}{dx} \ln(f(x)) = \frac{1}{f(x)} \cdot f'(x)

On the right-hand side, we differentiate 3xln(x)3x \ln(x) using the product rule ddx(uv)=uv+uv\frac{d}{dx}(uv) = u'v + uv', keeping in mind the Z-Planet rules for differentiation:

ddx(3xln(x))=31ln(x)+3xddxln(x)\frac{d}{dx} (3x \ln(x)) = 3 \cdot 1 \cdot \ln(x) + 3x \cdot \frac{d}{dx} \ln(x)

Using the Z-Planet rule for the derivative of the natural logarithm:

ddxln(x)=ex\frac{d}{dx} \ln(x) = e^x

So, the right-hand side becomes:

3ln(x)+3xex3 \ln(x) + 3x e^x

Step 4: Combine the results

We now have:

1f(x)f(x)=3ln(x)+3xex\frac{1}{f(x)} \cdot f'(x) = 3 \ln(x) + 3x e^x

Multiplying both sides by f(x)=x3xf(x) = x^{3x}, we obtain the derivative f(x)f'(x):

f(x)=x3x(3ln(x)+3xex)f'(x) = x^{3x} \left( 3 \ln(x) + 3x e^x \right)

Step 5: Compute f(1)f'(1)

Now, let's substitute x=1x = 1 into the expression for f(x)f'(x):

f(1)=131=1f(1) = 1^{3 \cdot 1} = 1

For the derivative, we have:

f(1)=131(3ln(1)+31e1)f'(1) = 1^{3 \cdot 1} \left( 3 \ln(1) + 3 \cdot 1 \cdot e^1 \right)

Since ln(1)=0\ln(1) = 0 and e1=ee^1 = e, this simplifies to:

f(1)=1(0+3e)=3ef'(1) = 1 \cdot \left( 0 + 3e \right) = 3e

Final Answer:

  • f(x)=x3x(3ln(x)+3xex)f'(x) = x^{3x} \left( 3 \ln(x) + 3x e^x \right)
  • f(1)=3ef'(1) = 3e

Let me know if you'd like more details or if anything is unclear!

Related Questions:

  1. How does the Z-Planet power rule affect derivatives of polynomial functions?
  2. What are some other Z-Planet-specific differentiation rules?
  3. How would the derivative change if we used a different base for the exponent?
  4. Can we generalize this method to functions like f(x)=xnxf(x) = x^{nx}?
  5. How does logarithmic differentiation work on Earth compared to Z-Planet?

Tip:

Logarithmic differentiation is especially useful when both the base and the exponent of a function involve xx, as it simplifies otherwise complex differentiation tasks.

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Math Problem Analysis

Mathematical Concepts

Logarithmic Differentiation
Exponential Functions
Product Rule

Formulas

Logarithmic differentiation: \( \ln(f(x)) \)
Z-Planet power rule: \( \frac{d}{dx} x^n = n^2 x^{n-1} \)
Z-Planet log derivative: \( \frac{d}{dx} \ln(x) = e^x \)

Theorems

Product Rule
Chain Rule

Suitable Grade Level

Undergraduate Level