Question

Find the derivative of
$y = \sqrt[3]{x}$
at the point where x = 27.​

Answer 1

Answer:

$\displaystyle y'(27) = \frac{1}{27}$

General Formulas and Concepts:

Pre-Algebra

Order of Operations: BPEMDAS

1. Brackets
2. Parenthesis
3. Exponents
4. Multiplication
5. Division
6. Addition
7. Subtraction

• Left to Right

Algebra I

• Exponential Rule [Root Rewrite]: $\displaystyle \sqrt[n]{x} = x^{\frac{1}{n}}$
• Exponential Rule [Rewrite]: $\displaystyle b^{-m} = \frac{1}{b^m}$

Calculus

Derivatives

Derivative Notation

Basic Power Rule

• f(x) = cxⁿ
• f’(x) = c·nxⁿ⁻¹

Step-by-step explanation:

Step 1: Define

y = ∛x

x = 27

Step 2: Differentiate

1. [Function] Rewrite [Exponential Rule - Root Rewrite]:                               $\displaystyle y = x^{\frac{1}{3}}$
2. Basic Power Rule:                                                                                         $\displaystyle y' = \frac{1}{3}x^{\frac{1}{3} - 1}$
3. Simplify:                                                                                                         $\displaystyle y' = \frac{1}{3}x^{-\frac{2}{3}}$
4. Rewrite [Exponential Rule - Rewrite]:                                                           $\displaystyle y' = \frac{1}{3x^{\frac{2}{3}}}$

Step 3: Solve

1. Substitute in x [Derivative]:                                                                           $\displaystyle y'(27) = \frac{1}{3(27)^{\frac{2}{3}}}$
2. Evaluate exponents:                                                                                     $\displaystyle y'(27) = \frac{1}{3(9)}$
3. Multiply:                                                                                                         $\displaystyle y'(27) = \frac{1}{27}$

Topic: AP Calculus AB/BC (Calculus I/II)

Unit: Derivatives

Book: College Calculus 10e