20. Zenzizenzicube is an obsolete word that represents the square of the square of a cube, In symbols, zenzizenzicube is written ((x) 2)?. a. Use the chain rule twice to find the derivative. b. Use the properties of exponents to first simplify the expression, and then find the derivative. a. The function is in the form y = f(n), where n = g(u) and u = h(x). This can also be written as flg[h(x)]]. Use your knowledge of the chain rule to determine the formula when using the chain rule twice. Choose the correct answer below. O A. =f'(g(h(x)]] •g'lh(x)] •h´(x) dx Ов. =f'(g(h(x)]] •gʻth(x)] dx OC. dy =f'(g(x)] •g'(x) • h'(x) dx OD. dy =f'(g(x)]•g'th(x)] •h'(x) dx Now determine the expressions for u, n, and y. What is the expression for u? u= h(x) = Determine the expression for n. n= g(u) = Next, determine the expression for y. y = f(n) = Now find the derivative of each expression. What is h'(x)? h'(x) = (Simplify your answer.) What is g'(u)? g'(u) = (Simplify your answer.) What is f'(n)? f'(n) = (Simplify your answer.) Substitute the results into the formula found for the chain rule and simplify. What is the result? The derivative of ((x³)²)² is (Simplify your answer.) b. Use the properties of exponents to simplify ((x))". ()?)* = (Simplify your answer.) What is the derivative of the result? The derivative is (Simplify your answer.) xlitemprod.pearsoncmg.com/api/v1/print/math

Elementary Algebra
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Author:Lynn Marecek, MaryAnne Anthony-Smith
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Chapter2: Solving Linear Equations And Inequalities
Section2.7: Solve Linear Inequalities
Problem 509E: Give an example from your life using the phrase ‘at most’.
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Certainly! Below is the transcription and explanation of the content for an educational website setting:

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**Topic: Zenzizenzicube**

Zenzizenzicube is an obsolete word representing the square of the square of a cube. In mathematical symbols, zenzizenzicube is written as \(\left( (x^3)^2 \right)^2\).

### Objectives:

a. Use the chain rule twice to find the derivative.
b. Use the properties of exponents to first simplify the expression, and then find the derivative.

---

**a. Chain Rule Method**

- The function is in the form \(y = f(n)\), where \(n = g(u)\) and \(u = h(x)\). This can also be written as \(f[g[h(x)]]\). Use your knowledge of the chain rule to determine the formula when using the chain rule twice. Choose the correct answer below.

1. **Options for the derivative using the chain rule:**
   - **A.** \(\frac{dy}{dx} = f'[g(h(x))] \cdot g'[h(x)] \cdot h'(x)\)
   - **B.** \(\frac{dy}{dx} = f'[g(h(x))] \cdot g'[h(x)]\)
   - **C.** \(\frac{dy}{dx} = f'[g(x)] \cdot h'(x)\)
   - **D.** \(\frac{dy}{dx} = f'[g(x)] \cdot g'[h(x)] \cdot h'(x)\)

**Steps to Determine the Expressions:**

- **Expression for \(u\):**  
  \(u = h(x) = \underline{\hspace{3cm}}\)

- **Determine the expression for \(n\):**  
  \(n = g(u) = \underline{\hspace{3cm}}\)

- **Determine the expression for \(y\):**  
  \(y = f(n) = \underline{\hspace{3cm}}\)

- **Finding the derivative of each expression:**

  - \(h'(x) = \underline{\hspace{3cm}}\) (Simplify your answer.)

  - \(g'(u) = \underline{\hspace{3cm}}\) (Simplify your answer.)

  - \(f'(n) = \underline
Transcribed Image Text:Certainly! Below is the transcription and explanation of the content for an educational website setting: --- **Topic: Zenzizenzicube** Zenzizenzicube is an obsolete word representing the square of the square of a cube. In mathematical symbols, zenzizenzicube is written as \(\left( (x^3)^2 \right)^2\). ### Objectives: a. Use the chain rule twice to find the derivative. b. Use the properties of exponents to first simplify the expression, and then find the derivative. --- **a. Chain Rule Method** - The function is in the form \(y = f(n)\), where \(n = g(u)\) and \(u = h(x)\). This can also be written as \(f[g[h(x)]]\). Use your knowledge of the chain rule to determine the formula when using the chain rule twice. Choose the correct answer below. 1. **Options for the derivative using the chain rule:** - **A.** \(\frac{dy}{dx} = f'[g(h(x))] \cdot g'[h(x)] \cdot h'(x)\) - **B.** \(\frac{dy}{dx} = f'[g(h(x))] \cdot g'[h(x)]\) - **C.** \(\frac{dy}{dx} = f'[g(x)] \cdot h'(x)\) - **D.** \(\frac{dy}{dx} = f'[g(x)] \cdot g'[h(x)] \cdot h'(x)\) **Steps to Determine the Expressions:** - **Expression for \(u\):** \(u = h(x) = \underline{\hspace{3cm}}\) - **Determine the expression for \(n\):** \(n = g(u) = \underline{\hspace{3cm}}\) - **Determine the expression for \(y\):** \(y = f(n) = \underline{\hspace{3cm}}\) - **Finding the derivative of each expression:** - \(h'(x) = \underline{\hspace{3cm}}\) (Simplify your answer.) - \(g'(u) = \underline{\hspace{3cm}}\) (Simplify your answer.) - \(f'(n) = \underline
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